Integral $\int_0^\infty dp \, \frac{p^5 \sin(p x) e^{-b p^2}}{p^4 + a^2}$: any clever ideas?

Question

I am trying to solve the following integral, with $a>0,$ $b>0$:

$I \equiv \int_0^\infty dp \, \frac{p^5 \sin(p x) e^{-b p^2}}{p^4 + a^2} $

By expanding the $\sin$, I get

$I = \sum_{n=1}^\infty \frac{x^{2n-1}}{(2n-1)!} \int_0^\infty dp \, \frac{p^{4+2n} e^{-b p^2}}{p^4 + a^2} \\ = \sum_{n=1}^\infty \frac{x^{2n-1}}{(2n-1)!}\Bigg\{ \frac{1}{2} b^{\frac{3}{2}-n} \Gamma (n-{3}/{2}) \, _1F_2\left(1;\frac{5}{4}-\frac{n}{2},\frac{7}{4}-\frac{n}{2};-\frac{1}{4} a^2 b^2\right) +\frac{1}{4} \pi a^{n-\frac{3}{2}} \left[\csc \left((2 \pi n+\pi )/4\right] \cos (a b) -\sec \left[ (2 \pi n+\pi )/4\right] \sin (a b)\right) \Bigg\}.$

Here $_1F_2$ is a hypergeometric function. The first summation can be done, so that we are left with

$I= \frac{\pi}{2a} \left[\cos (a b) \cos (x\sqrt{a/2}) \sinh (x\sqrt{a/2})+\sin (a b) \sin (x\sqrt{a/2}) \cosh (x\sqrt{a/2})\right] + \sum_{n=1}^\infty \frac{x^{2n-1}}{(2n-1)!} \frac{1}{2} b^{\frac{3}{2}-n} \Gamma (n-{3}/{2}) \, _1F_2\left(1;\frac{5}{4}-\frac{n}{2},\frac{7}{4}-\frac{n}{2};-\frac{1}{4} a^2 b^2\right).$

I am unable to find a closed form for the second summation.

Is there an easier way to solve $I$? Presumably one can apply the residue theorem, but I did not find a quick way. Any ideas would be appreciated!

Answer 1

We have $$\frac {p^5 e^{-b p^2} \sin x p} {p^4 + a^2} = p e^{-b p^2} \sin x p \left( 1 + \frac {i a} {2 (p^2 - i a)} - \frac {i a } {2 (p^2 + i a)} \right), \\ \frac d {db} \left( e^{i a b} \int_0^\infty \frac {p e^{-b p^2} \sin x p} {p^2 - i a} dp \right) = -e^{i a b} \int_0^\infty p e^{-b p^2} \sin x p \,dp,$$ and, after some calculations, $$\int_0^\infty \frac {p^5 e^{-b p^2} \sin x p} {p^4 + a^2} dp = F(a) + F(-a) + \frac {\sqrt \pi x e^{-x^2/(4 b)}} {4 b^{3/2}}, \\ F(a) = \frac {i \pi a e^{i a b}} 8 \left( e^{x \sqrt {i a}} \operatorname{erfc} \frac { 2 b \sqrt{i a} + x} {2 \sqrt b} - e^{-x \sqrt {i a}} \operatorname{erfc} \frac { 2 b \sqrt{i a} - x} {2 \sqrt b} \right).$$

Answer 2

Amazingly this integral can be solved in terms of well-known functions of mathematical physics. It is easy to see that $$I(a,b,x):=\int_0^\infty \frac{p^5\,\sin{(px)}}{p^4+a^2}\,e^{-b\,p^2} dp =\frac{1}{2} \frac{d}{db} \frac{d}{dx} \underbrace{\int_{-\infty}^\infty \frac{p^2\,\cos{(px)}}{p^4+a^2}\,e^{-b\,p^2} dp}_{:=J(a,b,x)}.$$ I won't do the derivatives, but present the formula for $J(a,b,x).$ Let $c=x/\sqrt{4b}.$ Then $$ J(a,b,x)=\frac{\pi}{2}\sqrt{b}\,e^{-c^2}\,Re\Big[\frac{1}{\sqrt{i\,a\,b}} \Big( \exp{\big( (\sqrt{i\,a\,b} - c)^2 \big)} \, \text{erfc}\big(\sqrt{i\,a\,b} - c\big) + $$ $$+ \exp{\big( (\sqrt{i\,a\,b} + c)^2 \big)} \, \text{erfc}\big(\sqrt{i\,a\,b} + c\big) \Big)\Big] $$ The 'erfc' is the complimentary error function. My proof is long and not rigorous so we'll wait a few days to see if anyone will present a proof. If not, I may return to it. However, I've tested the numerical integration vs. the closed form for a total of 1000 evaluations for 10 different $a,\, b, \, x.$ The differences were 0 to within machine precision. The tests were over positive $a,\, b, \, x$ each from 0.1 to 6.5 by increments of 0.6.