Laplace Convolution Integrals Involving Trigonometric and Hyperbolic Functions

Abstract

We calculate several finite integrals involving trigonometric and hyperbolic functions by applying the Laplace convolution theorem and known inverse Laplace transform formulas. As a consequence, we obtain a new integral representation of the Kelvin function $\mathrm{bei}\left(z\right)$, and a new reduction formula for a particular generalized hypergeometric function. In addition, we present several new inverse Laplace transform formulas that do not appear to have been reported in the existing literature.

1. Introduction

Although the Laplace transform is named after the astronomer and mathematician Pierre Simon, Marquis de Laplace, the modern version of this integral transform is due to Doetsch [1]. Nowadays, the Laplace transform and its inverse are commonly employed in the solution of differential, difference, and functional equations. By transforming differentiation and integration into algebraic operations, the Laplace transform simplifies many analytical problems, while its inverse enables the recovery of original functions and facilitates the derivation of explicit series and integral representations.

In addition, the Laplace Transform Method has become a significant technique for the numerical summation of slowly convergent series. Formally established by Gautschi and Milovanović [2], this method serves as an effective means of accelerating series convergence.

In this study, we are interested in the application of the Laplace transform to derive new integrals that are not reported in the most common tables, such as [3,4,5]. This approach has been extensively used by Apelblat in [6], who applied numerous properties of the Laplace transform. In particular, we focus on the application of the convolution theorem of the Laplace transform. Recently, in [7], the convolution theorem was applied to obtain new inverse Laplace transforms from known integrals. We also consider the latter approach in this paper, but we are mainly interested here in the reciprocal one, that is, in obtaining new integrals from known inverse Laplace transforms by applying the convolution theorem. This approach has been used by Apelblat in the context of Volterra functions [8], and in [9] for the calculation of integrals involving modified Bessel functions. Moreover, the calculation of new integrals using the convolution theorem is not restricted to the Laplace transform, but also applies to the Fourier transform, as in [10]. However, it seems that the convolution theorem has not been systematically exploited in order to compute what we have termed convolution integrals. It is worth noting that, in the case of Laplace convolution integrals, these integrals arise naturally in Volterra integral equations, which are used to model systems whose current state depends on all past states [11].

Since the number of these convolution integrals is quite large, we have structured the material as a sequence of papers, depending on the type of function in the integrand. This first paper is devoted to the Laplace convolution integrals involving trigonometric and hyperbolic functions.

This paper is organized as follows. In Section 2, we present the definition of the Laplace transform and its convolution theorem. We also define all the special functions that we will use throughout the paper, along with some of their properties. Section 3 is devoted to the derivation of new integrals involving trigonometric and hyperbolic functions by applying the Laplace convolution theorem to known inverse Laplace transforms, primarily from the standard tables in [12]. As a by-product of these calculations, we present a set of inverse Laplace transform formulas that do not seem to be reported in the existing literature. Finally, we present our conclusions in Section 4.

2. Preliminaries

2.1. Laplace Transform

Definition 1. 

Let $f\left(t\right)$ be a real- or complex-valued function of the variable $t>0$, and let s be a real or complex parameter. The Laplace transform of $f\left(t\right)$ is defined as [13] (Eqn. 1.1):

$$\begin{array}{ccc}\hfill F\left(s\right)& =& \mathcal{L}\left[f\left(t\right);s\right]={\int }_0^{\infty }{e}^{-st}f\left(t\right)dt,\hfill \\ & =& \underset{\tau \to \infty }{\lim }{\int }_0^{\tau }{e}^{-st}f\left(t\right)dt,\hfill \end{array}$$

whenever the limit exists as a finite number.

We recall that if $f\left(t\right)$ is piecewise continuous on $\left[0,\infty \right)$ and of exponential order $\alpha$, then the Laplace transform $\mathcal{L}\left[f\left(t\right);s\right]$ exists for $\Re \left(s\right)>\alpha$ and converges absolutely. In this case, we denote the inverse Laplace transform by

$$f\left(t\right)={\mathcal{L}}^{-1}\left[F\left(s\right);t\right].$$

Theorem 1 

(Convolution). If $f\left(t\right)={\mathcal{L}}^{-1}\left[F\left(s\right);t\right]$ and $g\left(t\right)={\mathcal{L}}^{-1}\left[G\left(s\right);t\right]$ are piecewise continuous on $\left[0,\infty \right)$ and of exponential order α, then [13] (Theorem 2.39)

$$\begin{array}{ccc}& & {\int }_0^t{\mathcal{L}}^{-1}\left[F\left(s\right);\tau \right]{\mathcal{L}}^{-1}\left[G\left(s\right);t-\tau \right]d\tau ={\mathcal{L}}^{-1}\left[F\left(s\right)G\left(s\right);t\right],\hfill \\ & & \Re \left(s\right)>\alpha ,t>0.\hfill \end{array}$$

(1)

Although the convolution Theorem (1) is stated for $t>0$, it is worth noting that we can relax this restriction to negative or complex values of t in many cases. Nevertheless, throughout this paper, we will consider $t>0$.

2.2. Special Functions

For $p\le q$, the following series converges for $z\in \mathbb{C}$ and defines an entire function called the generalized hypergeometric function [14] (Eqn. 16.2.1)

$${}_{p}F_q\left(\begin{array}{c}{a}_1,\dots ,a_p\\ b_1,\dots ,b_q\end{array};z\right)={\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{\left(a_1\right)}_k\cdots {\left(a_p\right)}_k}{{\left(b_1\right)}_k\cdots {\left(b_q\right)}_k}}{\displaystyle \frac{z^k}{k!}},$$

where ${\left(c\right)}_k$ with $k=0,1,2,\dots$ denotes the Pochhammer symbol [14] (Eqn. 5.2.4-5)

$${\left(c\right)}_k=\left\{\begin{array}{cc}c\left(c+1\right)\left(c+2\right)\cdots \left(c+k-1\right),\hfill & {\left(c\right)}_0=1,\hfill \\ {\displaystyle \frac{\mathsf{\Gamma }\left(c+k\right)}{\mathsf{\Gamma }\left(c\right)},}\hfill & c\ne 0,-1,-2,\dots \hfill \end{array}\right.$$

(2)

and $\mathsf{\Gamma }\left(z\right)$ denotes the gamma function [14] (Eqn. 5.2.1)

$$\mathsf{\Gamma }\left(z\right)={\int }_0^{\infty }{t}^{z-1}{e}^{-t}dt,\Re \left(z\right)>0.$$

(3)

The gamma function has the following property [15] (Eqn. 1.2.1)

$$\mathsf{\Gamma }\left(z+1\right)=z\mathsf{\Gamma }\left(z\right).$$

(4)

The beta function is defined as [14] (Eqn. 5.12.1)

$$\mathrm{B}\left(a,b\right)={\int }_0^1{t}^{a-1}{\left(1-t\right)}^{b-1}dt={\displaystyle \frac{\mathsf{\Gamma }\left(a\right)\mathsf{\Gamma }\left(b\right)}{\mathsf{\Gamma }\left(a+b\right)}},\Re \left(a\right),\Re \left(b\right)>0.$$

(5)

The exponential integral is defined as [15] (Eqn. 3.1.1)

$$\mathrm{Ei}\left(z\right)={\int }_{-\infty }^z{\displaystyle \frac{e^{-t}}{t}}dt,\left|\arg \left(-z\right)\right|<\pi ,$$

where the integration contour is taken along any path in the complex plane, avoiding the branch cut on the positive real axis.

For $z\in \mathbb{C}$, the error and the complementary error functions [16] (Eqns. 40:3:1&40:0:1) are defined as

$$\begin{array}{ccc}\hfill \mathrm{erf}\left(z\right)& =& {\displaystyle \frac{2}{\sqrt{\pi }}}{\int }_0^z\exp \left(-t^2\right)dt,\hfill \end{array}$$

(6)

$$\begin{array}{ccc}\hfill \mathrm{erfc}\left(z\right)& =& {\displaystyle \frac{2}{\sqrt{\pi }}}{\int }_z^{\infty }\exp \left(-t^2\right)dt=1-\mathrm{erf}\left(z\right).\hfill \end{array}$$

(7)

For $z\in \mathbb{C}$, the sine and hyperbolic sine integrals are defined as [14] (Sect. 6.2(ii))

$$\begin{array}{ccc}\hfill \mathrm{Si}\left(z\right)& =& {\int }_0^z{\displaystyle \frac{\sin t}{t}}dt,\hfill \end{array}$$

(8)

$$\begin{array}{ccc}\hfill \mathrm{Shi}\left(z\right)& =& {\int }_0^z{\displaystyle \frac{\sinh t}{t}}dt,\hfill \end{array}$$

(9)

and the cosine and hyperbolic cosine integrals are defined as

$$\begin{array}{ccc}\hfill \mathrm{Ci}\left(z\right)& =& \gamma +\log z+{\int }_0^z{\displaystyle \frac{\cos t-1}{t}}dt,\hfill \end{array}$$

(10)

$$\begin{array}{ccc}\hfill \mathrm{Chi}\left(z\right)& =& \gamma +\log z+{\int }_0^z{\displaystyle \frac{\cosh t-1}{t}}dt,\hfill \end{array}$$

(11)

where $\gamma$ denotes the Euler-Mascheroni constant. The following relations are satisfied for $\left|\arg z\right|<\pi /2$ [15] (Eqn. 3.3.6)

$$\begin{array}{ccc}\hfill \mathrm{Ci}\left(z\right)& =& {\displaystyle \frac{1}{2}}\left[\mathrm{Ei}\left(-z{e}^{i\pi /2}\right)+\mathrm{Ei}\left(-z{e}^{-i\pi /2}\right)\right],\hfill \end{array}$$

(12)

$$\begin{array}{ccc}\hfill \mathrm{Si}\left(z\right)& =& {\displaystyle \frac{\pi }{2}}-{\displaystyle \frac{1}{2i}}\left[\mathrm{Ei}\left(-z{e}^{i\pi /2}\right)-\mathrm{Ei}\left(-z{e}^{-i\pi /2}\right)\right].\hfill \end{array}$$

(13)

For $z\in \mathbb{C}$, the Fresnel integrals are defined as [14] (Eqns. 7.2.7&8)

$$\begin{array}{ccc}\hfill \mathrm{S}\left(z\right)& =& {\int }_0^z\sin \left({\displaystyle \frac{\pi t^2}{2}}\right)dt,\hfill \end{array}$$

(14)

$$\begin{array}{ccc}\hfill \mathrm{C}\left(z\right)& =& {\int }_0^z\cos \left({\displaystyle \frac{\pi t^2}{2}}\right)dt.\hfill \end{array}$$

(15)

The Bessel function of the first kind is defined as [15] (Eqn. 5.3.2)

$$J_{\nu }\left(z\right)={\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{\left(-1\right)}^k{\left(z/2\right)}^{\nu +2k}}{k!\mathsf{\Gamma }\left(k+\nu +1\right)}},\left|z\right|<\infty ,\left|\arg z\right|<\pi .$$

(16)

Similarly, the modified Bessel function is defined as [15] (Eqn. 5.7.1)

$$I_{\nu }\left(z\right)={\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{\left(z/2\right)}^{\nu +2k}}{k!\mathsf{\Gamma }\left(k+\nu +1\right)}},\left|z\right|<\infty ,\left|\arg z\right|<\pi .$$

(17)

For positive arguments $x>0$ and real order $\nu \in \mathbb{R}$, the Kelvin functions are defined as [14] (Eqn. 10.61.1)

$$\begin{array}{ccc}\hfill {\mathrm{ber}}_{\nu }\left(x\right)& =& \Re \left(e^{\nu \pi i}{J}_{\nu }\left(x{e}^{-i\pi /4}\right)\right),\hfill \end{array}$$

(18)

$$\begin{array}{ccc}\hfill {\mathrm{bei}}_{\nu }\left(x\right)& =& \Im \left(e^{\nu \pi i}{J}_{\nu }\left(x{e}^{-i\pi /4}\right)\right),\hfill \end{array}$$

(19)

and for other values of x, they are defined by analytic continuation. The following notation is usually adopted [16] (Eqn. 55:1:1)

$$\begin{array}{ccc}\hfill \mathrm{ber}\left(x\right)& =& {\mathrm{ber}}_0\left(x\right),\hfill \end{array}$$

(20)

$$\begin{array}{ccc}\hfill \mathrm{bei}\left(x\right)& =& {\mathrm{bei}}_0\left(x\right).\hfill \end{array}$$

(21)

Finally, for $x\in \mathbb{R}$, we denote the signum function as [16] (Eqn. 8:3:1)

$$\mathrm{sgn}\left(x\right)=\left\{\begin{array}{cc}1,& x>0,\\ 0,& x=0,\\ -1,& x<0.\end{array}\right.$$

3. Main Results

Theorem 2. 

For $a\in \mathbb{C}$, the following convolution integral holds true:

$${\int }_0^t{\displaystyle \frac{{\sinh }^2\left(a\sqrt{\tau }\right){\cos }^2\left(a\sqrt{t-\tau }\right)}{\sqrt{\tau \left(t-\tau \right)}}}d\tau ={\displaystyle \frac{\pi }{4}}\left[I_0\left(2a\sqrt{t}\right)-J_0\left(2a\sqrt{t}\right)\right].$$

(22)

Proof. 

Apply the Laplace convolution Theorem (1), taking into account the following inverse Laplace transforms [12] (Eqns. 2.3.5(7)&(8))

$$\begin{array}{ccc}\hfill {\mathcal{L}}^{-1}\left[{\displaystyle \frac{e^{a/s}\sinh \left(a/s\right)}{\sqrt{s}}};t\right]& =& -{\displaystyle \frac{{\sinh }^2\left(\sqrt{2at}\right)}{\sqrt{\pi t}}},\hfill \\ \hfill {\mathcal{L}}^{-1}\left[{\displaystyle \frac{e^{-a/s}\cosh \left(a/s\right)}{\sqrt{s}}};t\right]& =& {\displaystyle \frac{{\cos }^2\left(\sqrt{2at}\right)}{\sqrt{\pi t}}},\hfill \end{array}$$

and the trigonometric identity $\sinh 2\alpha =2\sinh \alpha \cosh \alpha$, to obtain

$$\begin{array}{ccc}& & -{\displaystyle \frac{1}{\pi }}{\int }_0^t{\displaystyle \frac{{\sinh }^2\left(\sqrt{2a\tau }\right){\cos }^2\left(\sqrt{2a\left(t-\tau \right)}\right)}{\sqrt{\tau \left(t-\tau \right)}}}d\tau \hfill \\ & =& {\mathcal{L}}^{-1}\left[{\displaystyle \frac{\sinh \left(a/s\right)\cosh \left(a/s\right)}{s}};t\right]={\displaystyle \frac{1}{2}}{\mathcal{L}}^{-1}\left[{\displaystyle \frac{\sinh \left(2a/s\right)}{s}};t\right].\hfill \end{array}$$

Now, apply the inverse Laplace transform formula for $\nu =1$ [12] (Eqn. 2.3.3(1))

$${\mathcal{L}}^{-1}\left[{\displaystyle \frac{\sinh \left(a/s\right)}{s^{\nu }}};t\right]=-{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{a}{t}}\right)}^{\left(1-\nu \right)/2}\left[I_{\nu -1}\left(2\sqrt{at}\right)-J_{\nu -1}\left(2\sqrt{at}\right)\right],$$

to complete the proof. □

Theorem 3. 

For $a\in \mathbb{C}$, the following convolution integral holds true:

$$\begin{array}{ccc}& & {\int }_0^t{\displaystyle \frac{\sinh \left(a\sqrt{\tau }\right)\sin \left(a\sqrt{\tau }\right)\cosh \left(a\sqrt{t-\tau }\right)\cos \left(a\sqrt{t-\tau }\right)}{\sqrt{\tau \left(t-\tau \right)}}}d\tau \hfill \\ & =& {\displaystyle \frac{\pi }{2}}\mathrm{bei}\left(2a\sqrt{t}\right).\hfill \end{array}$$

(23)

Proof. 

Apply the Laplace convolution Theorem (1), taking into account the inverse Laplace transforms [12] (Eqns. 2.4.1(2)-(3))

$$\begin{array}{ccc}\hfill {\mathcal{L}}^{-1}\left[{\displaystyle \frac{\sin \left(a/s\right)}{\sqrt{s}}};t\right]& =& {\displaystyle \frac{\sinh \left(\sqrt{2at}\right)\sin \left(\sqrt{2at}\right)}{\sqrt{\pi t}}},\hfill \end{array}$$

(24)

$$\begin{array}{ccc}\hfill {\mathcal{L}}^{-1}\left[{\displaystyle \frac{\cos \left(a/s\right)}{\sqrt{s}}};t\right]& =& {\displaystyle \frac{\cosh \left(\sqrt{2at}\right)\cos \left(\sqrt{2at}\right)}{\sqrt{\pi t}}},\hfill \end{array}$$

(25)

and the trigonometric identity $\sin 2\alpha =2\sin \alpha \cos \alpha$, to obtain

$$\begin{array}{ccc}& & {\displaystyle \frac{1}{\pi }}{\int }_0^t{\displaystyle \frac{\sinh \left(a\sqrt{\tau }\right)\sin \left(a\sqrt{\tau }\right)\cosh \left(a\sqrt{t-\tau }\right)\cos \left(a\sqrt{t-\tau }\right)}{\sqrt{\tau \left(t-\tau \right)}}}d\tau \hfill \\ & =& {\mathcal{L}}^{-1}\left[{\displaystyle \frac{\sin \left(a/s\right)\cos \left(a/s\right)}{s}};t\right]={\displaystyle \frac{1}{2}}{\mathcal{L}}^{-1}\left[{\displaystyle \frac{\sin \left(2a/s\right)}{s}};t\right].\hfill \end{array}$$

Now, apply the following inverse Laplace transform [12] (Eqn. 2.4.1(3))

$${\mathcal{L}}^{-1}\left[{\displaystyle \frac{\sin \left(a/s\right)}{s}};t\right]=\mathrm{bei}\left(2\sqrt{at}\right),$$

to complete the proof. □

Corollary 1. 

Taking $a={\displaystyle \frac{1}{2}}$ and performing the change of variable $\tau =z^2{u}^2$ in (23), we obtain the following integral representation of the $\mathrm{bei}\left(z\right)$ function for $z\in \mathbb{C}$

$$\mathrm{bei}\left(z\right)={\displaystyle \frac{4}{\pi }}{\int }_0^1{\displaystyle \frac{\sinh \left(\frac{zu}{2}\right)\sin \left(\frac{zu}{2}\right)\cosh \left(\frac{z}{2}\sqrt{1-u^2}\right)\cos \left(\frac{z}{2}\sqrt{1-u^2}\right)}{\sqrt{1-u^2}}}du,$$

(26)

which is analogous to the Poisson integral representation [17]

$$\mathrm{bei}\left(z\right)={\displaystyle \frac{2}{\pi }}{\int }_0^1{\displaystyle \frac{\sinh \left(\frac{zu}{\sqrt{2}}\right)\sin \left(\frac{zu}{\sqrt{2}}\right)}{\sqrt{1-u^2}}}du.$$

Theorem 4. 

For $a\in \mathbb{C}$, the following convolution integrals hold true:

$$\begin{array}{ccc}& & {\int }_0^t{\displaystyle \frac{\sinh \left(a\sqrt{\tau }\right)\sin \left(a\sqrt{\tau }\right)\sinh \left(a\sqrt{t-\tau }\right)\sin \left(a\sqrt{t-\tau }\right)}{\sqrt{\tau \left(t-\tau \right)}}}\hfill \\ & =& {\displaystyle \frac{\pi }{2}}\left[1-\mathrm{ber}\left(2a\sqrt{t}\right)\right].\hfill \\ & & {\int }_0^t{\displaystyle \frac{\cosh \left(a\sqrt{\tau }\right)\cos \left(a\sqrt{\tau }\right)\cosh \left(a\sqrt{t-\tau }\right)\cos \left(a\sqrt{t-\tau }\right)}{\sqrt{\tau \left(t-\tau \right)}}}\hfill \\ & =& {\displaystyle \frac{\pi }{2}}\left[1+\mathrm{ber}\left(2a\sqrt{t}\right)\right],\hfill \end{array}$$

Proof. 

Apply the Laplace convolution Theorem (1), taking into account (24), to obtain

$$\begin{array}{ccc}& & {\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\sin }^2\left(a/s\right)}{s}};t\right]\hfill \\ & =& {\int }_0^t{\displaystyle \frac{\sinh \left(\sqrt{2a\tau }\right)\sin \left(\sqrt{2a\tau }\right)\sinh \left(\sqrt{2a\left(t-\tau \right)}\right)\sin \left(\sqrt{2a\left(t-\tau \right)}\right)}{\pi \sqrt{\tau \left(t-\tau \right)}}}.\hfill \end{array}$$

(27)

Similarly, apply the convolution theorem, taking into account (25), to obtain

$$\begin{array}{ccc}& & {\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\cos }^2\left(a/s\right)}{s}};t\right]\hfill \\ & =& {\int }_0^t{\displaystyle \frac{\cosh \left(\sqrt{2a\tau }\right)\cos \left(\sqrt{2a\tau }\right)\cosh \left(\sqrt{2a\left(t-\tau \right)}\right)\cos \left(\sqrt{2a\left(t-\tau \right)}\right)}{\pi \sqrt{\tau \left(t-\tau \right)}}}.\hfill \end{array}$$

(28)

Note that, according to the trigonometric identity ${\sin }^2\alpha +{\cos }^2\alpha =1$, and the inverse Laplace transform given in [12] (Eqn. 2.1.1(1)), we have

$${\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\cos }^2\left(a/s\right)}{s}};t\right]+{\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\sin }^2\left(a/s\right)}{s}};t\right]={\mathcal{L}}^{-1}\left[{\displaystyle \frac{1}{s}};t\right]=1,$$

(29)

but, according to the trigonometric identity $\cos 2\alpha ={\cos }^2\alpha -{\sin }^2\alpha$, and the inverse Laplace transform given in [12] (Eqn. 2.4.1(3)), we have

$$\begin{array}{ccc}& & {\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\cos }^2\left(a/s\right)}{s}};t\right]-{\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\sin }^2\left(a/s\right)}{s}};t\right]\hfill \\ & =& {\mathcal{L}}^{-1}\left[{\displaystyle \frac{\cos \left(2a/s\right)}{s}};t\right]=\mathrm{ber}\left(2\sqrt{2at}\right).\hfill \end{array}$$

(30)

Solving the linear system of equations given in (29) and (30), we obtain

$$\begin{array}{ccc}\hfill {\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\sin }^2\left(a/s\right)}{s}};t\right]& =& {\displaystyle \frac{1-\mathrm{ber}\left(2\sqrt{2at}\right)}{2}},\hfill \end{array}$$

(31)

$$\begin{array}{ccc}\hfill {\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\cos }^2\left(a/s\right)}{s}};t\right]& =& {\displaystyle \frac{1+\mathrm{ber}\left(2\sqrt{2at}\right)}{2}}.\hfill \end{array}$$

(32)

Finally, match the results (27) with (31), and (28) with (32) to complete the proof. □

Remark 1. 

It is noteworthy that, as a by-product of the above proof, we have obtained the inverse Laplace transforms (31) and (32), which do not seem to be reported in the existing literature.

Theorem 5. 

For $\Re \left(\nu \right)>-1$ and $a\in \mathbb{C}$, the following convolution integral holds true:

$${\int }_0^t{\left[\tau \left(t-\tau \right)\right]}^{\nu }\sin \left(a\tau \right)\cos \left(a\left(t-\tau \right)\right)d\tau ={\displaystyle \frac{t^{2\nu +1}}{2}}\mathrm{B}\left(\nu +1,\nu +1\right)\sin \left(at\right).$$

Proof. 

Apply the Laplace convolution Theorem (1), taking into account the inverse Laplace transforms [12] (Eqns. 2.6.7(11)&(12))

$$\begin{array}{ccc}\hfill {\mathcal{L}}^{-1}\left[{\displaystyle \frac{\sin \left(\nu \arctan \left(a/s\right)\right)}{{\left(s^2+a^2\right)}^{\nu /2}}};t\right]& =& -{\displaystyle \frac{t^{\nu -1}}{\mathsf{\Gamma }\left(\nu \right)}}\sin \left(at\right),\hfill \end{array}$$

(33)

$$\begin{array}{ccc}\hfill {\mathcal{L}}^{-1}\left[{\displaystyle \frac{\cos \left(\nu \arctan \left(a/s\right)\right)}{{\left(s^2+a^2\right)}^{\nu /2}}};t\right]& =& {\displaystyle \frac{t^{\nu -1}}{\mathsf{\Gamma }\left(\nu \right)}}\cos \left(at\right),\hfill \end{array}$$

(34)

as well as the trigonometric identity $\sin 2\alpha =2\sin \alpha \cos \alpha$, to obtain

$$\begin{array}{ccc}& & {\displaystyle \frac{1}{{\mathsf{\Gamma }}^2\left(\nu \right)}}{\int }_0^t{\left[\tau \left(t-\tau \right)\right]}^{\nu -1}\sin \left(a\tau \right)\cos \left(a\left(t-\tau \right)\right)d\tau \hfill \\ & =& {\mathcal{L}}^{-1}\left[{\displaystyle \frac{\sin \left(\nu \arctan \left(a/s\right)\right)\cos \left(\nu \arctan \left(a/s\right)\right)}{{\left(s^2+a^2\right)}^{\nu }}};t\right]\hfill \\ & =& {\displaystyle \frac{1}{2}}{\mathcal{L}}^{-1}\left[{\displaystyle \frac{\sin \left(2\nu \arctan \left(a/s\right)\right)}{{\left(s^2+a^2\right)}^{\nu }}};t\right].\hfill \end{array}$$

Finally, apply again the inverse Laplace transform given in (33), and the definition of the beta function (5) to complete the proof. □

Theorem 6. 

For $\Re \left(\nu \right)>-1$ and $a\in \mathbb{C}$, the following convolution integrals hold true:

$$\begin{array}{ccc}& & {\int }_0^t{\left[\tau \left(t-\tau \right)\right]}^{\nu }\sin \left(a\tau \right)\sin \left(a\left(t-\tau \right)\right)d\tau \hfill \\ & =& {\displaystyle \frac{t^{\nu +1/2}}{2}}\left[{\displaystyle \frac{\sqrt{\pi }\mathsf{\Gamma }\left(\nu +1\right)}{{\left(2a\right)}^{\nu +1/2}}}{J}_{\nu +1/2}\left(at\right)-t^{\nu +1/2}\mathrm{B}\left(\nu +1,\nu +1\right)\cos \left(at\right)\right],\hfill \\ & & {\int }_0^t{\left[\tau \left(t-\tau \right)\right]}^{\nu }\cos \left(a\tau \right)\cos \left(a\left(t-\tau \right)\right)d\tau \hfill \\ & =& {\displaystyle \frac{t^{\nu +1/2}}{2}}\left[{\displaystyle \frac{\sqrt{\pi }\mathsf{\Gamma }\left(\nu +1\right)}{{\left(2a\right)}^{\nu +1/2}}}{J}_{\nu +1/2}\left(at\right)+t^{\nu +1/2}\mathrm{B}\left(\nu +1,\nu +1\right)\cos \left(at\right)\right].\hfill \end{array}$$

Proof. 

Apply the Laplace convolution Theorem (1), taking into account (33), to obtain

$$\begin{array}{ccc}& & {\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\sin }^2\left(\nu \arctan \left(a/s\right)\right)}{{\left(s^2+a^2\right)}^{\nu }}};t\right]\hfill \\ & =& {\displaystyle \frac{1}{{\mathsf{\Gamma }}^2\left(\nu \right)}}{\int }_0^t{\left[\tau \left(t-\tau \right)\right]}^{\nu }\sin \left(a\tau \right)\sin \left(a\left(t-\tau \right)\right)d\tau .\hfill \end{array}$$

(35)

Similarly, apply the convolution theorem, taking into account (25), to obtain

$$\begin{array}{ccc}& & {\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\cos }^2\left(\nu \arctan \left(a/s\right)\right)}{{\left(s^2+a^2\right)}^{\nu }}};t\right]\hfill \\ & =& {\displaystyle \frac{1}{{\mathsf{\Gamma }}^2\left(\nu \right)}}{\int }_0^t{\left[\tau \left(t-\tau \right)\right]}^{\nu }\cos \left(a\tau \right)\cos \left(a\left(t-\tau \right)\right)d\tau .\hfill \end{array}$$

(36)

Note that, according to the trigonometric identity ${\sin }^2\alpha +{\cos }^2\alpha =1$, and the inverse Laplace transform given in [12] (Eqn. 2.1.5(1)), we have

$$\begin{array}{ccc}& & {\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\sin }^2\left(\nu \arctan \left(a/s\right)\right)}{{\left(s^2+a^2\right)}^{\nu }}};t\right]+{\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\cos }^2\left(\nu \arctan \left(a/s\right)\right)}{{\left(s^2+a^2\right)}^{\nu }}};t\right]\hfill \end{array}$$

(37)

$$\begin{array}{ccc}& =& {\mathcal{L}}^{-1}\left[{\displaystyle \frac{1}{{\left(s^2+a^2\right)}^{\nu }}};t\right]={\displaystyle \frac{\sqrt{\pi }}{\mathsf{\Gamma }\left(\nu \right)}}{\left({\displaystyle \frac{t}{2a}}\right)}^{\nu -1/2}{J}_{\nu -1/2}\left(at\right),\hfill \end{array}$$

(38)

but, according to the trigonometric identity $\cos 2\alpha ={\cos }^2\alpha -{\sin }^2\alpha$, and the inverse Laplace transform given in (34), we have

$$\begin{array}{ccc}& & {\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\cos }^2\left(\nu \arctan \left(a/s\right)\right)}{{\left(s^2+a^2\right)}^{\nu }}};t\right]-{\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\sin }^2\left(\nu \arctan \left(a/s\right)\right)}{{\left(s^2+a^2\right)}^{\nu }}};t\right]\hfill \\ & =& {\mathcal{L}}^{-1}\left[{\displaystyle \frac{\cos \left(2\nu \arctan \left(a/s\right)\right)}{{\left(s^2+a^2\right)}^{\nu }}};t\right]={\displaystyle \frac{t^{2\nu -1}}{\mathsf{\Gamma }\left(2\nu \right)}}\cos \left(at\right).\hfill \end{array}$$

(39)

Solving the linear system of equations given in (37) and (39), we obtain

$$\begin{array}{ccc}& & {\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\sin }^2\left(\nu \arctan \left(a/s\right)\right)}{{\left(s^2+a^2\right)}^{\nu }}};t\right]\hfill \\ & =& {\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{\sqrt{\pi }{J}_{\nu -1/2}\left(at\right)}{\mathsf{\Gamma }\left(\nu \right)}}{\left({\displaystyle \frac{t}{2a}}\right)}^{\nu -1/2}-{\displaystyle \frac{t^{2\nu -1}}{\mathsf{\Gamma }\left(2\nu \right)}}\cos \left(at\right)\right],\hfill \end{array}$$

(40)

$$\begin{array}{ccc}& & {\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\cos }^2\left(\nu \arctan \left(a/s\right)\right)}{{\left(s^2+a^2\right)}^{\nu }}};t\right]\hfill \\ & =& {\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{\sqrt{\pi }{J}_{\nu -1/2}\left(at\right)}{\mathsf{\Gamma }\left(\nu \right)}}{\left({\displaystyle \frac{t}{2a}}\right)}^{\nu -1/2}+{\displaystyle \frac{t^{2\nu -1}}{\mathsf{\Gamma }\left(2\nu \right)}}\cos \left(at\right)\right].\hfill \end{array}$$

(41)

Finally, matching the results (35) with (40), and (36) with (41), taking into account the definition of the beta function (5), we complete the proof. □

Remark 2. 

It is noteworthy that, as a by-product of the above proof, we have obtained the inverse Laplace transforms (40) and (41), which do not seem to be reported in the existing literature.

Theorem 7. 

For $a\in \mathbb{C}$ and $\Re \left(\nu \right)>0$, the following convolution integral holds true:

$$\begin{array}{ccc}& & {\int }_0^t{\left[\tau \left(t-\tau \right)\right]}^{2\nu -1}\sin \left(\pi \nu +a\tau \right)\cos \left(\pi \nu +a\left(t-\tau \right)\right)d\tau \hfill \\ & =& {\displaystyle \frac{\mathrm{B}\left(2\nu ,2\nu \right)}{2}}{t}^{4\nu -1}\sin \left(2\pi \nu +at\right).\hfill \end{array}$$

Proof. 

Apply the Laplace convolution Theorem (1), taking into account the following inverse Laplace transforms [12] (Eqns. 2.6.7(1)-(2)):

$$\begin{array}{ccc}\hfill {\mathcal{L}}^{-1}\left[{\displaystyle \frac{\sin \left(\nu \arctan \left(as\right)\right)}{{\left(1+a^2{s}^2\right)}^{\nu /2}}};t\right]& =& {\displaystyle \frac{t^{\nu -1}}{a^{\nu }\mathsf{\Gamma }\left(\nu \right)}}\sin \left({\displaystyle \frac{\pi \nu }{2}}-{\displaystyle \frac{t}{a}}\right),\hfill \end{array}$$

(42)

$$\begin{array}{ccc}\hfill {\mathcal{L}}^{-1}\left[{\displaystyle \frac{\cos \left(\nu \arctan \left(as\right)\right)}{{\left(1+a^2{s}^2\right)}^{\nu /2}}};t\right]& =& {\displaystyle \frac{t^{\nu -1}}{a^{\nu }\mathsf{\Gamma }\left(\nu \right)}}\cos \left({\displaystyle \frac{\pi \nu }{2}}-{\displaystyle \frac{t}{a}}\right),\hfill \end{array}$$

(43)

as well as the trigonometric identity $\sin 2\alpha =2\sin \alpha \cos \alpha$, to obtain

$$\begin{array}{ccc}& & {\int }_0^t{\displaystyle \frac{{\left[\tau \left(t-\tau \right)\right]}^{\nu -1}}{a^{2\nu }{\mathsf{\Gamma }}^2\left(\nu \right)}}\sin \left({\displaystyle \frac{\pi \nu }{2}}-{\displaystyle \frac{\tau }{a}}\right)\cos \left({\displaystyle \frac{\pi \nu }{2}}-{\displaystyle \frac{t-\tau }{a}}\right)d\tau \hfill \\ & =& {\mathcal{L}}^{-1}\left[{\displaystyle \frac{\sin \left(\nu \arctan \left(as\right)\right)\cos \left(\nu \arctan \left(as\right)\right)}{{\left(1+a^2{s}^2\right)}^{\nu }}};t\right]\hfill \\ & =& {\displaystyle \frac{1}{2}}{\mathcal{L}}^{-1}\left[{\displaystyle \frac{\sin \left(2\nu \arctan \left(as\right)\right)}{{\left(1+a^2{s}^2\right)}^{\nu }}};t\right].\hfill \end{array}$$

Apply once again the inverse Laplace transform given in (33), and the definition of the beta function (5) to obtain

$$\begin{array}{ccc}& & {\int }_0^t{\left[\tau \left(t-\tau \right)\right]}^{\nu -1}\sin \left({\displaystyle \frac{\pi \nu }{2}}-{\displaystyle \frac{\tau }{a}}\right)\cos \left({\displaystyle \frac{\pi \nu }{2}}-{\displaystyle \frac{t-\tau }{a}}\right)d\tau \hfill \\ & =& {\displaystyle \frac{\mathrm{B}\left(\nu ,\nu \right)}{2}}{t}^{2\nu -1}\sin \left(\pi \nu -{\displaystyle \frac{t}{a}}\right).\hfill \end{array}$$

Finally, perform the substitutions $a\mapsto -1/a$ and $\nu \mapsto 2\nu$ to complete the proof. □

Remark 3. 

As an alternative proof of Theorems 5 and 7, we can split the corresponding integrals into two by using the trigonometric identity $2\sin \alpha \cos \beta =\sin \left(\alpha +\beta \right)+\sin \left(\alpha -\beta \right)$. The first resulting integral is computed by applying the definition of the beta function (5). The second resulting integral vanishes since the transformation $\tau \mapsto t-\tau$ converts the integral into minus itself.

Theorem 8. 

For $\Re \left(\nu \right)>0$ and $a\in \mathbb{C}$, the following convolution integrals hold true:

$$\begin{array}{ccc}& & {\int }_0^t{\left[\tau \left(t-\tau \right)\right]}^{2\nu -1}\sin \left(\pi \nu -a\tau \right)\sin \left(\pi \nu -a\left(t-\tau \right)\right)d\tau \hfill \\ & =& {\displaystyle \frac{1}{2}}\left[\sqrt{\pi }\mathsf{\Gamma }\left(2\nu \right){\left({\displaystyle \frac{t}{2a}}\right)}^{2\nu -1/2}{J}_{2\nu -1/2}\left(at\right)-t^{4\nu -1}\mathrm{B}\left(2\nu ,2\nu \right)\cos \left(2\pi \nu -at\right)\right],\hfill \\ & & {\int }_0^t{\left[\tau \left(t-\tau \right)\right]}^{\nu }\cos \left(\pi \nu -a\tau \right)\cos \left(\pi \nu -a\left(t-\tau \right)\right)d\tau \hfill \\ & =& {\displaystyle \frac{1}{2}}\left[\sqrt{\pi }\mathsf{\Gamma }\left(2\nu \right){\left({\displaystyle \frac{t}{2a}}\right)}^{2\nu -1/2}{J}_{2\nu -1/2}\left(at\right)+t^{4\nu -1}\mathrm{B}\left(2\nu ,2\nu \right)\cos \left(2\pi \nu -at\right)\right].\hfill \end{array}$$

Proof. 

Apply the Laplace convolution Theorem (1), taking into account the inverse Laplace transform (42), to obtain

$$\begin{array}{ccc}& & {\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\sin }^2\left(\nu \arctan \left(as\right)\right)}{{\left(1+a^2{s}^2\right)}^{\nu }}};t\right]\hfill \\ & =& {\int }_0^t{\displaystyle \frac{{\left[\tau \left(t-\tau \right)\right]}^{\nu -1}}{a^{2\nu }{\mathsf{\Gamma }}^2\left(\nu \right)}}\sin \left({\displaystyle \frac{\pi \nu }{2}}-{\displaystyle \frac{\tau }{a}}\right)\sin \left({\displaystyle \frac{\pi \nu }{2}}-{\displaystyle \frac{t-\tau }{a}}\right)d\tau .\hfill \end{array}$$

(44)

Similarly, apply the convolution theorem, taking into account (43), to obtain

$$\begin{array}{ccc}& & {\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\cos }^2\left(\nu \arctan \left(as\right)\right)}{{\left(1+a^2{s}^2\right)}^{\nu }}};t\right]\hfill \\ & =& {\int }_0^t{\displaystyle \frac{{\left[\tau \left(t-\tau \right)\right]}^{\nu -1}}{a^{2\nu }{\mathsf{\Gamma }}^2\left(\nu \right)}}\cos \left({\displaystyle \frac{\pi \nu }{2}}-{\displaystyle \frac{\tau }{a}}\right)\cos \left({\displaystyle \frac{\pi \nu }{2}}-{\displaystyle \frac{t-\tau }{a}}\right)d\tau .\hfill \end{array}$$

(45)

Note that, according to the trigonometric identity ${\sin }^2\alpha +{\cos }^2\alpha =1$, the inverse Laplace transform property [12] (Eqn. 1.1.1(4))

$${\mathcal{L}}^{-1}\left[F\left(as\right);t\right]={\displaystyle \frac{1}{a}}{\mathcal{L}}^{-1}\left[F\left(s\right);{\displaystyle \frac{t}{a}}\right],$$

and the inverse Laplace transform given in (38), we have

$$\begin{array}{ccc}& & {\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\sin }^2\left(\nu \arctan \left(as\right)\right)}{{\left(1+a^2{s}^2\right)}^{\nu }}};t\right]+{\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\cos }^2\left(\nu \arctan \left(as\right)\right)}{{\left(1+a^2{s}^2\right)}^{\nu }}};t\right]\hfill \\ & =& {\mathcal{L}}^{-1}\left[{\displaystyle \frac{1}{{\left(1+a^2{s}^2\right)}^{\nu }}};t\right]={\displaystyle \frac{1}{a}}{\mathcal{L}}^{-1}\left[{\displaystyle \frac{1}{{\left(1+s^2\right)}^{\nu }}};{\displaystyle \frac{t}{a}}\right]\hfill \\ & =& {\displaystyle \frac{\sqrt{\pi }}{a\mathsf{\Gamma }\left(\nu \right)}}{\left({\displaystyle \frac{t}{2a}}\right)}^{\nu -1/2}{J}_{\nu -1/2}\left({\displaystyle \frac{t}{a}}\right).\hfill \end{array}$$

(46)

On the other hand, according to the trigonometric identity $\cos 2\alpha ={\cos }^2\alpha -{\sin }^2\alpha$, and the inverse Laplace transform given in (43), we have

$$\begin{array}{ccc}& & {\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\cos }^2\left(\nu \arctan \left(as\right)\right)}{{\left(1+a^2{s}^2\right)}^{\nu }}};t\right]-{\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\sin }^2\left(\nu \arctan \left(as\right)\right)}{{\left(1+a^2{s}^2\right)}^{\nu }}};t\right]\hfill \\ & =& {\mathcal{L}}^{-1}\left[{\displaystyle \frac{\cos \left(2\nu \arctan \left(as\right)\right)}{{\left(1+a^2{s}^2\right)}^{\nu }}};t\right]={\displaystyle \frac{t^{2\nu -1}}{a^{2\nu }\mathsf{\Gamma }\left(2\nu \right)}}\cos \left(\pi \nu -{\displaystyle \frac{t}{a}}\right).\hfill \end{array}$$

(47)

Solving the linear system of equations given in (46) and (47), we obtain

$$\begin{array}{ccc}& & {\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\sin }^2\left(\nu \arctan \left(as\right)\right)}{{\left(1+a^2{s}^2\right)}^{\nu }}};t\right]\hfill \\ & =& {\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{\sqrt{\pi }}{a\mathsf{\Gamma }\left(\nu \right)}}{\left({\displaystyle \frac{t}{2a}}\right)}^{\nu -1/2}{J}_{\nu -1/2}\left({\displaystyle \frac{t}{a}}\right)-{\displaystyle \frac{t^{2\nu -1}}{a^{2\nu }\mathsf{\Gamma }\left(2\nu \right)}}\cos \left(\pi \nu -{\displaystyle \frac{t}{a}}\right)\right],\hfill \end{array}$$

(48)

$$\begin{array}{ccc}& & {\mathcal{L}}^{-1}\left[{\displaystyle \frac{{\cos }^2\left(\nu \arctan \left(as\right)\right)}{{\left(1+a^2{s}^2\right)}^{\nu }}};t\right]\hfill \\ & =& {\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{\sqrt{\pi }}{a\mathsf{\Gamma }\left(\nu \right)}}{\left({\displaystyle \frac{t}{2a}}\right)}^{\nu -1/2}{J}_{\nu -1/2}\left({\displaystyle \frac{t}{a}}\right)+{\displaystyle \frac{t^{2\nu -1}}{a^{2\nu }\mathsf{\Gamma }\left(2\nu \right)}}\cos \left(\pi \nu -{\displaystyle \frac{t}{a}}\right)\right].\hfill \end{array}$$

(49)

Finally, match the results (44) with (48), and (45) with (49), taking into account the definition of the beta function (5) and performing the substitutions $a\mapsto 1/a$, $\nu \mapsto 2\nu$, to complete the proof. □

Remark 4. 

It is noteworthy that, as a by-product of the above proof, we have obtained the inverse Laplace transforms (48) and (49), which do not seem to be reported in the existing literature.

Theorem 9. 

For $a\in \mathbb{R}$, the following convolution integral holds true:

$${\int }_0^t{\displaystyle \frac{\sin \left(\frac{a}{\tau }\right)\cos \left(\frac{a}{t-\tau }\right)}{\sqrt{\tau \left(t-\tau \right)}}}d\tau ={\displaystyle \frac{\pi }{2}}\mathrm{sgn}\left(a\right)\left[\mathrm{C}\left(2\sqrt{{\displaystyle \frac{2\left|a\right|}{\pi t}}}\right)-\mathrm{S}\left(2\sqrt{{\displaystyle \frac{2\left|a\right|}{\pi t}}}\right)\right].$$

(50)

Proof. 

Consider $a>0$ and apply the Laplace convolution Theorem (1), taking into account the inverse Laplace transforms [12] (Eqns. 2.4.2(2))

$$\begin{array}{ccc}\hfill {\mathcal{L}}^{-1}\left[{\displaystyle \frac{\exp \left(-a\sqrt{s}\right)}{\sqrt{s}}}\sin \left(a\sqrt{s}\right);t\right]& =& {\displaystyle \frac{1}{\sqrt{\pi t}}}\sin \left({\displaystyle \frac{a^2}{2t}}\right),\hfill \end{array}$$

(51)

$$\begin{array}{ccc}\hfill {\mathcal{L}}^{-1}\left[{\displaystyle \frac{\exp \left(-a\sqrt{s}\right)}{\sqrt{s}}}\cos \left(a\sqrt{s}\right);t\right]& =& {\displaystyle \frac{1}{\sqrt{\pi t}}}\cos \left({\displaystyle \frac{a^2}{2t}}\right),\hfill \end{array}$$

(52)

as well as the trigonometric identity $\sin 2\alpha =2\sin \alpha \cos \alpha$, to obtain

$$\begin{array}{ccc}& & {\displaystyle \frac{1}{\pi }}{\int }_0^t{\displaystyle \frac{\sin \left(\frac{a^2}{2\tau }\right)\cos \left(\frac{a^2}{2\left(t-\tau \right)}\right)}{\sqrt{\tau \left(t-\tau \right)}}}d\tau \hfill \\ & =& {\mathcal{L}}^{-1}\left[{\displaystyle \frac{\exp \left(-a\sqrt{s}\right)}{s}}\sin \left(a\sqrt{s}\right)\cos \left(a\sqrt{s}\right);t\right]\hfill \\ & =& {\displaystyle \frac{1}{2}}{\mathcal{L}}^{-1}\left[{\displaystyle \frac{\exp \left(-a\sqrt{s}\right)}{s}}\sin \left(2a\sqrt{s}\right);t\right].\hfill \end{array}$$

Now apply the inverse Laplace transform [12] (Eqns. 2.4.2(3))

$${\mathcal{L}}^{-1}\left[{\displaystyle \frac{\exp \left(-a\sqrt{s}\right)}{s}}\sin \left(a\sqrt{s}\right);t\right]=\mathrm{C}\left({\displaystyle \frac{a}{\sqrt{\pi t}}}\right)-\mathrm{S}\left({\displaystyle \frac{a}{\sqrt{\pi t}}}\right),$$

and perform the substitution $a\mapsto \sqrt{2a}$ to obtain for $a>0$,

$${\int }_0^t{\displaystyle \frac{\sin \left(\frac{a}{\tau }\right)\cos \left(\frac{a}{t-\tau }\right)}{\sqrt{\tau \left(t-\tau \right)}}}d\tau ={\displaystyle \frac{\pi }{2}}\left[\mathrm{C}\left(2\sqrt{{\displaystyle \frac{2a}{\pi t}}}\right)-\mathrm{S}\left(2\sqrt{{\displaystyle \frac{2a}{\pi t}}}\right)\right].$$

(53)

Finally, extend the result to $a\in \mathbb{R}$, taking into account that the integral in (53) is odd with respect to a. □

Remark 5. 

Note that the integral given in (50) can be recast as follows:

$${\int }_0^t{\displaystyle \frac{\sin \left(\frac{4at}{z^2}\right)}{\sqrt{t^2-z^2}}}dz={\displaystyle \frac{\pi }{2}}\mathrm{sgn}\left(a\right)\left[\mathrm{C}\left(2\sqrt{{\displaystyle \frac{2\left|a\right|}{\pi t}}}\right)-\mathrm{S}\left(2\sqrt{{\displaystyle \frac{2\left|a\right|}{\pi t}}}\right)\right].$$

(54)

To this end, notice that

$$I={\int }_0^t{\displaystyle \frac{\sin \left(\frac{a}{\tau }\right)\cos \left(\frac{a}{t-\tau }\right)}{\sqrt{\tau \left(t-\tau \right)}}}d\tau ={\int }_0^t{\displaystyle \frac{\sin \left(\frac{a}{t-\tau }\right)\cos \left(\frac{a}{\tau }\right)}{\sqrt{\tau \left(t-\tau \right)}}}d\tau .$$

Thus, applying the trigonometric identity $\sin \left(\alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta$, we obtain

$$\begin{array}{ccc}\hfill I& =& {\displaystyle \frac{1}{2}}{\int }_0^t{\displaystyle \frac{\sin \left(\frac{at}{\tau \left(t-\tau \right)}\right)}{\sqrt{\tau \left(t-\tau \right)}}}d\tau \hfill \\ & =& {\int }_0^{t/2}{\displaystyle \frac{\sin \left(\frac{at}{\tau \left(t-\tau \right)}\right)}{2\sqrt{\tau \left(t-\tau \right)}}}d\tau +{\int }_{t/2}^t{\displaystyle \frac{\sin \left(\frac{at}{\tau \left(t-\tau \right)}\right)}{2\sqrt{\tau \left(t-\tau \right)}}}d\tau .\hfill \end{array}$$

Finally, perform the change of variables $\tau ={\displaystyle \frac{1}{2}}\left(t\pm \sqrt{t^2-z^2}\right)$ taking the positive sign + for the first integral and the negative sign − for the second integral to obtain the desired result.

Theorem 10. 

For $a\in \mathbb{R}$, the following convolution integrals hold true:

$$\begin{array}{ccc}& & {\int }_0^t{\displaystyle \frac{\sin \left(\frac{a}{\tau }\right)\sin \left(\frac{a}{t-\tau }\right)}{\sqrt{\tau \left(t-\tau \right)}}}d\tau \hfill \\ & =& {\displaystyle \frac{\pi }{2}}\left[\mathrm{C}\left(2\sqrt{{\displaystyle \frac{2\left|a\right|}{\pi t}}}\right)+\mathrm{S}\left(2\sqrt{{\displaystyle \frac{2\left|a\right|}{\pi t}}}\right)-\mathrm{erf}\left(\sqrt{{\displaystyle \frac{2\left|a\right|}{t}}}\right)\right],\hfill \end{array}$$

(55)

$$\begin{array}{ccc}& & {\int }_0^t{\displaystyle \frac{\cos \left(\frac{a}{\tau }\right)\cos \left(\frac{a}{t-\tau }\right)}{\sqrt{\tau \left(t-\tau \right)}}}d\tau \hfill \\ & =& {\displaystyle \frac{\pi }{2}}\left[2-\mathrm{C}\left(2\sqrt{{\displaystyle \frac{2\left|a\right|}{\pi t}}}\right)-\mathrm{S}\left(2\sqrt{{\displaystyle \frac{2\left|a\right|}{\pi t}}}\right)-\mathrm{erf}\left(\sqrt{{\displaystyle \frac{2\left|a\right|}{t}}}\right)\right].\hfill \end{array}$$

(56)

Proof. 

Apply the Laplace convolution Theorem (1), taking into account the inverse Laplace transform (51), to obtain

$${\mathcal{L}}^{-1}\left[{\displaystyle \frac{\exp \left(-2a\sqrt{s}\right)}{s}}{\sin }^2\left(a\sqrt{s}\right);t\right]={\displaystyle \frac{1}{\pi }}{\int }_0^t{\displaystyle \frac{\sin \left(\frac{a^2}{2\tau }\right)\sin \left(\frac{a^2}{2\left(t-\tau \right)}\right)}{\sqrt{\tau \left(t-\tau \right)}}}d\tau .$$

(57)

Similarly, apply the convolution theorem, taking into account (52), to obtain

$${\mathcal{L}}^{-1}\left[{\displaystyle \frac{\exp \left(-2a\sqrt{s}\right)}{s}}{\cos }^2\left(a\sqrt{s}\right);t\right]={\displaystyle \frac{1}{\pi }}{\int }_0^t{\displaystyle \frac{\cos \left(\frac{a^2}{2\tau }\right)\cos \left(\frac{a^2}{2\left(t-\tau \right)}\right)}{\sqrt{\tau \left(t-\tau \right)}}}d\tau .$$

(58)

Note that, according to the trigonometric identity ${\sin }^2\alpha +{\cos }^2\alpha =1$, and the inverse Laplace transform given in [12] (Eqn. 2.2.1(16)), we have

$$\begin{array}{ccc}& & {\mathcal{L}}^{-1}\left[{\displaystyle \frac{\exp \left(-2a\sqrt{s}\right)}{s}}{\sin }^2\left(a\sqrt{s}\right);t\right]+{\mathcal{L}}^{-1}\left[{\displaystyle \frac{\exp \left(-2a\sqrt{s}\right)}{s}}{\cos }^2\left(a\sqrt{s}\right);t\right]\hfill \\ & =& {\mathcal{L}}^{-1}\left[{\displaystyle \frac{\exp \left(-2a\sqrt{s}\right)}{s}};t\right]=\mathrm{erfc}\left({\displaystyle \frac{a}{\sqrt{t}}}\right).\hfill \end{array}$$

(59)

On the other hand, according to the trigonometric identity $\cos 2\alpha ={\cos }^2\alpha -{\sin }^2\alpha$, and the inverse Laplace transform [12] (Eqns. 2.4.2(3)), we have

$$\begin{array}{ccc}& & {\mathcal{L}}^{-1}\left[{\displaystyle \frac{\exp \left(-2a\sqrt{s}\right)}{s}}{\cos }^2\left(a\sqrt{s}\right);t\right]-{\mathcal{L}}^{-1}\left[{\displaystyle \frac{\exp \left(-2a\sqrt{s}\right)}{s}}{\sin }^2\left(a\sqrt{s}\right);t\right]\hfill \\ & =& {\mathcal{L}}^{-1}\left[{\displaystyle \frac{\exp \left(-2a\sqrt{s}\right)}{s}}\cos \left(2a\sqrt{s}\right);t\right]=1-\mathrm{C}\left({\displaystyle \frac{2a}{\sqrt{\pi t}}}\right)-\mathrm{S}\left({\displaystyle \frac{2a}{\sqrt{\pi t}}}\right).\hfill \end{array}$$

(60)

Solving the system of equations given in (59) and (60) and taking into account the definition of the complementary error function (7), we obtain for $a>0$

$$\begin{array}{ccc}& & {\mathcal{L}}^{-1}\left[{\displaystyle \frac{\exp \left(-2a\sqrt{s}\right)}{s}}{\sin }^2\left(a\sqrt{s}\right);t\right]\hfill \\ & =& {\displaystyle \frac{1}{2}}\left[\mathrm{C}\left({\displaystyle \frac{2a}{\sqrt{\pi t}}}\right)+\mathrm{S}\left({\displaystyle \frac{2a}{\sqrt{\pi t}}}\right)-\mathrm{erf}\left({\displaystyle \frac{a}{\sqrt{t}}}\right)\right],\hfill \end{array}$$

(61)

$$\begin{array}{ccc}& & {\mathcal{L}}^{-1}\left[{\displaystyle \frac{\exp \left(-2a\sqrt{s}\right)}{s}}{\cos }^2\left(a\sqrt{s}\right);t\right]\hfill \\ & =& 1-{\displaystyle \frac{1}{2}}\left[\mathrm{C}\left({\displaystyle \frac{2a}{\sqrt{\pi t}}}\right)+\mathrm{S}\left({\displaystyle \frac{2a}{\sqrt{\pi t}}}\right)+\mathrm{erf}\left({\displaystyle \frac{a}{\sqrt{t}}}\right)\right].\hfill \end{array}$$

(62)

Match the results (57) with (61), and (58) with (62), and perform the substitution $a\mapsto \sqrt{2a}$. Finally, extend the result to $a\in \mathbb{R}$, taking into account that the integrals in (55) and (56) are even with respect to a. □

Remark 6. 

It is noteworthy that, as a by-product of the above proof, we have obtained the inverse Laplace transforms (61) and (62), which do not seem to be reported in the existing literature.

Theorem 11. 

For $\Re \left(\nu \right)>-1$ and $a\in \mathbb{C}$, the following convolution integral holds true:

$$\begin{array}{ccc}& & {\int }_0^t{\displaystyle \frac{{\left(t-\tau \right)}^{\nu }}{\tau }}\sinh \left(a\tau \right)\sin \left(a\tau \right)d\tau \hfill \\ & =& {\displaystyle \frac{a^2{t}^{\nu +2}}{\left(\nu +2\right)\left(\nu +1\right)}}{}_{2}F_5\left(\begin{array}{c}1,{\displaystyle \frac{1}{2}}\\ {\displaystyle \frac{3}{2}},{\displaystyle \frac{\nu +3}{4}},{\displaystyle \frac{\nu +4}{4}},{\displaystyle \frac{\nu +5}{4}},{\displaystyle \frac{\nu +6}{4}}\end{array};-{\displaystyle \frac{a^4{t}^4}{64}}\right).\hfill \end{array}$$

(63)

Proof. 

Apply the Laplace convolution Theorem (1), taking into account the inverse Laplace transforms [12] (Eqns. 2.6.4(13)&2.1.1(1))

$$\begin{array}{ccc}\hfill {\mathcal{L}}^{-1}\left[\arctan \left({\displaystyle \frac{a}{s^2}}\right);t\right]& =& {\displaystyle \frac{2}{t}}\sinh \left(\sqrt{{\displaystyle \frac{a}{2}}}t\right)\sin \left(\sqrt{{\displaystyle \frac{a}{2}}}t\right),\hfill \\ \hfill {\mathcal{L}}^{-1}\left[{\displaystyle \frac{1}{s^{\nu }}};t\right]& =& {\displaystyle \frac{t^{\nu -1}}{\mathsf{\Gamma }\left(\nu \right)}},\hfill \end{array}$$

which yields

$$\begin{array}{ccc}& & {\displaystyle \frac{2}{\mathsf{\Gamma }\left(\nu \right)}}{\int }_0^t{\displaystyle \frac{{\left(t-\tau \right)}^{\nu -1}}{\tau }}\sinh \left(\sqrt{{\displaystyle \frac{a}{2}}}\tau \right)\sin \left(\sqrt{{\displaystyle \frac{a}{2}}}\tau \right)d\tau \hfill \\ & =& {\mathcal{L}}^{-1}\left[{\displaystyle \frac{1}{s^{\nu }}}\arctan \left({\displaystyle \frac{a}{s^2}}\right);t\right]\hfill \\ & =& {\displaystyle \frac{a{t}^{\nu +1}}{\mathsf{\Gamma }\left(\nu +2\right)}}{}_{2}F_5\left(\begin{array}{c}1,{\displaystyle \frac{1}{2}}\\ {\displaystyle \frac{3}{2}},{\displaystyle \frac{\nu +2}{4}},{\displaystyle \frac{\nu +3}{4}},{\displaystyle \frac{\nu +4}{4}},{\displaystyle \frac{\nu +5}{4}}\end{array};-{\displaystyle \frac{a^2{t}^4}{256}}\right),\hfill \end{array}$$

where the inverse Laplace transform [12] (Eqn. 2.6.4(21)) has been used. Finally, perform the substitutions $a\mapsto 2a^2$, $\nu \mapsto \nu +1$, and apply the recursive property of the gamma function (4) to complete the proof. □

Theorem 12. 

For $z\in \mathbb{C}$, the following reduction formula holds true:

$${}_{1}F_4\left(\begin{array}{c}{\displaystyle \frac{1}{2}}\\ {\displaystyle \frac{3}{2}},{\displaystyle \frac{3}{2}},{\displaystyle \frac{3}{4}},{\displaystyle \frac{5}{4}}\end{array};z\right)={\displaystyle \frac{\mathrm{Chi}\left(4z^{1/4}\right)-\mathrm{Ci}\left(4z^{1/4}\right)}{8\sqrt{z}}}.$$

(64)

Proof. 

First, observe that for $\nu =0$, (63) reduces to

$${\int }_0^t\sinh \left(a\tau \right)\sin \left(a\tau \right){\displaystyle \frac{d\tau }{\tau }}={\displaystyle \frac{a^2{t}^2}{2}}{}_{1}F_4\left(\begin{array}{c}{\displaystyle \frac{1}{2}}\\ {\displaystyle \frac{3}{2}},{\displaystyle \frac{3}{2}},{\displaystyle \frac{3}{4}},{\displaystyle \frac{5}{4}}\end{array};-{\displaystyle \frac{a^4{t}^4}{64}}\right).$$

(65)

However,

$${\int }_0^t\sinh \left(a\tau \right)\sin \left(a\tau \right){\displaystyle \frac{d\tau }{\tau }}={\displaystyle \frac{1}{2}}\left[I\left(a\right)+I\left(-a\right)\right],$$

(66)

where we define

$$I\left(a\right)={\int }_0^t\exp \left(a\tau \right)\sin \left(a\tau \right){\displaystyle \frac{d\tau }{\tau }}=\Im {\int }_0^t\exp \left(\left(1+i\right)a\tau \right){\displaystyle \frac{d\tau }{\tau }}.$$

Since [3] (Eqn. 1.3.1(12))

$$\int {\displaystyle \frac{\exp \left(at\right)}{t}}dt=\mathrm{Ei}\left(at\right),a\ne 0,$$

we have

$$I\left(a\right)=\Im {\left[\mathrm{Ei}\left(\left(1+i\right)a\tau \right)\right]}_{\tau =0}^{\tau =t}.$$

Apply the property [18] (Sect. 5. Eqn. 6)

$$\Im \left(z\right)={\displaystyle \frac{z-\overline{z}}{2i}},$$

and the asymptotic formula [16] (Eqn. 37.9.4)

$$\mathrm{Ei}\left(t\right)\approx \log t+t+\gamma ,t\to 0^{+},$$

to obtain

$$I\left(a\right)={\displaystyle \frac{1}{2i}}\left[\mathrm{Ei}\left(\left(1+i\right)at\right)-\mathrm{Ei}\left(\left(1-i\right)at\right)\right]-{\displaystyle \frac{\pi }{4}},$$

thus, (66) reduces to

$$\begin{array}{ccc}& & {\int }_0^t\sinh \left(a\tau \right)\sin \left(a\tau \right){\displaystyle \frac{d\tau }{\tau }}={\displaystyle \frac{\pi }{4}}\hfill \\ & & -{\displaystyle \frac{i}{2}}\left[\mathrm{Ei}\left(\left(1+i\right)at\right)+\mathrm{Ei}\left(-\left(1+i\right)at\right)-\mathrm{Ei}\left(\left(1-i\right)at\right)-\mathrm{Ei}\left(-\left(1-i\right)at\right)\right].\hfill \end{array}$$

(67)

Insert (67) in (65) and take into account (12) to obtain

$$\begin{array}{ccc}& & a^2{t}^2{}_{1}F_4\left(\begin{array}{c}{\displaystyle \frac{1}{2}}\\ {\displaystyle \frac{3}{2}},{\displaystyle \frac{3}{2}},{\displaystyle \frac{3}{4}},{\displaystyle \frac{5}{4}}\end{array};-{\displaystyle \frac{a^4{t}^4}{64}}\right)\hfill \\ & =& {\displaystyle \frac{\pi }{4}}+{\displaystyle \frac{i}{2}}\left[\mathrm{Ci}\left(i\sqrt{2}at{e}^{-i\pi /4}\right)-\mathrm{Ci}\left(\sqrt{2}at{e}^{-i\pi /4}\right)\right].\hfill \end{array}$$

(68)

However, according to (10) and (11), it is not difficult to derive the following relation:

$$\mathrm{Ci}\left(i\omega \right)=\mathrm{Chi}\left(\omega \right)+{\displaystyle \frac{i\pi }{2}},$$

(69)

thus, performing the change of variables $z=-{\displaystyle \frac{a^4{t}^4}{64}}$ in (68) and taking into account (69), we complete the proof. □

4. Conclusions

We have computed several finite integrals of trigonometric and hyperbolic functions using the Laplace convolution integral in Theorems 2–11. These integrals do not seem to be reported in the most common tables of integrals. Moreover, we have verified that they are not computable with the aid of computer algebra systems. In addition, we have derived new inverse Laplace transform formulas in (31), (32), (40), (41), (48), (49), (61), (62).

As an application of some of these new convolution integrals, we have obtained a new integral representation of a Kelvin function in (26) and a new reduction formula for a generalized hypergeometric function in (64). All numerical checks were carried out using Mathematica (version 13.2). The corresponding Mathematica notebook is available at URL https://shorturl.at/yrLz0 (accessed on 3 May 2026).