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# Trigonometric Integral

$\large \int_{0}^{\pi} (\sin (x))^{m-1} e^{inx} \ \mathrm{d}x = \dfrac{\pi}{2^{m-1}} \ \dfrac{e^{i n/2}}{m \operatorname{B} \left( \dfrac{1}{2} (m+n+1) , \dfrac{1}{2} (m-n+1) \right)} \ ; \ \Re(m) > 0$

Prove the above identity.

Notation :

• $$i$$ denotes the complex unit iota ; $$i = \sqrt{-1}$$

• $$\operatorname{B} (a,b)$$ denotes the Beta Function.

This is a part of the set Formidable Series and Integrals.

Note by Ishan Singh
6 months ago