Seeking the derivation of the Fourier Sine Transform of $x^{2\nu}(x^2+a^2)^{-\mu-1}$

Question

In this answer on math.stackexchange.com the Fourier Sine Transform of $x^{2\nu}(x^2+a^2)^{-\mu-1}$ is given in terms of the generalized hypergeometric function: $$\frac{1}{2}a^{2\nu-2\mu}\frac{\Gamma(1+\nu)\Gamma(\mu-\nu)}{\Gamma(\mu+1)}y \:_1\text{F}_2(\nu+1;\nu+1-\mu,3/2;a^2y^2/4)\:+\:4^{\nu-\mu-1}\sqrt{\pi}\frac{\Gamma(\nu-\mu)}{\Gamma(\mu-\nu+3/2)}y^{2\mu-2\nu+1}\:_1\text{F}_2(\mu+1;\mu-\nu+3/2,\mu-\nu+1;a^2y^2/4). $$ In particular, the integral $\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx$ is expressed in terms of the error functions: $$\frac{\pi}{2\sqrt{2}\:e}\left(-e^2\text{erfc}(1)+\text{erf}(1) +1\right).$$

However, no derivation is provided there. I am seeking a derivation of the above expressions.

Answer 1

We follow Nemo's suggestion in the comment and derive an expression for $$f(x):=\frac{\sin(x)}{1+x^2}=\sum_{k=0}^\infty \lambda(k) x^k$$ to use Ramanujan's master theorem. We can do this in 2 ways.

1) Direct expansion: $$f(x) = \sum_{m=0}^\infty (-1)^m\frac{x^{2m+1}}{(2m+1)!}\sum_{n=0}^\infty (-x^2)^n = \sum_{p=0}^\infty (-1)^px^{2p+1}\sum_{m=0}^p\frac{1}{(2m+1)!}$$ Apparently, we need a meromorphic function $\xi(x)$ satisfying the functional equation $$\xi(z+1)=\xi(z)+\frac1{\Gamma\big(2(z+1)\big)}$$ for which the inner summation as a function of $p$ would be a discrete case. What is this $\xi(z)$?

2)

$f(x)$ satisfies the following ODE $$(1+x^2)\,f''+4x\,f'-(3+x^2)\,f=0,\;\;f(0)=0,\; f'(0)=1.$$ The Frobenius method gives \begin{align} \phi(2k)&=0 \\ \phi(1)&=1 \\ \phi(3)=-\phi(1)&= -1 \\ \phi(k+2)+(k^2+3k-3)\phi(k)-k(k-1)\phi(k-2) &= 0 \end{align}

(to be continued)