Reducing a class of two-dimensional integrals to one-dimension with application to Gaussian Transforms

Quantum theory is awash in multidimensional integrals that contain exponentials in the integration variables, their inverses, and inverse polynomials of those variables. The present paper introduces a means to reduce pairs of such integrals to one dimension when the integrand contains powers times an arbitrary function of xy/(x+y) multiplying various combinations of exponentials. In some cases these exponentials arise directly from transition-amplitudes involving products of plane waves, hydrogenic wave functions, Yukawa and/or Coulomb potentials. In other cases these exponentials arise from Gaussian transforms of such functions.


Introduction
Prudnikov, Brychkov, and Marichev [1] provide a means to reduce a class of two-dimensional integrals involving exponentials and a general function of the integration variables in the specific configuration f xy x+y to a one-dimensional integral involving that same function of the new integration variable f (t), x + y f xy x + y e −px−qy dx dy As may be seen with comparison to the specific cases, such as entry No. 3.1.3.5 on the same page, the coefficient is in error. This reduction should read ∞ 0ˆ∞ 0 1 √ x + y f xy x + y e −px−qy dx dy In the present paper we extend this reduction technique to a class of integrals that arise in Gaussian transforms of atomic, molecular, and optical transition amplitudes [2,3] in which the above exponentials on the left-hand side may contain more complicated forms, e − a x − b y −c xy/(x+y)−h y/(x+y)−j/(x+y)−px−qy , and one may also have positive and negative powers of x and y.
We will begin with a more general form R 2 ( n , m, ν, a, b, c, h, j, p, q) =ˆ∞ The reduction is facilitated by a simultaneous change of variables to x + y , t = sy √ x + y (4) ∂φ ∂s ∂ψ ∂t − ∂ψ ∂s where the last line is the Jacobian determinant of the transformation. We retain the integration limits over [0, ∞]. Then R 2 ( n , m, ν, a, b, c, h, j, p, q) then If we set {n = 0, m = 0, ν = 1}, the second factor in the denominator of (11) goes to unity, allowing us to do the integral [4], and we recover the corrected integral (2) from Prudnikov, Brychkov, and Marichev [1] R 2 ( 0 , 0, 1, 0, 0, 0, 0, 0, p, q) =ˆ∞ 0ˆ∞ 0 1 √ x + y f xy x + y e −px−qy dx dy A second form may be obtained by setting in the first form, giving Additional applications of (13) will give larger powers of x and y in the denominator with powers of x + y in the numerator. Applications of the inverse of (13) will give powers of x and y in the numerator with larger powers of x + y in the denominator. One may also do the integral if we create even powers for the last factor in the numerator of (11) and expand, such as setting {n = −1, m = 1, ν = 0}, which gives dissimilar powers of x and y [5], The transformation (13) can be applied to this and all subsequent integrals.

Integrals with n + m + 2ν = 4
We obtain a new set of integral reductions by finding values of n, m, and ν that set the last factor in the numerator of (11) to unity.

Integrals yielding Macdonald functions
One can also set {n = 1, m = 1, ν = 1} to set the s 2 + t term of (8) to unity to obtain [6] R 2 ( 1 , 1, 1, 0, 0, 0, 0, 0, p, q) We can instead integrate [5] the u-form with the last factor in the numerator of (11) set to one with {n = 1, m = 1, ν = 1}. Since we also have, but this may also be obtained from (19) by using (13). We can also set the last factor in the numerator of (11) to unity with {n = 4, m = 0, ν = 0} to obtain a set of integral reductions that have unlike powers of the original coordinates [6] , It is to be understood that each of the integral reductions in this paper are valid only for those functions f that are convergent in the final integral. In the above equation we have made this restriction more explicit with the notation f 2 (t) that indicates that the integrals do not converge for terms in a powers series representation of f for powers less than 2. But if we set then we can rewrite R 2 ( 4 , 0, 0, 0, 0, 0, 0, 0, p, q) = R 2 (0, −4, 4, 0, 0, 0, 0, 0, p, q) a form that converges for any function g that can be expanded in a powers series (without the negative powers of a Laurant series). For n = 0, m = 4, ν = 0} we simply interchange p ↔ q in the above. Similarly, for {n = 5, n = 1, ν = −1} we employ (22) to obtain One may continue on in this fashion with more extreme powers such as {n = 7, n = −1, ν = −1}, by using the identity to obtain R 2 ( 7 , −1, −1, 0, 0, 0, 0, 0, p, q) = R 2 (1, −7, 5, 0, 0, 0, 0, 0, p, q)

Integrals with m + ν = 2
We obtain a new set of integral reductions by finding values of m and ν that move the second factor in the denominator of (11) to the numerator with unit power. The ratio of this term with the other term left in the denominator reduces nicely [5], allowing integrals of nonzero powers of the numerator of (11) to be done. We already saw the results for We have checked that this series of reductions all converge for any function f that can be expanded in a powers series through {n = −9, m = 1, ν = 1}.

Integral reductions for inverse powers in the exponentials
Instead of the positive powers in the exponential of the last section, let us take the reverse and set p = q = h = j = 0, where we have split off e −c xy/(x+y) from f xy x+y to explicate how this is different from simply making the replacement . In Eq. (7) we apply a partial fraction decomposition in the exponential of The last term in the integrand is turned into a square if m = ν = 3 and the middle term is flipped upside-down if n ≥ 1 so letting {n = 1, m = ν = 3} gives [7,8] and letting {n = m = ν = 3} gives [8,9] If instead we set the last term to unity with m = ν = 2 one gets for The extension to larger values of n is straightforward. Finally, we note that none of the above are well defined for a = b. Since the complicating exponential in the second line of (32) goes to unity in this case, we may write the general form This is most easily proved by setting a = b and h = 0 in Eq. (60) of section 5.

Set of integral reductions for more complicated exponentials
A entirely different set of integral reductions may be crafted that include both positive and negative powers in the exponentials. Using the same partial fraction decomposition as in Eq. (31) gives and we may well have convergence problems for b > a unless c is large enough. Note that the exponential e −c xy/(x+y) multiplying f xy x+y is transformed as e −ct multiplying f (t) as one would expect of a factor that can be folded into the definition of f . If we then specialize this integral to the case where q = 0, one may complete the square in the latter exponential and change variables Of the four possible solutions, we choose the one with Then with we have [5] R 2 ( n , m, ν, a, b, c, 0, 0, p, 0) If we set {m = 4 − ν}, the last factor of (44) goes to unity, giving us some hope of doing the integral. But even the simplest of those, with {n = 3 − ν} [5] has conditions that cannot be met as t becomes very small unless √ pt = √ a − b or one of these coefficients in the original exponents is an imaginary number.
So let us return to the integral in terms of x and y to form a self-consistent version of the former condition by setting p = (a − b)(x + y) 2 / x 2 y 2 . The utility of this value is not apparent in terms of the original variables (where We again specialize this integral to the case where q = 0 and complete the square in the latter exponential and change variables Of the four possible solutions, we choose the one with Then with we have [5] R 2 ( n , m, ν, a, b, c, p, 0) (50) If we set {m = 4 − ν} the second factor in the denominator of (50) goes to unity, as does the last factor in the numerator if we then set {n = 3 − ν}, giving [5] The powers of ν can be folded into a redefinition of f. If we instead retain the last factor in the numerator with unit power by setting {n = 1 − ν}, after reducing we obtain [5]

≤0
One can also move the second factor in the denominator of (50) into the numerator with unit power by setting {m = 6 − ν}. We then set {n = 1 − ν} to eliminate the other term in the numerator, and reduce the resultant quotient to obtain [5] ] Again, the powers of ν can be folded into a redefinition of f.
One may also move the second factor in the denominator of (50) into the numerator with a power of two by setting {m = 8 − ν}. Again eliminate the other term in the numerator with {n = 1−ν}, and reduce the resultant quotient somewhat. Mathematica 7 was unable to do this integral but Mathematica 9 could, with Finally, one can also move the second factor in the denominator of (50) into the numerator with unit power by setting {m = 6 − ν} while allowing the other term in the numerator to appear with unit power by setting {n = −1 − ν}, and reduce the resultant quotient, one obtains If we change variables to with we have The general result, good for (m + ν) > 2, (n + ν) > 2 is Whenever m = n ± 2M , where M is a non-negative integer, we may simplify this with [12] that can be readily integrated whenever n + ν = 2 (A + 1) is an even integer. Even when ν alone is an even integer, and M is a non-negative integer, we may nevertheless find an integrable form with

Application
Probability amplitudes in atomic physics and variational wave functions for many-electron atoms involve products of Yukawa or (their special case) Coulomb potentials and hydrogenic orbitals that may be derived from Yukawa potentials via derivatives. Consider, then, the case of two Yukawa potentials centered on different positions S η10η20 The author has given an analytically reduced form for multidimensional integrals over any number of such products in terms of Gaussian transforms, generally with one remaining integral for each atomic center in the original integrals. [13] Such a reduction in the present case from three to two integral dimensions is in no way dramatic, but does provide a soluble example for the utility of the (further) reduction formulae of this section. One may write down the final form for the above integral using the notation formalism in that paper, but the reduction may instead be easily found by completing the square in the Gaussian transform S η10η20 where we have changed variables from x 1 to x ′ 1 = x 1 − ρ2 ρ1+ρ2 x 2 with unit Jacobian. Then the spatial integral may be done [14] What is left is S η10η20 where in the last line we have taken our function to be and in the present case have µ = 3/2. We could also have set µ = 0 with n = m = 1 and ν = 3. Then using (62) we have which is indeed the correct result. Suppose instead of a Yukawa potential in η 1 we have a hydrogenic 1s wave function, that may be had by differentiation: One may take the limit η 2 → η 1 to obtain Eq. (49) from a previous paper [13] that used a very different integration method, S η2η20 1s,1 (0, ; 0, 6 Set of integral reductions for inverse powers inverse inverse binomials in the exponentials

The transformation
We may use arbitrary values of a and b in integrals of exponentials that contain both inverse powers and −j/(x + y) . We obtain If we again change variables to with ds = − 2s we have This is integrable for even values of n + ν ≥ 4 if we set the second factor in the r integral to unity with m = −ν + 4 . One may also do a binomial expansion of (1 − rt) for even values of m + ν > 4. This is done by completing the square in the exponential and setting with unit Jacobian.

Application
Fourier transforms of products of Yukawa or Coulomb potentials and hydrogenic orbitals are more difficult than (65). Consider, for instance, the case of the Fourier Transform of a product of two Yukawa potentials centered on different positions S η10η20 The procedure is as in Section 5.2 above, but with two extra terms in the exponential after completing the square: As a check one can instead change variables to τ = ρ 2 ρ 1 + ρ 2 = y x + y in (78) to give [6] S η10η20 1 ( k ; 0, x 2 ) = π 1/2ˆ1 One can show numerically that these two integrals are equal. Cheshire [15] reduced the related integral (his eq. (19)) I 1 = η which matches the present result after substituting the specialized values for η 1 and η 2 . If η 1 = η 2 we have In the limits k → 0 and η 2 = η 1 we can analytically integrate either form to give S η10η10 1 (0, ; 0, x 2 ) = 2πe −x2η1 η 1 (85)

Acknowledgment
I would like to thank Professor Ray A. Mayer, of Reed College, for creating a very different proof of (2) for me, and to his colleague Professor Nicholas Wheeler for the introduction.