The integral is as:

$$\int_0^\infty\sin\left(x\right)\times x^{n}\;\mathbb{d}x$$ Where $n\in\mathfrak{R}$.

I was able to find the convergence of this integral using Dirichlet's test, which implied $-2<n<0$.

But the hard part is to find the value of this integral for me. Wolfram alpha gives the answer as:

Which is very close to gamma function. So i tried to assume integral of form $$\int_0^\infty e^{-ix}\times x^n\;\mathbb{d}x$$

So that i would take the $-\mathfrak{Im}$ part of this integral. But I still have no idea how to evaluate this.


marked as duplicate by Simply Beautiful Art, Connor Harris, Sil, José Carlos Santos calculus Aug 16 '18 at 19:28

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    $\begingroup$ Use integration via parts thats all by LIATE rule $\endgroup$ – Archis Welankar Mar 6 '16 at 17:28
  • $\begingroup$ Sorry i kinda don't understand let us assume there's no bound for n. Even then we can only go upto integral part of n, the fraction part will create trouble. $\endgroup$ – Mann Mar 6 '16 at 17:44