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Question Number 161899 by HongKing last updated on 23/Dec/21

$$\mathrm{Find}:$$$$\mathbf{\Omega }=\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}\frac{1}{{(1+{\mathrm{x}}^{2\mathbf{n}})}^2}\mathrm{dx};\mathrm{n}\in \mathbb{Z}$$

Answered by Ar Brandon last updated on 25/Dec/21

$$\mathrm{\Omega }={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{{(1+x^{2n})}^2}=2{\int }_0^{\mathrm{\infty }}\frac{dx}{{(1+x^{2n})}^2}=\frac{1}{n}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{1}{2n}-1}}{{(1+u)}^2}du$$$$=\frac{1}{n}\beta (\frac{1}{2n},2-\frac{1}{2n})=\frac{1}{n}\cdot \mathrm{\Gamma }\left(\frac{1}{2n}\right)\mathrm{\Gamma }(2-\frac{1}{2n})$$$$=\frac{1}{n}(1-\frac{1}{2n})\mathrm{\Gamma }\left(\frac{1}{2n}\right)\mathrm{\Gamma }(1-\frac{1}{2n})$$$$=\frac{2n-1}{2n^2}\cdot \frac{\pi }{\sin \left(\frac{\pi }{2n}\right)}$$$$n\ne \frac{1}{2(k+1)},k\in \mathbb{Z}\mathrm{\setminus }\{-1\}$$

Answered by mathmax by abdo last updated on 25/Dec/21

$$\mathrm{{\rm Y}}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^{2\mathrm{n}}+1)}^2}\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^{2\mathrm{n}}+{\mathrm{a}}^{2\mathrm{n}}}(\mathrm{a}>0)\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-2\mathrm{n}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{a}}^{2\mathrm{n}-1}}{{({\mathrm{x}}^{2\mathrm{n}}+{\mathrm{a}}^{2\mathrm{n}})}^2}\mathrm{dx}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(1\right)=-2\mathrm{n}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^{2\mathrm{n}}+1)}^2}$$$$\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^{2\mathrm{n}}+1)}^2}=-\frac{1}{2\mathrm{n}}{\mathrm{f}}^{{}^{\prime }}\left(1\right)$$$$\mathrm{on}\mathrm{f}\left(\mathrm{a}\right){=}_{\mathrm{x}=\mathrm{at}}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{adt}}{{\mathrm{a}}^{2\mathrm{n}}{\mathrm{t}}^{2\mathrm{n}}+{\mathrm{a}}^{2\mathrm{n}}}=\frac{2}{{\mathrm{a}}^{2\mathrm{n}-1}}{\int }_0^{+\mathrm{\infty }}\frac{\mathrm{dt}}{{\mathrm{t}}^{2\mathrm{n}}+1}$$$${=}_{\mathrm{t}={\mathrm{z}}^{\frac{1}{2\mathrm{n}}}}\frac{2}{{\mathrm{a}}^{2\mathrm{n}-1}}{\int }_0^{\mathrm{\infty }}\frac{1}{\mathrm{z}+1}\frac{1}{2\mathrm{n}}{\mathrm{z}}^{\frac{1}{2\mathrm{n}}-1}\mathrm{dz}$$$$=\frac{1}{{\mathrm{na}}^{2\mathrm{n}-1}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{z}}^{\frac{1}{2\mathrm{n}}-1}}{\mathrm{z}+1}=\frac{1}{{\mathrm{na}}^{2\mathrm{n}-1}}.\frac{\pi }{\sin \left(\frac{\pi }{2\mathrm{n}}\right)}=\frac{\pi }{\mathrm{nsin}\left(\frac{\pi }{2\mathrm{n}}\right)}{\mathrm{a}}^{-2\mathrm{n}+1}$$$$\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=(-2\mathrm{n}+1).\frac{\pi }{\mathrm{nsin}\left(\frac{\pi }{2\mathrm{n}}\right)}{\mathrm{a}}^{-2\mathrm{n}}\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(1\right)=\frac{\pi (-2\mathrm{n}+1)}{\mathrm{nsin}\left(\frac{\pi }{2\mathrm{n}}\right)}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^{2\mathrm{n}}+1)}^2}=-\frac{1}{2\mathrm{n}}\times \frac{\pi (-2\mathrm{n}+1)}{\mathrm{nsin}\left(\frac{\pi }{2\mathrm{n}}\right)}$$$$=\frac{\pi (2\mathrm{n}-1)}{{2\mathrm{n}}^2\sin \left(\frac{\pi }{2\mathrm{n}}\right)}$$

Commented by HongKing last updated on 25/Dec/21

$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{cool}\mathrm{my}\mathrm{dear}\mathrm{Sir}$$

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