Integral $\int_0^\infty\frac{\sin x}{x^\alpha}dx=\frac{\sqrt\pi\;\Gamma\big(1-\frac{\alpha}{2}\big)}{2^\alpha\; \Gamma\big(\frac{1+\alpha}{2}\big)}$

Asked Jul 21 '21 at 16:12

Active Jul 21 '21 at 16:21

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Lemma 1 in Szekely G J , Rizzo M L.(2005) has such equation:

$$\int_0^\infty\frac{\sin x}{x^\alpha}dx=\frac{\sqrt\pi\;\Gamma\big(1-\frac{\alpha}{2}\big)}{2^\alpha\; \Gamma\big(\frac{1+\alpha}{2}\;\big)},\quad\alpha\in(0,2).$$ I don't know how to prove this equation, please give me some hints or reference.

edited Jul 21 '21 at 16:21

Thomas Andrews

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asked Jul 21 '21 at 16:12

John

Deform the contour to the imaginary axis. – Mark Viola Jul 21 '21 at 16:16
1

Here is how to ask a good question. Maybe lengthen the question a bit more with some research on how to solve it. – Тyma Gaidash Jul 21 '21 at 16:28
See any of the answers in 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 $\alpha\in (0,2)$. and use the Legendre duplication formula to convert the closed form given in the linked question to yours. – projectilemotion Jul 21 '21 at 16:42

Integral $\int_0^\infty\frac{\sin x}{x^\alpha}dx=\frac{\sqrt\pi\;\Gamma\big(1-\frac{\alpha}{2}\big)}{2^\alpha\; \Gamma\big(\frac{1+\alpha}{2}\big)}$

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