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.

  • $\begingroup$ 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

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 $\xi(z)$?


$f(x)$ 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)

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.