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Question Number 131405 by bramlexs22 last updated on 04/Feb/21

$$$$$$\dots \dots \mathrm{super}\mathrm{cooles}\mathrm{Integral}\text{iddots}\text{iddots}$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{{(1+x^2)}^2}=?$$

Commented by bramlexs22 last updated on 04/Feb/21

$$$$$$\to \to ⌣\to \to \mathrm{ok}\mathrm{alle}\mathrm{antworten}\mathrm{sind}\mathrm{super}$$

Commented by Dwaipayan Shikari last updated on 04/Feb/21

$$ForGeneral{\int }_0^{\mathrm{\infty }}\frac{dx}{{(1+x^n)}^n}=\frac{\mathrm{\Gamma }(n-\frac{1}{n})\mathrm{\Gamma }\left(\frac{1}{n}\right)}{n!}n\ne 0,1$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{{(1+x^2)}^2}=\frac{\mathrm{\Gamma }\left(\frac{3}{2}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{2!}=\frac{\pi }{4}$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{{(1+x^3)}^3}=\frac{\mathrm{\Gamma }\left(\frac{2}{3}\right)\mathrm{\Gamma }\left(\frac{1}{3}\right)}{6}.\frac{5}{3}.\frac{2}{3}=\frac{10\pi }{27\sqrt{3}}$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{{(1+x^4)}^4}=\frac{231\pi }{256\sqrt{2}}$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{{(1+x^5)}^5}=\frac{2054\sqrt{2}}{9375\sqrt{5-\sqrt{5}}}\pi$$$$...$$

Answered by mnjuly1970 last updated on 04/Feb/21

$$\varphi \stackrel{x=\frac{1}{t}}{=}{\int }_0^{\mathrm{\infty }}\frac{1}{{(1+\frac{1}{t^2})}^2}\ast \frac{dt}{t^2}={\int }_0^{\mathrm{\infty }}\frac{t^4}{{(1+t^2)}^2.t^2}dt$$$$={\int }_0^{\mathrm{\infty }}\frac{t^2+1-1}{{(1+t^2)}^2}dt=\frac{\pi }{2}-\varphi$$$$\therefore \varphi =\frac{\pi }{4}....✓$$

Answered by Ar Brandon last updated on 04/Feb/21

$${\mathrm{x}}^2=\mathrm{u}\Rightarrow 2\mathrm{xdx}=\mathrm{du}$$$$\mathcal{I}=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{u}}^{-\frac{1}{2}}}{{(1+\mathrm{u})}^2}\mathrm{du}=\frac{1}{2}\beta (\frac{1}{2},\frac{3}{2})$$$$=\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{3}{2}\right)}{2\mathrm{\Gamma }\left(2\right)}=\frac{\pi }{4}$$

Answered by mnjuly1970 last updated on 04/Feb/21

$$$$$$\varphi ={\int }_0^{\mathrm{\infty }}\frac{dx}{{(1+x^2)}^2}\stackrel{x^2=t}{=}\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{-1}{2}}}{{(1+t)}^2}dt$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{1}{2}-1}}{{(1+t)}^2}dt=\frac{1}{2}\beta (\frac{1}{2},\frac{3}{2})$$$$=\frac{1}{2}\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{3}{2}\right)}{\mathrm{\Gamma }\left(2\right)}=\frac{1}{4}{\mathrm{\Gamma }}^2\left(\frac{1}{2}\right)=\frac{\pi }{4}..✓$$

Answered by mathmax by abdo last updated on 04/Feb/21

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{(1+{\mathrm{x}}^2)}^2}\Rightarrow \mathrm{I}{=}_{\mathrm{x}=\mathrm{tant}}{\int }_0^{\frac{\pi }{2}}\frac{(1+{\tan }^2\mathrm{t})}{{(1+{\tan }^2\mathrm{t})}^2}\mathrm{dt}={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dt}}{1+{\tan }^2\mathrm{t}}$$$$={\int }_0^{\frac{\pi }{2}}{\cos }^2\mathrm{t}\mathrm{dt}={\int }_0^{\frac{\pi }{2}}\frac{1+\cos \left(2\mathrm{t}\right)}{2}\mathrm{dt}=\frac{\pi }{4}+\frac{1}{4}{\left[\sin \left(2\mathrm{t}\right)\right]}_0^{\frac{\pi }{2}}=\frac{\pi }{4}+0\Rightarrow$$$$\mathrm{I}=\frac{\pi }{4}$$

Answered by mathmax by abdo last updated on 04/Feb/21

$$\mathrm{parametric}\mathrm{method}\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^2+{\mathrm{a}}^2}\mathrm{with}\mathrm{a}>0\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-{\int }_0^{\mathrm{\infty }}\frac{2\mathrm{a}}{{({\mathrm{x}}^2+{\mathrm{a}}^2)}^2}\mathrm{dx}\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^2+{\mathrm{a}}^2)}^2}=-\frac{1}{2\mathrm{a}}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^2+1)}^2}=-\frac{1}{2}{\mathrm{f}}^{{}^{\prime }}\left(1\right)\mathrm{we}\mathrm{have}\mathrm{f}\left(\mathrm{a}\right){=}_{\mathrm{x}=\mathrm{az}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{adz}}{{\mathrm{a}}^2(1+{\mathrm{z}}^2)}$$$$=\frac{1}{\mathrm{a}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dz}}{1+{\mathrm{z}}^2}=\frac{\pi }{2\mathrm{a}}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-\frac{\pi }{{2\mathrm{a}}^2}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(1\right)=-\frac{\pi }{2}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^2+1)}^2}=-\frac{1}{2}(-\frac{\pi }{2})=\frac{\pi }{4}$$

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