Integral $\int_0^\infty dp \, \frac{p^5 \sin(p x) e^{-b p^2}}{p^4 + a^2}$: any clever ideas? 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!
 A: 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.
A: 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,
$$\int_0^\infty \frac {p^5 e^{-b p^2} \sin x p} {p^4 + a^2} dp =
F(a) + F(-a) + \frac {\sqrt \pi x e^{-x^2/(4 b)}} {4 b^{3/2}}, \\
F(a) = \frac {i \pi a e^{i a b}} 8 \left(
 e^{x \sqrt {i a}} \operatorname{erfc} \frac { 2 b \sqrt{i a} + x} {2 \sqrt b} -
 e^{-x \sqrt {i a}} \operatorname{erfc} \frac { 2 b \sqrt{i a} - x} {2 \sqrt b}
\right).$$
