0

Proof of the identity $\int_0^{+\infty}\frac{\sin(x)}{x^\alpha}dx=\frac{\Gamma(\alpha/2)\Gamma(1-\alpha/2)}{2\Gamma(\alpha)}$ for $\alpha\in (0,2)$.

In the above post, in the 2nd answer (given by mark viola) - in the 2nd part - "ALTERNATIVE METHODOLOGY: CONTOUR INTEGRATION" equation no 8 goes to zero as $\varepsilon\to0^+$ only if $\alpha <1$ but we need to prove the identity for $\alpha\in (0,2)$. Is the proof incomplete then? If it's incomplete then how to extend the proof upto $\alpha < 2$.

Gary
  • 31,845
shanrrg
  • 41
  • 2
    Use analytic continuation in $\alpha$. – Gary Feb 26 '23 at 07:06
  • 2
    We could simply use the analytical continuation approach. The proof is rigorous for $\alpha\in(0;1)$, and the answer (and the initial integral) are well defined for $\alpha\in(0;2)$. – Svyatoslav Feb 26 '23 at 07:07
  • 1
    Note that after arriving at $(8)$, I proceeded to integrate by parts. Then, I took the limit as $\varepsilon\to 0^+$, which leads to a result that is valid for $\alpha\in (0,2)$ – Mark Viola Mar 02 '23 at 18:59
  • Ah. Got it. thanks – shanrrg Mar 19 '23 at 06:04

0 Answers0