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This is regarding the answer given by @bertrand87 in this post Integral $\int_0^\infty x^{-\alpha} \sin(x)dx.$

$$\left\{\mathscr{M} \sin\right\}(s) = \int_{0}^{\infty} x^{s-1} \sin(x) dx = \Gamma(s)\sin\left(\frac{s\pi}{2}\right)$$

Now, the above identity is valid for $s>0$ as the gamma integral converges for $s>0$.

Subsituing $s=1−a$, we have that

$$\int_{0}^{\infty} x^{-a} \sin(x) dx = \Gamma(1-a)\sin\left(\frac{\pi}{2}-\frac{a\pi}{2}\right) = \Gamma(1-a)\cos\left(\frac{a\pi}{2}\right)$$ which is valid for $a<1$.

However, it is mentioned in the answer that the above identity is valid for $a\in(0,1)\cup(1,2)$. How?

Christian E. Ramirez
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shanrrg
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    As far as I know, $$\int_{0}^{\infty} x^{s-1} \sin(x) dx = \Gamma(s)\sin\left(\frac{s\pi}{2}\right)$$ is valid if $ -1<\Re(s)<1$ – Claude Leibovici Feb 25 '23 at 08:10
  • Since that identity was derived using Ramanujan's master theorem ,what then is the range for s, in which RMT is valid? I assumed it valid for s>0 only , which is why I had asked this question. – shanrrg Feb 25 '23 at 12:01

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