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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.

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  • $\begingroup$ Derive differential equation satisfied by this Fourier transform as a function of $ay$, and show that this is equation for generalized hypergeomeric function dlmf.nist.gov/16.8. Then write down linear combination of independent solutions and find coefficients considering series expansion around $ay=0$. $\endgroup$ – Martin Nicholson Feb 6 '18 at 13:31
  • $\begingroup$ @Nemo: It seems easier said than done. The expression is a sum of two terms each of which is a product of some special functions and $\:_1\text{F}_2$. The two F's belong to distinct rather than a single generalized hypergeometric ODE's. You need to separate the terms first, which itself is a task unsolved. $\endgroup$ – Hans Feb 6 '18 at 17:30
  • $\begingroup$ compare the two linearly independent solutions of Gauss' hypergeometric equation dlmf.nist.gov/15.10.E2 . They seemingly belong to different hypergeometric ODE's. $\endgroup$ – Martin Nicholson Feb 6 '18 at 18:08
  • $\begingroup$ There is also another method by using Ramanujan's master theorem to calculate Mellin transform of $f(x)=\frac{\sin xy}{(x^2+1)^{\mu+1}}=\sum_{n=0}^\infty \lambda(n)(-x)^n$. So I just told you 2 different methods how to solve this problem. $\endgroup$ – Martin Nicholson Feb 6 '18 at 18:14
  • $\begingroup$ @Nemo: Thank you. The Ramanujan's master theorem method looks promising. I will try that. But I am still at a loss with your suggestion regarding Gauss' hypergeometric equation. Are you suggesting that each of the integrals in the question actually does satisfy one ODE or just that the right-hand side expression does not preclude the left-hand side from satisfying one single ODE? $\endgroup$ – Hans Feb 6 '18 at 18:55
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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)

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  • $\begingroup$ The part marked 2) is not needed. $\xi$ can be written as a hypergeometric function with parameters dependent on $z$, so that when $z=-p$ it reduces to a finite sum. $\endgroup$ – Martin Nicholson Feb 8 '18 at 9:11
  • $\begingroup$ By the way the wikipedia article is not complete. There is formula 11.19.4 from Hardy's book "Ramanujan, Twelve lectures suggested by his life and work" that will be more useful. $\endgroup$ – Martin Nicholson Feb 8 '18 at 10:17

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