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Question Number 73484 by abdomathmax last updated on 13/Nov/19

$$decomposeinsideC\left(x\right)thefraction$$$$F\left(x\right)=\frac{1}{{(x^2+1)}^n}$$$$calculate{\int }_0^{\mathrm{\infty }}F\left(x\right)dx$$

Answered by mind is power last updated on 13/Nov/19

$$F\left(x\right)=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{{(x^2+1)}^n}$$$$=i\pi Res(\frac{1}{{(x^2+1)}^n},i)$$$$=i\pi .\frac{1}{(n-1)!}\frac{d^{n-1}}{{dx}^{n-1}}.{(x-i)}^n.\frac{1}{{(x-i)}^n{(x+i)}^n}\underset{x=i}{\mid }$$$$f\left(x\right)=\frac{1}{{(x+i)}^n},f^{\left(k\right)}=\frac{{(-1)}^k.n.(n+1)...(n+k-1)}{{(x+i)}^{n+k}}$$$$f^{(n-)}\left(i\right)=\frac{{(-1)}^{n-1}.n.....(2n-2)}{{\left(2i\right)}^{2n-1}}=-2i\frac{n....(2n-2)}{4^n}=-2i\frac{(2n-2)!}{4^n.(n-1)!}$$$$=-2i(n-1)!.C_{2n-2}^{n-1}$$$$F\left(x\right)=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{{(x^2+1)}^n}=.\frac{i\pi }{(n-1)!}.-2i(n-1)C_{2n-1}^{n-1}.\frac{1}{4^n}=\pi \frac{C_{2n-2}^{n-1}}{2^{2n-1}}$$$$2ndMethode$$$${\int }_0^{+\mathrm{\infty }}\frac{dx}{(x^2+1)}={\int }_0^{\frac{\pi }{2}}\frac{d\left(tg\left(t\right)\right)}{{(1+{tg}^2\left(t\right))}^n}={\int }_0^{\frac{\pi }{2}}\frac{(1+{tg}^2\left(t\right))dt}{{(1+{tg}^2\left(t\right))}^n}$$$$={\int }_0^{\frac{\pi }{2}}\frac{dt}{{(1+{tg}^2\left(t\right))}^{n-1}}={\int }_0^{\frac{\pi }{2}}{cos}^{2n-2}\left(t\right)dt=W_{2n-2}..walise$$$$or,useB(x,y)=2{\int }_0^{\frac{\pi }{2}}{cos}^{2x-1}\left(t\right){sin}^{2y-1}\left(t\right)dt$$$$B(n-\frac{1}{2},\frac{1}{2})=2{\int }_0^{\frac{\pi }{2}}.{cos}^{2n-2}\left(t\right)dt=\frac{\mathrm{\Gamma }(n-\frac{1}{2}).\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }\left(n\right)}$$$$$$$$\mathrm{\Gamma }\left(\frac{1}{2}\right)=\sqrt{\pi }$$$$\mathrm{\Gamma }(n-\frac{1}{2})=\mathrm{\Gamma }\left(\frac{2n-1}{2}\right)=\mathrm{\Gamma }(\frac{2n-3}{2}+1)=\frac{(2n-3)}{2}......\frac{1}{2}.\sqrt{\pi }$$$$=\frac{(2n-3).....1.}{2^{n-1}}\sqrt{\pi }...n⩾2$$$$B(n-\frac{1}{2},\frac{1}{2})=\frac{\sqrt{\pi }}{2^{n-1}}.(2n-3)....\left(1\right).\sqrt{\pi }.\frac{1}{(n-1)!}$$$$=\frac{\pi .(2n-3)....1.....2.4.....(2n-2)}{2^{n-1}(n-1)!.2^{n-1}.(n-1)!}=\pi \frac{C_{2n-2}^{n-1}}{2^{2n-2}}=2{\int }_0^{\frac{\pi }{2}}{cos}^{2n-2}\left(t\right)dt$$$$\Rightarrow {\int }_0^{\frac{\pi }{2}}{cos}^{2n-2}\left(t\right)dt=\frac{\pi }{2}\frac{C_{2n-2}^{n-1}}{2^{2n-2}}=\pi \frac{C_{2n-2}^{n-1}}{2^{2n-1}}$$$$$$$$$$$$$$$$$$

Commented by abdomathmax last updated on 17/Nov/19

$$thsnkxsir.$$

Commented by mind is power last updated on 17/Nov/19

$${\mathrm{y}}^{\prime }\mathrm{re}\mathrm{welcom}$$

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