Evaluating Some Improper Sine and Cosine Integrals

Abstract

We consider four infinite collections of improper integrals involving the sine and cosine functions and provide explicit values for the integrals in each of the collections. The methods used are elementary and involve the sine integral function and the cosine integral function. All needed results are included in this paper, but the details are a little tedious at times.

1. Introduction

Several papers (see [1,2,3,4]) have determined the explicit values of collections of improper integrals that involve the sine and cosine functions, along with their Maclaurin series. The article [3] extends the computations of [2] using results from complex analysis, while the article [4] proves some of these same results using more elementary methods. A recent problem in Amer. Math. Monthly (see [5]) involved a different type of improper integral containing sine and cosine functions. The purpose of this paper is to greatly expand the type of integral in [5]. For positive integers q and n, consider the two collections of improper integrals

$${\int }_0^{\infty }{\displaystyle \frac{{(x^q-{sin}^{q}x)}^n}{x^a}}dx\mathrm{and}{\int }_0^{\infty }{\displaystyle \frac{{(1-{cos}^{q}x)}^n}{x^a}}dx,$$

where the values of a are positive integers chosen so that the improper integrals exist. In this paper, we prove the following two facts and provide explicit values for each collection of integrals.

• If the integrands are even functions, then the values of the integrals are rational multiples of $\pi$.
• If the integrands are odd functions, then the values of the integrals are linear combinations of logarithms of integers, using rational numbers for the coefficients, plus a rational number.

For some simple examples, computer algebra systems easily generate the following values:

$$\begin{array}{cc}\hfill {\int }_0^{\infty }{\displaystyle \frac{{(x^3-{sin}^{3}x)}^2}{x^8}}dx& ={\displaystyle \frac{409}{1680}}\pi ;\hfill \\ \hfill {\int }_0^{\infty }{\displaystyle \frac{{(x^3-{sin}^{3}x)}^2}{x^9}}dx& =-{\displaystyle \frac{73}{168}}+{\displaystyle \frac{73}{105}}ln2+{\displaystyle \frac{81}{280}}ln3;\hfill \\ \hfill {\int }_0^{\infty }{\displaystyle \frac{{(x^3-{sin}^{3}x)}^3}{x^{11}}}dx& ={\displaystyle \frac{4263359}{20643840}}\pi ;\hfill \\ \hfill {\int }_0^{\infty }{\displaystyle \frac{{(1-{cos}^{4}x)}^3}{x^3}}dx& =-{\displaystyle \frac{9}{32}}ln2+{\displaystyle \frac{45}{32}}ln3-{\displaystyle \frac{75}{256}}ln5;\hfill \\ \hfill {\int }_0^{\infty }{\displaystyle \frac{{(1-{cos}^{4}x)}^3}{x^4}}dx& ={\displaystyle \frac{43}{128}}\pi ;\hfill \\ \hfill {\int }_0^{\infty }{\displaystyle \frac{{(1-{cos}^{11}x)}^2}{x^3}}dx& ={\displaystyle \frac{79013}{4096}}ln2-{\displaystyle \frac{62667}{65536}}ln3-{\displaystyle \frac{2356475}{1048576}}ln5-{\displaystyle \frac{2401245}{1048576}}ln7-{\displaystyle \frac{123783}{1048576}}ln11.\hfill \end{array}$$

When the exponents are larger, computer algebra systems have more trouble evaluating the integrals. For integers $q\ge 1$, $n\ge 1$, and $1\le r\le n$, we verify that

$$\begin{array}{cc}\hfill {\int }_0^{\infty }{\displaystyle \frac{{(1-{cos}^{q}x)}^n}{x^{2r}}}dx& ={(-1)}^r\pi \sum _{k=1}^n\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^k}{2^{qk}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(qk-2j)}^{2r-1}}{(2r-1)!}};\hfill \\ \hfill {\int }_0^{\infty }{\displaystyle \frac{{(x^q-{sin}^{q}x)}^n}{x^{qn+2r}}}dx& ={(-1)}^r\pi \sum _{k=1}^n\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^{j+k}}{2^{qk}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(qk-2j)}^{qk+2r-1}}{(qk+2r-1)!}}.\hfill \end{array}$$

Note that the integrands for all of these integrals are even functions and the values of the integrals are rational multiples of $\pi$. For integers $q\ge 1$, $n\ge 2$, and $1\le r\le n-1$, we verify that

$$\begin{array}{cc}\hfill {\int }_0^{\infty }{\displaystyle \frac{{(1-{cos}^{q}x)}^n}{x^{2r+1}}}dx& =2{(-1)}^{r+1}\sum _{k=1}^n\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^k}{2^{qk}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(qk-2j)}^{2r}}{\left(2r\right)!}}ln(qk-2j);\hfill \\ \hfill {\int }_0^{\infty }{\displaystyle \frac{{(x^q-{sin}^{q}x)}^n}{x^{qn+2r+1}}}dx& =2{(-1)}^{r+1}\sum _{k=1}^n\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^{j+k}}{2^{qk}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(qk-2j)}^{qk+2r}}{(qk+2r)!}}\left(ln(qk-2j)-h_{qk+2r}\right);\hfill \end{array}$$

where $h_n$ represents the nth harmonic number. The integrands for these collections of integrals are odd functions and the values of the integrals satisfy the conditions listed above for item (2).

2. Some Preliminary Formulas

To evaluate the integrals in these collections, we make use of the sine integral function $\mathrm{Si}\left(x\right)$ and the cosine integral function $\mathrm{Ci}\left(x\right)$. These functions are defined by

$$\mathrm{Si}\left(x\right)={\int }_0^x{\displaystyle \frac{sint}{t}}dt\mathrm{and}\mathrm{Ci}\left(x\right)=-{\int }_x^{\infty }{\displaystyle \frac{cost}{t}}dt.$$

It can be shown that (see Jameson [6] for one of many references)

$$\mathrm{Ci}\left(x\right)=\gamma +lnx-{\int }_0^x{\displaystyle \frac{1-cost}{t}}dt,$$

where $\gamma$ is Euler’s constant. From these definitions, it readily follows that

$${\mathrm{Si}}^{\prime }\left(x\right)={\displaystyle \frac{sinx}{x}},\underset{a\to 0^{+}}{lim}\mathrm{Si}\left(a\right)=0,\underset{b\to \infty }{lim}\mathrm{Si}\left(b\right)={\displaystyle \frac{\pi }{2}},$$

and

$${\mathrm{Ci}}^{\prime }\left(x\right)={\displaystyle \frac{cosx}{x}},\underset{a\to 0^{+}}{lim}\mathrm{Ci}\left(a\right)=-\infty ,\underset{b\to \infty }{lim}\mathrm{Ci}\left(b\right)=0.$$

Finally, for any positive number r, we find that (since $(1-cost)/t$ is a continuous function)

$$\begin{array}{cc}\hfill \underset{a\to 0^{+}}{lim}\left(\mathrm{Ci}\left(ra\right)-\mathrm{Ci}\left(a\right)\right)& =\underset{a\to 0^{+}}{lim}\left(\left(\gamma +ln\left(ra\right)-{\int }_0^{ra}{\displaystyle \frac{1-cost}{t}}dt\right)-\left(\gamma +lna-{\int }_0^a{\displaystyle \frac{1-cost}{t}}dt\right)\right)\hfill \\ \hfill & =\underset{a\to 0^{+}}{lim}\left(lnr-{\int }_a^{ra}{\displaystyle \frac{1-cost}{t}}dt\right)=lnr.\hfill \end{array}$$

These facts will be useful in verifying the values of our integrals.

To find antiderivatives for our integrands, the following two sequences of functions appear. For each positive integer n, let

$$U_n\left(x\right)=\sum _{k=1}^n{\displaystyle \frac{{(-1)}^{k+1}(2k-1)!}{x^{2k}}}\mathrm{and}{V}_n\left(x\right)=\sum _{k=1}^n{\displaystyle \frac{{(-1)}^k(2k-2)!}{x^{2k-1}}}.$$

We also let $U_0\left(x\right)=0=V_0\left(x\right)$. The first few explicit expressions for these functions are

$$\begin{array}{ccc}{U}_1\left(x\right)={\displaystyle \frac{1}{x^2}};\hfill & \hfill & V_1\left(x\right)=-{\displaystyle \frac{1}{x}};\hfill \\ U_2\left(x\right)={\displaystyle \frac{1}{x^2}}-{\displaystyle \frac{6}{x^4}};\hfill & \hfill & V_2\left(x\right)=-{\displaystyle \frac{1}{x}}+{\displaystyle \frac{2}{x^3}};\hfill \\ U_3\left(x\right)={\displaystyle \frac{1}{x^2}}-{\displaystyle \frac{6}{x^4}}+{\displaystyle \frac{120}{x^6}};\hfill & \hfill & V_3\left(x\right)=-{\displaystyle \frac{1}{x}}+{\displaystyle \frac{2}{x^3}}-{\displaystyle \frac{24}{x^5}}.\hfill \end{array}$$

It is easy to verify that $V_n^{\prime }\left(x\right)=U_n\left(x\right)$ for all $n\ge 0$. In addition, we find that

$$U_n^{\prime }\left(x\right)+V_{n+1}\left(x\right)=\sum _{k=1}^n{\displaystyle \frac{{(-1)}^{k+2}\left(2k\right)!}{x^{2k+1}}}+\sum _{k=0}^n{\displaystyle \frac{{(-1)}^{k+1}\left(2k\right)!}{x^{2k+1}}}=-{\displaystyle \frac{1}{x}}$$

for all $n\ge 0$.

For the integrands that are even functions, we require the following lemma. Note that the integrands for the integrals in the lemma are even functions and that for each $n\ge 0$, the functions

$$U_n\left(x\right)sinx-V_n\left(x\right)cosx\mathrm{and}{U}_n\left(x\right)sinx-V_{n+1}\left(x\right)cosx$$

are odd functions defined for all nonzero values of x.

Lemma 1. 

For each nonnegative integer n and positive real number x, we have

$$\begin{array}{cc}\hfill {\int }_x^{\infty }{\displaystyle \frac{sint}{t^{2n+1}}}dt& ={\displaystyle \frac{{(-1)}^{n+1}}{\left(2n\right)!}}\left(U_n\left(x\right)sinx-V_n\left(x\right)cosx+\mathrm{Si}\left(x\right)-{\displaystyle \frac{\pi }{2}}\right);\hfill \\ \hfill {\int }_x^{\infty }{\displaystyle \frac{cost}{t^{2n+2}}}dt& ={\displaystyle \frac{{(-1)}^n}{(2n+1)!}}\left(U_n\left(x\right)sinx-V_{n+1}\left(x\right)cosx+\mathrm{Si}\left(x\right)-{\displaystyle \frac{\pi }{2}}\right).\hfill \end{array}$$

Proof. 

To prove the result for the integral involving the sine function, it is sufficient to verify that

$$\begin{array}{cc}\hfill \underset{x\to \infty }{lim}\left(U_n\left(x\right)sinx-V_n\left(x\right)cosx+\mathrm{Si}\left(x\right)-{\displaystyle \frac{\pi }{2}}\right)& =0;\hfill \\ \hfill {\displaystyle \frac{{(-1)}^{n+1}}{\left(2n\right)!}}·{\displaystyle \frac{d}{dx}}\left(U_n\left(x\right)sinx-V_n\left(x\right)cosx+\mathrm{Si}\left(x\right)-{\displaystyle \frac{\pi }{2}}\right)& =-{\displaystyle \frac{sinx}{x^{2n+1}}};\hfill \end{array}$$

for all $n\ge 0$. It is easy to establish the value of the limit. For the derivative, we use the properties of the functions $U_n$ and $V_n$ to obtain

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dx}}\left(U_n\left(x\right)sinx-V_n\left(x\right)cosx+\mathrm{Si}\left(x\right)-{\displaystyle \frac{\pi }{2}}\right)& =U_n\left(x\right)cosx+U_n^{\prime }\left(x\right)sinx+V_n\left(x\right)sinx-V_n^{\prime }\left(x\right)cosx+{\displaystyle \frac{sinx}{x}}\hfill \\ \hfill & =\left(U_n^{\prime }\left(x\right)+V_n\left(x\right)+{\displaystyle \frac{1}{x}}\right)sinx\hfill \\ \hfill & =-\left(V_{n+1}\left(x\right)-V_n\left(x\right)\right)sinx\hfill \\ \hfill & =-{\displaystyle \frac{{(-1)}^{n+1}\left(2n\right)!}{x^{2n+1}}}sinx,\hfill \end{array}$$

the desired derivative for all $n\ge 0$.

The proof for the cosine integral is similar (using integration by parts is another option, but this approach is a bit more tedious). Since the appropriate limit is also trivial, we need to verify that

$${\displaystyle \frac{{(-1)}^n}{(2n+1)!}}·{\displaystyle \frac{d}{dx}}\left(U_n\left(x\right)sinx-V_{n+1}\left(x\right)cosx+\mathrm{Si}\left(x\right)-{\displaystyle \frac{\pi }{2}}\right)=-{\displaystyle \frac{cosx}{x^{2n+2}}}$$

for all $n\ge 0$. Taking the derivative, we find that

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dx}}(U_n\left(x\right)sinx& - V_{n+1}\left(x\right)cosx+\mathrm{Si}\left(x\right)-{\displaystyle \frac{\pi }{2}})\hfill \\ \hfill & =U_n\left(x\right)cosx+U_n^{\prime }\left(x\right)sinx+V_{n+1}\left(x\right)sinx-V_{n+1}^{\prime }\left(x\right)cosx+{\displaystyle \frac{sinx}{x}}\hfill \\ \hfill & =\left(U_n\left(x\right)-U_{n+1}\left(x\right)\right)cosx+\left(U_n^{\prime }\left(x\right)+V_{n+1}\left(x\right)+{\displaystyle \frac{1}{x}}\right)sinx\hfill \\ \hfill & =-{\displaystyle \frac{{(-1)}^{n+2}(2n+1)!}{x^{2n+2}}}cosx,\hfill \end{array}$$

which is the expected value for the derivative for all $n\ge 0$. □

Let q and n be positive integers. Since the function ${(x^q-{sin}^{q}x)}^n$ behaves like $x^{qn}$ for large values of x and like $x^{qn+2n}$ for values of x near 0, the improper integrals in the sine collection exist when the exponent a in the denominator satisfies $qn+1<a<qn+2n+1$. Since a is an integer, it follows that $qn+2\le a\le qn+2n$. Similarly, the function ${(1-{cos}^{q}x)}^n$ behaves like 1 for large values of x and like $x^{2n}$ for values of x near 0. Hence, the improper integrals in the cosine collection exist when the exponent a in the denominator satisfies $1<a<2n+1$. Since a is an integer, it follows that $2\le a\le 2n$. We assume these facts in the results that follow.

We will make extensive use of the following four trigonometric identities (see [7]):

$$\begin{array}{ccc}\hfill {sin}^{2k-1}x& ={\displaystyle \frac{{(-1)}^{k+1}}{2^{2k-2}}}\sum _{j=0}^{k-1}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{2k-1}{j}}\right)sin\left((2k-1-2j)x\right);\hfill & \hfill \mathrm{OS}\mathrm{identity}\\ \hfill {sin}^{2k}x& ={\displaystyle \frac{{(-1)}^k}{2^{2k-1}}}\sum _{j=0}^{k-1}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{j}}\right)cos\left((2k-2j)x\right)+{\displaystyle \frac{1}{2^{2k}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right);\hfill & \hfill \mathrm{ES}\mathrm{identity}\\ \hfill {cos}^{2k-1}x& ={\displaystyle \frac{1}{2^{2k-2}}}\sum _{j=0}^{k-1}\left({\displaystyle \genfrac{}{}{0pt}{}{2k-1}{j}}\right)cos\left((2k-1-2j)x\right);\hfill & \hfill \mathrm{OC}\mathrm{identity}\\ \hfill {cos}^{2k}x& ={\displaystyle \frac{1}{2^{2k-1}}}\sum _{j=0}^{k-1}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{j}}\right)cos\left((2k-2j)x\right)+{\displaystyle \frac{1}{2^{2k}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right);\hfill & \hfill \mathrm{EC}\mathrm{identity}\end{array}$$

valid for all positive integers k. The fact that these identities change slightly depending on whether the exponent is even or odd complicates some of our calculations.

3. Integrals with Even Integrands

Since the computations are easier when the exponents on sine and cosine are even, we begin by considering one of these options.

Theorem 1. 

If q and n are positive integers, then

$${\int }_0^{\infty }{\displaystyle \frac{{(x^{2q}-{sin}^{2q}x)}^n}{x^{2qn+2r}}}dx={(-1)}^r\pi \sum _{k=1}^n\sum _{j=0}^{qk-1}{\displaystyle \frac{{(-1)}^{k+j}}{2^{2qk}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2qk}{j}}\right){\displaystyle \frac{{(2qk-2j)}^{2qk+2r-1}}{(2qk+2r-1)!}}$$

for each integer r that satisfies $1\le r\le n$.

Proof. 

For a given set of appropriate values for q, n, and r, let $A\left(x\right)$ represent the function

$$A\left(x\right)={\int }_x^{\infty }{\displaystyle \frac{{(t^{2q}-{sin}^{2q}t)}^n}{t^{2qn+2r}}}dt.$$

We want to find the value of $\underset{x\to 0^{+}}{lim}A\left(x\right)$. Using the ES identity for ${sin}^{2qk}t$, we find that

$$\begin{array}{cc}\hfill & {\displaystyle \frac{{(t^{2q}-{sin}^{2q}t)}^n}{t^{2qn+2r}}}={\displaystyle \frac{1}{t^{2qn+2r}}}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\left(-{sin}^{2q}t\right)}^k{\left(t^{2q}\right)}^{n-k}=\sum _{k=0}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{sin}^{2qk}t}{t^{2qk+2r}}}\hfill \\ \hfill & ={\displaystyle \frac{1}{t^{2r}}}+\sum _{k=1}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{1}{t^{2qk+2r}}}\left({\displaystyle \frac{{(-1)}^{qk}}{2^{2qk-1}}}\sum _{j=0}^{qk-1}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{2qk}{j}}\right)cos\left((2qk-2j)t\right)+{\displaystyle \frac{1}{2^{2qk}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2qk}{qk}}\right)\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{t^{2r}}}+\sum _{k=1}^n\sum _{j=0}^{qk-1}{\displaystyle \frac{{(-1)}^{qk+k+j}}{2^{2qk-1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2qk}{j}}\right){\displaystyle \frac{cos\left((2qk-2j)t\right)}{t^{2qk+2r}}}+\sum _{k=1}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2qk}{qk}}\right){\displaystyle \frac{{(-1)}^k}{2^{2qk}{t}^{2qk+2r}}}.\hfill \end{array}$$

Since Lemma 1 (with $n=qk+r-1$) yields

$$\begin{array}{cc}\hfill {\int }_x^{\infty }& {\displaystyle \frac{cos\left((2qk-2j)t\right)}{t^{2qk+2r}}}dt={\left((2qk-2j)\right)}^{2qk+2r-1}{\int }_{(2qk-2j)x}^{\infty }{\displaystyle \frac{cosu}{u^{2qk+2r}}}du\hfill \\ \hfill & ={\left((2qk-2j)\right)}^{2qk+2r-1}{\displaystyle \frac{{(-1)}^{qk+r-1}}{(2qk+2r-1)!}}(U_{qk+r-1}\left((2qk-2j)x\right)sin\left((2qk-2j)x\right)\hfill \\ \hfill & -V_{qk+r}\left((2qk-2j)x\right)cos\left((2qk-2j)x\right)+\mathrm{Si}\left((2qk-2j)x\right)-{\displaystyle \frac{\pi }{2}}),\hfill \end{array}$$

it follows that

$$\begin{array}{cc}\hfill A\left(x\right)& ={\displaystyle \frac{1}{(2r-1)x^{2r-1}}}+\sum _{k=1}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2qk}{qk}}\right){\displaystyle \frac{{(-1)}^k}{2^{2qk}(2qk+2r-1)x^{2qk+2r-1}}}\hfill \\ \hfill & +\sum _{k=1}^n\sum _{j=0}^{qk-1}{\displaystyle \frac{{(-1)}^{k+j+r-1}}{2^{2qk-1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2qk}{j}}\right){\displaystyle \frac{{(2qk-2j)}^{2qk+2r-1}}{(2qk+2r-1)!}}(U_{qk+r-1}\left((2qk-2j)x\right)sin\left((2qk-2j)x\right)\hfill \\ \hfill & -V_{qk+r}\left((2qk-2j)x\right)cos\left((2qk-2j)x\right)+\mathrm{Si}\left((2qk-2j)x\right)-{\displaystyle \frac{\pi }{2}}).\hfill \end{array}$$

Since the improper integral exists, we know that $A\left(x\right)$ has a limit as $x\to 0^{+}$. The terms not involving the $\mathrm{Si}$ function and the constant term ${\displaystyle \frac{1}{2}}\pi$ form a Laurent series for an odd function. Since the limit as $x\to 0^{+}$ exists, all of the terms involving negative exponents must cancel, and there is no constant term since the Laurent series function is odd. Knowing that $\underset{x\to 0^{+}}{lim}\mathrm{Si}\left(x\right)=0$, it follows that the value of $\underset{x\to 0^{+}}{lim}A\left(x\right)$ is

$$\pi \sum _{k=1}^n\sum _{j=0}^{qk-1}{\displaystyle \frac{{(-1)}^{k+j+r}}{2^{2qk}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2qk}{j}}\right){\displaystyle \frac{{(2qk-2j)}^{2qk+2r-1}}{(2qk+2r-1)!}}.$$

This completes the proof. □

In the proof of the previous theorem, we found an explicit expression for the function $A\left(x\right)$. However, as indicated in the proof, we only need the multipliers of the ${\displaystyle \frac{1}{2}}\pi$ terms; all of the other terms contribute nothing to the final value of $\underset{x\to 0^{+}}{lim}A\left(x\right)$. We denote these multipliers by the symbol ${\int }_{\mathrm{\Pi }}f\left(t\right)dt$ for appropriate functions f. In particular, using Lemma 1, we find that

$$\begin{array}{ccc}\hfill {\int }_{\mathrm{\Pi }}{\displaystyle \frac{sin\left(\theta t\right)}{t^{2a+1}}}dt& ={\theta }^{2a}{\int }_{\mathrm{\Pi }}{\displaystyle \frac{sint}{t^{2a+1}}}dt={(-1)}^a{\displaystyle \frac{{\theta }^{2a}}{\left(2a\right)!}};\hfill & \hfill \left(S_e\mathrm{equation}\right)\\ \hfill {\int }_{\mathrm{\Pi }}{\displaystyle \frac{cos\left(\theta t\right)}{t^{2a}}}dt& ={\theta }^{2a-1}{\int }_{\mathrm{\Pi }}{\displaystyle \frac{cost}{t^{2a}}}dt={(-1)}^a{\displaystyle \frac{{\theta }^{2a-1}}{(2a-1)!}};\hfill & \hfill \left(C_e\mathrm{equation}\right)\end{array}$$

where $\theta$ is a positive integer and a is a positive integer for the cosine integral and a nonnegative integer for the sine integral. With these facts and notational conventions, we are able to prove the following two general results without writing out the full expression for the function $A\left(x\right)$ explicitly. As a result, the proofs involve much less clutter. If for other purposes this function is needed, the proofs below can be expanded to determine an explicit expression for $A\left(x\right)$.

Theorem 2. 

If q and n are positive integers, then

$${\int }_0^{\infty }{\displaystyle \frac{{(x^q-{sin}^{q}x)}^n}{x^{qn+2r}}}dx={(-1)}^r\pi \sum _{k=1}^n\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^{j+k}}{2^{qk}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(qk-2j)}^{qk+2r-1}}{(qk+2r-1)!}}$$

for each integer r that satisfies $1\le r\le n$.

Proof. 

For a given set of appropriate values for q, n, and r, let $A\left(x\right)$ represent the function

$$A\left(x\right)={\int }_x^{\infty }{\displaystyle \frac{{(t^q-{sin}^{q}t)}^n}{t^{qn+2r}}}dt.$$

We want to find the value of $\underset{x\to 0^{+}}{lim}A\left(x\right)$. Since

$${\int }_{\mathrm{\Pi }}{\displaystyle \frac{{(t^q-{sin}^{q}t)}^n}{t^{qn+2r}}}dt={\int }_{\mathrm{\Pi }}\sum _{k=1}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{sin}^{qk}t}{t^{qk+2r}}}dt,$$

it is sufficient to find the value of each of the expressions ${\displaystyle {\int }_{\mathrm{\Pi }}\frac{{sin}^{qk}t}{t^{qk+2r}}dt.}$ Using the ES identity when $qk$ is even and the $C_e$ equation, we find that

$$\begin{array}{cc}\hfill {\int }_{\mathrm{\Pi }}{\displaystyle \frac{{sin}^{qk}t}{t^{qk+2r}}}dt& ={\displaystyle \frac{{(-1)}^{qk/2}}{2^{qk-1}}}\sum _{j=0}^{(qk/2)-1}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\int }_{\mathrm{\Pi }}{\displaystyle \frac{cos\left((qk-2j)t\right)}{t^{qk+2r}}}dt\hfill \\ \hfill & ={\displaystyle \frac{{(-1)}^{qk/2}}{2^{qk-1}}}\sum _{j=0}^{(qk/2)-1}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){(-1)}^{(qk/2)+r}{\displaystyle \frac{{(qk-2j)}^{qk+2r-1}}{(qk+2r-1)!}}\hfill \\ \hfill & =\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^{j+r}}{2^{qk-1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(qk-2j)}^{qk+2r-1}}{(qk+2r-1)!}}.\hfill \end{array}$$

Similarly, when $qk$ is odd, we use the OS identity and the $S_e$ equation to obtain

$$\begin{array}{cc}\hfill {\int }_{\mathrm{\Pi }}{\displaystyle \frac{{sin}^{qk}t}{t^{qk+2r}}}dt& ={\displaystyle \frac{{(-1)}^{(qk+3)/2}}{2^{qk-1}}}\sum _{j=0}^{(qk-1)/2}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\int }_{\mathrm{\Pi }}{\displaystyle \frac{sin\left((qk-2j)t\right)}{t^{qk+2r}}}dt\hfill \\ \hfill & ={\displaystyle \frac{{(-1)}^{(qk+3)/2}}{2^{qk-1}}}\sum _{j=0}^{(qk-1)/2}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){(-1)}^{\left(\right(qk-1)/2)+r}{\displaystyle \frac{{(qk-2j)}^{qk+2r-1}}{(qk+2r-1)!}}\hfill \\ \hfill & =\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^{j+r}}{2^{qk-1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(qk-2j)}^{qk+2r-1}}{(qk+2r-1)!}}.\hfill \end{array}$$

It follows that

$${\int }_{\mathrm{\Pi }}{\displaystyle \frac{{(t^q-{sin}^{q}t)}^n}{t^{qn+2r}}}dt=\sum _{k=1}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^{j+r}}{2^{qk-1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(qk-2j)}^{qk+2r-1}}{(qk+2r-1)!}}$$

and thus

$$\underset{x\to 0^{+}}{lim}A\left(x\right)=\pi \sum _{k=1}^n\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^{k+j+r}}{2^{qk}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(qk-2j)}^{qk+2r-1}}{(qk+2r-1)!}}.$$

This completes the proof. □

The recent Monthly problem (see [5]) corresponds to one of the integrals in Theorem 2 by choosing with $q=2$ and $r=1$. Using similar reasoning, we are able to establish the values of the cosine integrals for even integrands.

Theorem 3. 

If q and n are positive integers, then

$${\int }_0^{\infty }{\displaystyle \frac{{(1-{cos}^{q}x)}^n}{x^{2r}}}dx={(-1)}^r\pi \sum _{k=1}^n\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^k}{2^{qk}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(qk-2j)}^{2r-1}}{(2r-1)!}}$$

for each integer r that satisfies $1\le r\le n$.

Proof. 

For a given set of appropriate values for q, n, and r, let $A\left(x\right)$ represent the function

$$A\left(x\right)={\int }_x^{\infty }{\displaystyle \frac{{(1-{cos}^{q}t)}^n}{t^{2r}}}dt.$$

We want to find the value of $\underset{x\to 0^{+}}{lim}A\left(x\right)$. Since

$${\int }_{\mathrm{\Pi }}{\displaystyle \frac{{(1-{cos}^{q}t)}^n}{t^{2r}}}dt={\int }_{\mathrm{\Pi }}\sum _{k=1}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{cos}^{qk}t}{t^{2r}}}dt,$$

it is sufficient to find the value of each of the expressions ${\displaystyle {\int }_{\mathrm{\Pi }}\frac{{cos}^{qk}t}{t^{2r}}dt}$. Using the EC identity when $qk$ is even and the $C_e$ equation, we find that

$$\begin{array}{cc}\hfill {\int }_{\mathrm{\Pi }}{\displaystyle \frac{{cos}^{qk}t}{t^{2r}}}dt& ={\displaystyle \frac{1}{2^{qk-1}}}\sum _{j=0}^{(qk/2)-1}\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\int }_{\mathrm{\Pi }}{\displaystyle \frac{cos\left((qk-2j)t\right)}{t^{2r}}}dt\hfill \\ \hfill & ={\displaystyle \frac{1}{2^{qk-1}}}\sum _{j=0}^{(qk/2)-1}\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){(-1)}^r{\displaystyle \frac{{(qk-2j)}^{2r-1}}{(2r-1)!}}\hfill \\ \hfill & =\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^r}{2^{qk-1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(qk-2j)}^{2r-1}}{(2r-1)!}}.\hfill \end{array}$$

Similarly, when $qk$ is odd, we obtain

$$\begin{array}{cc}\hfill {\int }_{\mathrm{\Pi }}{\displaystyle \frac{{cos}^{qk}t}{t^{2r}}}dt& ={\displaystyle \frac{1}{2^{qk-1}}}\sum _{j=0}^{(qk-1)/2}\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\int }_{\mathrm{\Pi }}{\displaystyle \frac{cos\left((qk-2j)t\right)}{t^{2r}}}dt\hfill \\ \hfill & ={\displaystyle \frac{1}{2^{qk-1}}}\sum _{j=0}^{(qk-1)/2}\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){(-1)}^r{\displaystyle \frac{{(qk-2j)}^{2r-1}}{(2r-1)!}}\hfill \\ \hfill & =\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^r}{2^{qk-1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(qk-2j)}^{2r-1}}{(2r-1)!}}.\hfill \end{array}$$

It follows that

$${\int }_{\mathrm{\Pi }}{\displaystyle \frac{{(1-{cos}^{q}t)}^n}{t^{2r}}}dt=\sum _{k=1}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^r}{2^{qk-1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(qk-2j)}^{2r-1}}{(2r-1)!}}$$

and, thus,

$$\underset{x\to 0^{+}}{lim}A\left(x\right)=\pi \sum _{k=1}^n\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^{k+r}}{2^{qk}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(qk-2j)}^{2r-1}}{(2r-1)!}}.$$

This completes the proof. □

4. Integrals with Odd Integrands

We now consider the integrals that have odd integrands. The antiderivatives in these cases involve the $\mathrm{Ci}$ function. The next four lemmas provide the necessary results to evaluate these integrals.

Lemma 2. 

For each nonnegative integer n, we have

$$\begin{array}{cc}\hfill {\int }_x^{\infty }{\displaystyle \frac{cost}{t^{2n+1}}}dt& ={\displaystyle \frac{{(-1)}^{n+1}}{\left(2n\right)!}}\left(U_n\left(x\right)cosx+V_n\left(x\right)sinx+\mathrm{Ci}\left(x\right)\right);\hfill \\ \hfill {\int }_x^{\infty }{\displaystyle \frac{sint}{t^{2n+2}}}dt& ={\displaystyle \frac{{(-1)}^{n+1}}{(2n+1)!}}\left(U_n\left(x\right)cosx+V_{n+1}\left(x\right)sinx+\mathrm{Ci}\left(x\right)\right).\hfill \end{array}$$

Proof. 

To prove the result for the integral involving the cosine function, it is sufficient to verify that

$$\begin{array}{cc}\hfill \underset{x\to \infty }{lim}\left(U_n\left(x\right)cosx+V_n\left(x\right)sinx+\mathrm{Ci}\left(x\right)\right)& =0;\hfill \\ \hfill {\displaystyle \frac{{(-1)}^{n+1}}{\left(2n\right)!}}·{\displaystyle \frac{d}{dx}}\left(U_n\left(x\right)cosx+V_n\left(x\right)sinx+\mathrm{Ci}\left(x\right)\right)& =-{\displaystyle \frac{cosx}{x^{2n+1}}};\hfill \end{array}$$

for all $n\ge 0$. It is easy to verify the value of the limit. For the derivative, we note that (refer to the properties of the functions $U_n$ and $V_n$ established earlier)

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dx}}\left(U_n\left(x\right)cosx+V_n\left(x\right)sinx+\mathrm{Ci}\left(x\right)\right)& =-U_n\left(x\right)sinx+U_n^{\prime }\left(x\right)cosx+V_n\left(x\right)cosx+V_n^{\prime }\left(x\right)sinx+{\displaystyle \frac{cosx}{x}}\hfill \\ \hfill & =\left(U_n^{\prime }\left(x\right)+V_n\left(x\right)+{\displaystyle \frac{1}{x}}\right)cosx\hfill \\ \hfill & =-\left(V_{n+1}\left(x\right)-V_n\left(x\right)\right)cosx\hfill \\ \hfill & =-{\displaystyle \frac{{(-1)}^{n+1}\left(2n\right)!}{x^{2n+1}}}cosx;\hfill \end{array}$$

this gives the desired derivative for all $n\ge 0$.

The proof for the sine result is similar. Since the limit is elementary, we need to verify that

$${\displaystyle \frac{{(-1)}^{n+1}}{(2n+1)!}}·{\displaystyle \frac{d}{dx}}\left(U_n\left(x\right)cosx+V_{n+1}\left(x\right)sinx+\mathrm{Ci}\left(x\right)\right)=-{\displaystyle \frac{sinx}{x^{2n+2}}}$$

for all $n\ge 0$. For this function, we have

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dx}}(U_n\left(x\right)cosx& + V_{n+1}\left(x\right)sinx+\mathrm{Ci}\left(x\right))\hfill \\ \hfill & =-U_n\left(x\right)sinx+U_n^{\prime }\left(x\right)cosx+V_{n+1}\left(x\right)cosx+V_{n+1}^{\prime }\left(x\right)sinx+{\displaystyle \frac{cosx}{x}}\hfill \\ \hfill & =\left(U_{n+1}\left(x\right)-U_n\left(x\right)\right)sinx+\left(U_n^{\prime }\left(x\right)+V_{n+1}\left(x\right)+{\displaystyle \frac{1}{x}}\right)cosx\hfill \\ \hfill & ={\displaystyle \frac{{(-1)}^{n+2}(2n+1)!}{x^{2n+2}}}sinx,\hfill \end{array}$$

which is the expected value for the derivative for all $n\ge 0$. □

Let $h_k=\sum _{j=1}^{k}1/j$ and $o_k=\sum _{j=1}^{k}1/(2j-1)$ for each positive integer k, and let $h_0=0=o_0$. Note that

$$o_k=\sum _{j=1}^k{\displaystyle \frac{1}{2j-1}}=\sum _{j=1}^{2k}{\displaystyle \frac{1}{j}}-{\displaystyle \frac{1}{2}}\sum _{j=1}^k{\displaystyle \frac{1}{j}}=h_{2k}-{\displaystyle \frac{1}{2}}{h}_k$$

for all $k\ge 1$. It follows that $o_k+{\displaystyle \frac{1}{2}}{h}_k=h_{2k}$ and $o_{k+1}+{\displaystyle \frac{1}{2}}{h}_k=h_{2k+1}$ for all $k\ge 0$.

Lemma 3. 

Let n be a positive integer.

(a)

For each positive integer k that satisfies $1\le k\le 2n+1$, the coefficient for the constant term in the Laurent series for the function $U_k\left(x\right)cosx$ is $-{\displaystyle \frac{1}{2}}{h}_k$.

(b)

For each positive integer k that satisfies $1\le k\le 2n+2$, the coefficient for the constant term in the Laurent series for the function $V_k\left(x\right)sinx$ is $-o_k$.

(c)

For each positive integer k that satisfies $1\le k\le 2n+1$, the coefficient for the constant term in the Laurent series for the function $U_k\left(x\right)cosx+V_k\left(x\right)sinx$ is $-h_{2k}$.

(d)

For each positive integer k that satisfies $1\le k\le 2n+1$, the coefficient for the constant term in the Laurent series for the function $U_k\left(x\right)cosx+V_{k+1}\left(x\right)sinx$ is $-h_{2k+1}$.

Proof. 

For appropriate values of n and k as given for part (a), the constant term for $U_k\left(x\right)cosx$ satisfies

$$\left(\sum _{j=1}^k{\displaystyle \frac{{(-1)}^{j+1}(2j-1)!}{x^{2j}}}\right)\left(\sum _{m=0}^{2n+1}{\displaystyle \frac{{(-1)}^m}{\left(2m\right)!}}{x}^{2m}\right)=\sum _{j=1}^k{\displaystyle \frac{{(-1)}^{j+1}(2j-1)!{(-1)}^j}{\left(2j\right)!}}=-{\displaystyle \frac{1}{2}}\sum _{j=1}^k{\displaystyle \frac{1}{j}}=-{\displaystyle \frac{1}{2}}{h}_k.$$

Note that we have just used a finite portion of the Maclaurin series for $cosx$ as the terms involving higher powers generate polynomial terms in the product. Similarly, the constant term for $V_k\left(x\right)sinx$ satisfies

$$\left(\sum _{j=1}^k{\displaystyle \frac{{(-1)}^j(2j-2)!}{x^{2j-1}}}\right)\left(\sum _{m=0}^{2n+1}{\displaystyle \frac{{(-1)}^m}{(2m+1)!}}{x}^{2m+1}\right)=\sum _{j=1}^k{\displaystyle \frac{{(-1)}^j(2j-2)!{(-1)}^{j-1}}{(2j-1)!}}=-\sum _{j=1}^k{\displaystyle \frac{1}{2j-1}}=-o_k$$

for appropriate values of n and k as given for part (b). Adding the results for parts (a) and (b) then gives parts (c) and (d). □

It is easy to see that the constant terms that appear in the Laurent series for the functions

$$U_k\left(x\right)cosx+V_k\left(x\right)sinx\mathrm{and}{U}_k\left(x\right)cosx+V_{k+1}\left(x\right)sinx$$

are the same as the constant terms that appear in the functions

$$U_k\left(rx\right)cos\left(rx\right)+V_k\left(rx\right)sin\left(rx\right)\mathrm{and}{U}_k\left(rx\right)cos\left(rx\right)+V_{k+1}\left(rx\right)sin\left(rx\right),$$

respectively, for any positive integer r. This fact will be used in the proofs of the following results.

Lemma 4. 

Let m be a positive integer. For a given positive integer w, consider the function

$$B\left(x\right)={\displaystyle \frac{P\left(x\right)}{x^{2w}}}+\sum _{k=1}^m{a}_k\left(U_{p_k}\left(r_{k}x\right)cos\left(r_{k}x\right)+V_{p_k}\left(r_{k}x\right)sin\left(r_{k}x\right)+\mathrm{Ci}\left(r_{k}x\right)\right),$$

where the $p_k$ and $r_k$ values are positive integers, the $a_k$ values are real numbers, and P is a polynomial of degree less than $2w$ that is an even function. If $\underset{x\to 0^{+}}{lim}B\left(x\right)$ exists, then

$$\sum _{k=1}^m{a}_k=0\mathrm{and}\underset{x\to 0^{+}}{lim}B\left(x\right)=\sum _{k=1}^m{a}_k\left(ln\left(r_k\right)-h_{2p_k}\right).$$

Proof. 

The function $B\left(x\right)$ can be expressed as $L\left(x\right)+{\displaystyle \sum _{k=1}^m}{a}_k\mathrm{Ci}\left(r_{k}x\right)$, where $L\left(x\right)$ is a Laurent series for an even function. Since the $\mathrm{Ci}$ expressions involve the function $lnx$, they cannot have any impact (in the sense of affecting the existence of a limit) on the Laurent series terms for x near 0. In other words, the constant term that appears in $\underset{x\to 0^{+}}{lim}L\left(x\right)$ is not affected by the $\mathrm{Ci}$ functions. By part (c) of Lemma 3, the value of this constant is $-{\displaystyle \sum _{k=1}^m}{a}_k{h}_{2p_k}$. Recall that

$$\mathrm{Ci}\left(x\right)=\gamma +lnx-\mathrm{Cin}\left(x\right),\mathrm{where}\mathrm{Cin}\left(x\right)={\int }_0^x{\displaystyle \frac{1-cost}{t}}dt.$$

It is clear that $\underset{x\to 0^{+}}{lim}\mathrm{Cin}\left(rx\right)=0$ for any $r>0$. We then note that

$$\begin{array}{cc}\hfill \sum _{k=1}^m{a}_k\mathrm{Ci}\left(r_{k}x\right)& =\sum _{k=1}^m{a}_k\left(\gamma +ln\left(r_{k}x\right)-\mathrm{Cin}\left(r_{k}x\right)\right)\hfill \\ \hfill & =\gamma \sum _{k=1}^m{a}_k+\sum _{k=1}^m{a}_{k}ln\left(r_k\right)+lnx\sum _{k=1}^m{a}_k-\sum _{k=1}^m{a}_k\mathrm{Cin}\left(r_{k}x\right).\hfill \end{array}$$

In order for this particular function to have a limit as $x\to 0^{+}$, we must have ${\displaystyle \sum _{k=1}^m}{a}_k=0$, giving us a limit of ${\displaystyle \sum _{k=1}^m}{a}_{k}ln\left(r_k\right)$. Putting the two parts together, we find that ${\displaystyle \underset{x\to 0^{+}}{lim}B\left(x\right)=\sum _{k=1}^m{a}_k\left(ln\left(r_k\right)-h_{2p_k}\right)}$. □

Lemma 5. 

Let m be a positive integer. For a given positive integer w, consider the function

$$B\left(x\right)={\displaystyle \frac{P\left(x\right)}{x^{2w}}}+\sum _{k=1}^m{a}_k\left(U_{p_k}\left(r_{k}x\right)cos\left(r_{k}x\right)+V_{p_k+1}\left(r_{k}x\right)sin\left(r_{k}x\right)+\mathrm{Ci}\left(r_{k}x\right)\right),$$

where the $p_k$ and $r_k$ values are positive integers, the $a_k$ values are real numbers, and P is a polynomial of degree less than $2w$ that is an even function. If $\underset{x\to 0^{+}}{lim}B\left(x\right)$ exists, then

$$\sum _{k=1}^m{a}_k=0\mathrm{and}\underset{x\to 0^{+}}{lim}B\left(x\right)=\sum _{k=1}^m{a}_k\left(ln\left(r_k\right)-h_{2p_k+1}\right).$$

Proof. 

The proof is almost identical to the proof of Lemma 4. □

As we did for the integrals with even integrands, we first focus on a somewhat simpler case to help clarify the process for odd integrands.

Theorem 4. 

If $q\ge 1$ and $n\ge 2$ are positive integers, then

$${\int }_0^{\infty }{\displaystyle \frac{{(x^{2q}-{sin}^{2q}x)}^n}{x^{2qn+2r+1}}}dx=2{(-1)}^{r+1}\sum _{k=1}^n\sum _{j=0}^{qk-1}{\displaystyle \frac{{(-1)}^{k+j}}{2^{2qk}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2qk}{j}}\right){\displaystyle \frac{{(2qk-2j)}^{2qk+2r}}{(2qk+2r)!}}\left(ln\left(2qk-2j\right)-h_{2qk+2r}\right)$$

for each integer r that satisfies $1\le r\le n-1$.

Proof. 

For a given set of appropriate values for q, n, and r, let $A\left(x\right)$ represent the function

$$A\left(x\right)={\int }_x^{\infty }{\displaystyle \frac{{(t^{2q}-{sin}^{2q}t)}^n}{t^{2qn+2r+1}}}dt.$$

We want to find the value of $\underset{x\to 0^{+}}{lim}A\left(x\right)$. Using the ES identity, we find that (as in the proof of Theorem 1)

$${\displaystyle \frac{{(t^{2q}-{sin}^{2q}t)}^n}{t^{2qn+2r+1}}}={\displaystyle \frac{1}{t^{2r+1}}}+\sum _{k=1}^n\sum _{j=0}^{qk-1}{\displaystyle \frac{{(-1)}^{qk+k+j}}{2^{2qk-1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2qk}{j}}\right){\displaystyle \frac{cos\left((2qk-2j)t\right)}{t^{2qk+2r+1}}}+\sum _{k=1}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2qk}{qk}}\right){\displaystyle \frac{{(-1)}^k}{2^{2qk}{t}^{2qk+2r+1}}}.$$

To simplify the notation a bit, let ${\theta }_{j,k}=2qk-2j$. Using Lemma 2, we see that

$$\begin{array}{cc}\hfill {\int }_x^{\infty }{\displaystyle \frac{cos\left({\theta }_{j,k}t\right)}{t^{2qk+2r+1}}}dt& ={\theta }_{j,k}^{2qk+2r}{\int }_{{\theta }_{j,k}x}^{\infty }{\displaystyle \frac{cosu}{u^{2qk+2r+1}}}du\hfill \\ \hfill & ={\theta }_{j,k}^{2qk+2r}{\displaystyle \frac{{(-1)}^{qk+r+1}}{(2qk+2r)!}}\left(U_{qk+r}\left({\theta }_{j,k}x\right)cos\left({\theta }_{j,k}x\right)+V_{qk+r}\left({\theta }_{j,k}x\right)sin\left({\theta }_{j,k}x\right)+\mathrm{Ci}\left({\theta }_{j,k}x\right)\right).\hfill \end{array}$$

It follows that

$$\begin{array}{cc}\hfill A\left(x\right)& ={\displaystyle \frac{1}{\left(2r\right)x^{2r}}}+\sum _{k=1}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2qk}{qk}}\right){\displaystyle \frac{{(-1)}^k}{2^{2qk}(2qk+2r)x^{2qk+2r}}}+\hfill \\ \hfill & \sum _{k=1}^n\sum _{j=0}^{qk-1}{\displaystyle \frac{{(-1)}^{qk+k+j}}{2^{2qk-1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2qk}{j}}\right){\displaystyle \frac{{\theta }_{j,k}^{2qk+2r}{(-1)}^{qk+r+1}}{(2qk+2r)!}}(U_{qk+r}\left({\theta }_{j,k}x\right)cos\left({\theta }_{j,k}x\right)\hfill \\ \hfill & +V_{qk+r}\left({\theta }_{j,k}x\right)sin\left({\theta }_{j,k}x\right)+\mathrm{Ci}\left({\theta }_{j,k}x\right))\hfill \\ \hfill & ={\displaystyle \frac{1}{\left(2r\right)x^{2r}}}+\sum _{k=1}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2qk}{qk}}\right){\displaystyle \frac{{(-1)}^k}{2^{2qk}(2qk+2r)x^{2qk+2r}}}+\hfill \\ \hfill & \sum _{k=1}^n\sum _{j=0}^{qk-1}{\displaystyle \frac{{(-1)}^{k+j+r+1}}{2^{2qk-1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2qk}{j}}\right){\displaystyle \frac{{\theta }_{j,k}^{2qk+2r}}{(2qk+2r)!}}(U_{qk+r}\left({\theta }_{j,k}x\right)cos\left({\theta }_{j,k}x\right)\hfill \\ \hfill & +V_{qk+r}\left({\theta }_{j,k}x\right)sin\left({\theta }_{j,k}x\right)+\mathrm{Ci}\left({\theta }_{j,k}x\right)).\hfill \end{array}$$

Referring to Lemma 4, we find that

$$\underset{x\to 0^{+}}{lim}A\left(x\right)=\sum _{k=1}^n\sum _{j=0}^{qk-1}{\displaystyle \frac{{(-1)}^{k+j+r+1}}{2^{2qk-1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2qk}{j}}\right){\displaystyle \frac{{\theta }_{j,k}^{2qk+2r}}{(2qk+2r)!}}\left(ln\left({\theta }_{j,k}\right)-h_{2qk+2r}\right).$$

Inserting the values for the ${\theta }_{j,k}$ terms, we obtain

$$\underset{x\to 0^{+}}{lim}A\left(x\right)=2{(-1)}^{r+1}\sum _{k=1}^n\sum _{j=0}^{qk-1}{\displaystyle \frac{{(-1)}^{k+j}}{2^{2qk}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2qk}{j}}\right){\displaystyle \frac{{(2qk-2j)}^{2qk+2r}}{(2qk+2r)!}}\left(ln\left(2qk-2j\right)-h_{2qk+2r}\right).$$

This gives the stated value of the integral. □

In the proof of Theorem 4, we found an explicit expression for the function $A\left(x\right)$. However, as indicated in the proof, we only need to find the constant term that appears in the Laurent series portion of the antiderivative and the coefficients for terms of the form $\mathrm{Ci}\left(rx\right)$. We denote these values by ${\int }_{\mathcal{L}}f\left(t\right)dt$ for appropriate functions f. By Lemmas 3–5 and Theorem 4, we find that

$$\begin{array}{ccc}\hfill {\int }_{\mathcal{L}}{\displaystyle \frac{cos\left(\theta t\right)}{t^{2a+1}}}dt& ={\displaystyle \frac{{(-1)}^{a+1}{\theta }^{2a}}{\left(2a\right)!}}\left(ln\theta -h_{2a}\right);\hfill & \hfill \left(C_o\right)\\ \hfill {\int }_{\mathcal{L}}{\displaystyle \frac{sin\left(\theta t\right)}{t^{2a+2}}}dt& ={\displaystyle \frac{{(-1)}^{a+1}{\theta }^{2a+1}}{(2a+1)!}}\left(ln\theta -h_{2a+1}\right);\hfill & \hfill \left(S_o\right)\end{array}$$

when $\theta$ and a are positive integers.

Theorem 5. 

If $q\ge 1$ and $n\ge 2$ are positive integers, then

$${\int }_0^{\infty }{\displaystyle \frac{{(x^q-{sin}^{q}x)}^n}{x^{qn+2r+1}}}dx=2{(-1)}^{r+1}\sum _{k=1}^n\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^{j+k}}{2^{qk}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(qk-2j)}^{qk+2r}}{(qk+2r)!}}\left(ln(qk-2j)-h_{qk+2r}\right)$$

for each integer r that satisfies $1\le r\le n-1$.

Proof. 

For a given set of appropriate values for q, n, and r, let $A\left(x\right)$ represent the function

$$A\left(x\right)={\int }_x^{\infty }{\displaystyle \frac{{(t^q-{sin}^{q}t)}^n}{t^{qn+2r+1}}}dt.$$

We want to find the value of $\underset{x\to 0^{+}}{lim}A\left(x\right)$. Since

$${\int }_{\mathcal{L}}{\displaystyle \frac{{(t^q-{sin}^{q}t)}^n}{t^{qn+2r+1}}}dt={\int }_{\mathcal{L}}\sum _{k=1}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{sin}^{qk}t}{t^{qk+2r+1}}}dt,$$

it is sufficient to find the value of each of the expressions ${\displaystyle {\int }_{\mathcal{L}}\frac{{sin}^{qk}t}{t^{qk+2r+1}}dt.}$ Using the ES identity when $qk$ is even and the $C_o$ equation, we find that

$$\begin{array}{cc}\hfill {\int }_{\mathcal{L}}{\displaystyle \frac{{sin}^{qk}t}{t^{qk+2r+1}}}dt& ={\displaystyle \frac{{(-1)}^{qk/2}}{2^{qk-1}}}\sum _{j=0}^{(qk/2)-1}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\int }_{\mathcal{L}}{\displaystyle \frac{cos\left((qk-2j)t\right)}{t^{qk+2r+1}}}dt\hfill \\ \hfill & ={\displaystyle \frac{{(-1)}^{qk/2}}{2^{qk-1}}}\sum _{j=0}^{(qk/2)-1}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(-1)}^{(qk/2)+r+1}{(qk-2j)}^{qk+2r}}{(qk+2r)!}}\left(ln(qk-2j)-h_{qk+2r}\right)\hfill \\ \hfill & =\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^{j+r+1}}{2^{qk-1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(qk-2j)}^{qk+2r}}{(qk+2r)!}}\left(ln(qk-2j)-h_{qk+2r}\right).\hfill \end{array}$$

Similarly, when $qk$ is odd, we use the OS identity and the $S_o$ equation to obtain

$$\begin{array}{cc}\hfill {\int }_{\mathcal{L}}{\displaystyle \frac{{sin}^{qk}t}{t^{qk+2r+1}}}dt& ={\displaystyle \frac{{(-1)}^{(qk+3)/2}}{2^{qk-1}}}\sum _{j=0}^{(qk-1)/2}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\int }_{\mathcal{L}}{\displaystyle \frac{sin\left((qk-2j)t\right)}{t^{qk+2r+1}}}dt\hfill \\ \hfill & ={\displaystyle \frac{{(-1)}^{(qk+3)/2}}{2^{qk-1}}}\sum _{j=0}^{(qk-1)/2}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){(-1)}^{\left(\right(qk-1)/2)+r+1}{\displaystyle \frac{{(qk-2j)}^{qk+2r}}{(qk+2r)!}}\left(ln(qk-2j)-h_{qk+2r}\right)\hfill \\ \hfill & =\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^{j+r+1}}{2^{qk-1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(qk-2j)}^{qk+2r}}{(qk+2r)!}}\left(ln(qk-2j)-h_{qk+2r}\right).\hfill \end{array}$$

It follows that the value of $\underset{x\to 0^{+}}{lim}A\left(x\right)$ is

$${\int }_{\mathcal{L}}{\displaystyle \frac{{(t^q-{sin}^{q}t)}^n}{t^{qn+2r+1}}}dt=\sum _{k=1}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^{j+r+1}}{2^{qk-1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(qk-2j)}^{qk+2r}}{(qk+2r)!}}\left(ln(qk-2j)-h_{qk+2r}\right),$$

which is an equivalent form of the value given in the statement of the theorem. □

Theorem 6. 

If $q\ge 1$ and $n\ge 2$ are positive integers, then

$${\int }_0^{\infty }{\displaystyle \frac{{(1-{cos}^{q}x)}^n}{x^{2r+1}}}dx=2{(-1)}^{r+1}\sum _{k=1}^n\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^k}{2^{qk}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(qk-2j)}^{2r}}{\left(2r\right)!}}ln(qk-2j)$$

for each integer r that satisfies $1\le r\le n-1$.

Proof. 

For a given set of appropriate values for q, n, and r, let $A\left(x\right)$ represent the function

$$A\left(x\right)={\int }_x^{\infty }{\displaystyle \frac{{(1-{cos}^{q}t)}^n}{t^{2r+1}}}dt.$$

We want to find the value of $\underset{x\to 0^{+}}{lim}A\left(x\right)$. Since

$${\int }_{\mathcal{L}}{\displaystyle \frac{{(1-{cos}^{q}t)}^n}{t^{2r+1}}}dt={\int }_{\mathcal{L}}\sum _{k=1}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{cos}^{qk}t}{t^{2r+1}}}dt,$$

it is sufficient to find the value of each of the expressions ${\displaystyle {\int }_{\mathcal{L}}\frac{{cos}^{qk}t}{t^{2r+1}}dt}$. Using the EC identity when $qk$ is even and the $C_o$ equation, we find that

$$\begin{array}{cc}\hfill {\int }_{\mathcal{L}}{\displaystyle \frac{{cos}^{qk}t}{t^{2r+1}}}dt& ={\displaystyle \frac{1}{2^{qk-1}}}\sum _{j=0}^{(qk/2)-1}\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\int }_{\mathcal{L}}{\displaystyle \frac{cos\left((qk-2j)t\right)}{t^{2r+1}}}dt\hfill \\ \hfill & ={\displaystyle \frac{1}{2^{qk-1}}}\sum _{j=0}^{(qk/2)-1}\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(-1)}^{r+1}{(qk-2j)}^{2r}}{\left(2r\right)!}}\left(ln(qk-2j)-h_{2r}\right)\hfill \\ \hfill & =\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^{r+1}}{2^{qk-1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(qk-2j)}^{2r}}{\left(2r\right)!}}\left(ln(qk-2j)-h_{2r}\right).\hfill \end{array}$$

Similarly, when $qk$ is odd, we obtain

$$\begin{array}{cc}\hfill {\int }_{\mathcal{L}}{\displaystyle \frac{{cos}^{qk}t}{t^{2r+1}}}dt& ={\displaystyle \frac{1}{2^{qk-1}}}\sum _{j=0}^{(qk-1)/2}\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\int }_{\mathcal{L}}{\displaystyle \frac{cos\left((qk-2j)t\right)}{t^{2r+1}}}dt\hfill \\ \hfill & ={\displaystyle \frac{1}{2^{qk-1}}}\sum _{j=0}^{(qk-1)/2}\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(-1)}^{r+1}{(qk-2j)}^{2r}}{\left(2r\right)!}}\left(ln(qk-2j)-h_{2r}\right)\hfill \\ \hfill & =\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^{r+1}}{2^{qk-1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(qk-2j)}^{2r}}{\left(2r\right)!}}\left(ln(qk-2j)-h_{2r}\right).\hfill \end{array}$$

Referring to Lemma 4 and noting that the constant $h_{2r}$ appears in all of the terms, we find that the term $h_{2r}$ disappears when all the terms are added together. It follows that the value of $\underset{x\to 0^{+}}{lim}A\left(x\right)$ is

$${\int }_{\mathcal{L}}{\displaystyle \frac{{(1-{cos}^{q}t)}^n}{t^{2r+1}}}dt=\sum _{k=1}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^{r+1}}{2^{qk-1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(qk-2j)}^{2r}}{\left(2r\right)!}}ln(qk-2j),$$

which is an equivalent form of the value given in the statement of the theorem. □

Corollary 1. 

The identities

$$\begin{array}{cc}\hfill \sum _{k=1}^n\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^{j+k}}{2^{qk}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){\displaystyle \frac{{(qk-2j)}^{qk+2r}}{(qk+2r)!}}& =0;\hfill \\ \hfill \sum _{k=1}^n\sum _{j=0}^{\lfloor (qk-1)/2\rfloor }{\displaystyle \frac{{(-1)}^k}{2^{qk}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{qk}{j}}\right){(qk-2j)}^{2r}& =0;\hfill \end{array}$$

are valid for integers q, n, and r that satisfy $q\ge 1$, $n\ge 2$, and $1\le r\le n-1$.

Proof. 

These two results follow from the last expressions for the functions $A\left(x\right)$ in the proofs of Theorems 5 and 6, along with the first part of the conclusion of Lemma 4. It is not clear how to prove these binomial identities using other methods. □

In Theorems 5 and 6, the arguments for the logarithm function that appear in the values of the integrals are $qk-2j$ for $1\le k\le n$ and $0\le j\le \lfloor (qk-1)/2\rfloor$. Given positive integers q and n, let $S(n,q)$ denote the set $\{lnp:2\le p\le qn,p\mathrm{is}\mathrm{prime}\}$. Using Theorem 6, it is easy to verify that the value of

$${\int }_0^{\infty }{\displaystyle \frac{{(1-{cos}^{q}x)}^n}{x^{2r+1}}}dx$$

belongs to the span of

• $S(n,q/2)$ using rational numbers for the scalars when q is even;
• $S(n,q)$ using rational numbers for the scalars when q is odd and n is odd;
• $S(n-1,q)$ using rational numbers for the scalars when q is odd and n is even.

An analogous result is valid for the integrals

$${\int }_0^{\infty }{\displaystyle \frac{{(x^{2q}-{sin}^{2q}x)}^n}{x^{2qn+2r+1}}}dx$$

by replacing the set $S(n,q)$ with the set $\left\{1\right\}\cup S(n,q)$.

We note in passing that the integrals in this paper can also be evaluated using the Laplace Transform Operator Method (see [1]). It is enlightening to compare the formal values of the sine integrals

$$\begin{array}{ccc}{\int }_{\mathrm{\Pi }}{\displaystyle \frac{sin\left(\theta t\right)}{t^{2a+1}}}dt={\displaystyle \frac{{(-1)}^a{\theta }^{2a}}{\left(2a\right)!}};\hfill & \hfill & {\int }_{\mathcal{L}}{\displaystyle \frac{sin\left(\theta t\right)}{t^{2a+2}}}dt={\displaystyle \frac{{(-1)}^{a+1}{\theta }^{2a+1}}{(2a+1)!}}\left(ln\theta -h_{2a+1}\right);\hfill \end{array}$$

and the cosine integrals

$$\begin{array}{ccc}{\int }_{\mathrm{\Pi }}{\displaystyle \frac{cos\left(\theta t\right)}{t^{2a}}}dt={\displaystyle \frac{{(-1)}^a{\theta }^{2a-1}}{(2a-1)!}};\hfill & \hfill & {\int }_{\mathcal{L}}{\displaystyle \frac{cos\left(\theta t\right)}{t^{2a+1}}}dt={\displaystyle \frac{{(-1)}^{a+1}{\theta }^{2a}}{\left(2a\right)!}}\left(ln\theta -h_{2a}\right);\hfill \end{array}$$

presented in this paper with the limits that appear in Corollary 9 of [1].

5. Conclusions

In this paper, we have found explicit values for two large collections of improper integrals. We are not aware of any particular references for these integrals. The five lemmas presented in the paper can be useful for finding the values of other collections of similar integrals.