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Question Number 169797 by mathocean1 last updated on 09/May/22

$$Calculateforn\in {\mathbb{N}}^{\ast }:{\int }_0^{+\mathrm{\infty }}\frac{dt}{{(t^2+1)}^n}$$$$(Notice:1=(1+t^2)-t^2)$$

Answered by floor(10²Eta[1]) last updated on 09/May/22

$$\mathrm{t}=\mathrm{tg}\theta \Rightarrow \mathrm{dt}={\sec }^2\theta \mathrm{d}\theta$$$${\int }_0^{\pi /2}\frac{{\sec }^2\mathrm{d}\theta }{{\sec }^{2\mathrm{n}}\theta }={\int }_0^{\pi /2}{\cos }^{2\mathrm{n}-2}\theta \mathrm{d}\theta ,\mathrm{use}\mathrm{reduction}\mathrm{formula}.$$

Answered by Mathspace last updated on 09/May/22

$$u_n={\int }_0^{\mathrm{\infty }}\frac{dt}{{(t^2+1)}^n}\Rightarrow$$$$wehaveB(x,y)={\int }_0^{\mathrm{\infty }}\frac{t^{x-1}}{{(1+t)}^{x+y}}dt$$$$so{\int }_0^{\mathrm{\infty }}\frac{dt}{{(1+t^2)}^n}{=}_{t=\sqrt{x}}\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{x^{-\frac{1}{2}}}{{(1+x)}^n}dx$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{1}{2}-1}}{{(1+t)}^{n+\frac{1}{2}-\frac{1}{2}}}dt$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{1}{2}-1}}{{(1+t)}^{\frac{1}{2}+n-\frac{1}{2}}}dt$$$$=\frac{1}{2}B(\frac{1}{2},n-\frac{1}{2})$$$$=\frac{1}{2}\times \frac{\mathrm{\Gamma }\left(\frac{1}{2}\right).\mathrm{\Gamma }(n-\frac{1}{2})}{\mathrm{\Gamma }\left(n\right)}$$$$=\frac{\sqrt{\pi }}{2(n-1)!}\mathrm{\Gamma }(n-\frac{1}{2})$$$$=\frac{\sqrt{\pi }}{2}\times \frac{(n-\frac{3}{2})!}{(n-1)!}(n>0)$$

Answered by Mathspace last updated on 09/May/22

$$residusmethod$$$${\int }_0^{\mathrm{\infty }}\frac{dt}{{(t^2+1)}^n}=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dt}{{(t^2+1)}^n}$$$$\phi \left(z\right)=\frac{1}{{(z^2+1)}^n}\Rightarrow \phi \left(z\right)=\frac{1}{{(z-i)}^n{(z+i)}^n}$$$${\int }_R\phi dz=2i\pi Res(\phi ,i)$$$$Res(\phi ,i)={lim}_{z\to i}\frac{1}{(n-1)!}{\left\{{(z-i)}^n\phi \left(z\right)\right\}}^{(n-1)}$$$$={lim}_{z\to i}\frac{1}{(n-1)!}{\left\{{(z+i)}^{-n}\right\}}^{(n-1)}$$$${wehave{(z+i)}^p\}}^{\left(1\right)}=p{(z+i)}^{p-1}$$$${...{(z+i)}^p\}}^{\left(k\right)}=p(p-1)...(p-k+1){(z+i)}^{p-k}$$$${{(z+i)}^{-n}\}}^{(n-1)}=(-n)(-n-1)...(-n-n+1+1){(z+i)}^{-n-n+1}$$$$={(-1)}^{n-1}n(n+1)....(2n+2){(z+i)}^{-2n+1}$$$$.....$$

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