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Question Number 39711 by math khazana by abdo last updated on 10/Jul/18

$$calculate{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^n}{{(1+x^2)}^n}dxwithnnatralintegr$$

Commented by maxmathsup by imad last updated on 11/Jul/18

$$let{A}_n={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^{n}dx}{{(1+x^2)}^n}andlet\phi \left(z\right)=\frac{z^n}{{(1+z^2)}^n}$$$$\phi \left(z\right)=\frac{z^n}{{(z-i)}^n{(z+i)}^n}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,i)but$$$$Res(\phi ,i)={lim}_{z\to i}\frac{1}{(n-1)!}{\left\{{(z-i)}^n\phi \left(z\right)\right\}}^{(n-1)}$$$$={lim}_{z\to i}\frac{1}{(n-1)!}{\left\{x^n{(z+i)}^{-n}\right\}}^{(n-1)}and$$$${\left\{x^n{(z+i)}^{-n}\right\}}^{(n-1)}=\sum _{k=0}^{n-1}{C}_{n-1}^k{\left\{{(z+i)}^{-n}\right\}}^{\left(k\right)}{\left(x^n\right)}^{(n-1-k)}but$$$${{(z+i)}^{-n}\}}^{\left(k\right)}={(-1)}^{k}n(n+1)...(n+k-1){(z+i)}^{-n-k}$$$${\left(x^n\right)}^{\left(p\right)}=n(n-1)...(n-p+1)x^{n-p}\Rightarrow {\left(x^n\right)}^{(n-1-k)}=n(n-1)...(k+2)x^{k+1}\Rightarrow$$$$Res(\phi ,i)=\frac{1}{(n-1)!}\sum _{k=0}^{n-1}{C}_{n-1}^k{(-1)}^{k}n(n+1)...(n+k-1){\left(2i\right)}^{-n-k}(n-1)...(k+2)i^{k+1}$$$$A_n=2i\pi Res(\phi ,i)$$

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