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Question Number 144875 by lapache last updated on 30/Jun/21

Answered by Dwaipayan Shikari last updated on 30/Jun/21

$$\frac{1}{n!}{\int }_0^{\mathrm{\infty }}{x}^n\sin \left(x\right)e^{-x}dx$$$$=\frac{1}{2in!}{\int }_0^{\mathrm{\infty }}{x}^n{e}^{-(1-i)x}-x^n{e}^{-(1+i)x}dx$$$$=\frac{1}{2in!}(\frac{n!}{{(1-i)}^{n+1}}-\frac{n!}{{(1+i)}^{n+1}})$$$$=\frac{2^{\frac{-n+1}{2}}}{2i}(e^{\frac{i\pi (n+1)}{4}}-e^{\frac{-i\pi (n+1)}{4}})=2^{-\frac{n+1}{2}}\sin \left(\frac{\pi }{4}(n+1)\right)$$

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