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Question Number 115404 by EvoneAkashi last updated on 25/Sep/20

$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{1}{1+x^4}dx=$$

Answered by mathmax by abdo last updated on 25/Sep/20

$$\mathrm{let}\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{1+{\mathrm{x}}^4}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}={\mathrm{t}}^{\frac{1}{4}}$$$$\Rightarrow \mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{1}{1+\mathrm{t}}\times \frac{1}{4}{\mathrm{t}}^{\frac{1}{4}-1}\mathrm{dt}=\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{4}-1}}{1+\mathrm{t}}\mathrm{dt}\mathrm{but}\mathrm{we}\mathrm{know}$$$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{a}-1}}{1+\mathrm{t}}\mathrm{dt}=\frac{\pi }{\sin \left(\pi \mathrm{a}\right)}\mathrm{with}0<\mathrm{a}<1\Rightarrow \mathrm{I}=\frac{1}{4}.\frac{\pi }{\sin \left(\frac{\pi }{4}\right)}$$$$=\frac{\pi }{4.\frac{\sqrt{2}}{2}}=\frac{\pi }{2\sqrt{2}}$$

Answered by Dwaipayan Shikari last updated on 25/Sep/20

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

Answered by TANMAY PANACEA last updated on 25/Sep/20

$$\frac{1}{2}\int \frac{2\times \frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx$$$$\frac{1}{2}\int \frac{d(x-\frac{1}{x})-d(x+\frac{1}{x})}{x^2+\frac{1}{x^2}}$$$$\frac{1}{2}\int \frac{d(x-\frac{1}{x})}{{(x-\frac{1}{x})}^2+2}-\frac{1}{2}\int \frac{d(x+\frac{1}{x})}{{(x+\frac{1}{x})}^2-2}$$$$nowuseformulapls..Tanmay$$$$$$$$$$$$$$$$$$

Answered by mathmax by abdo last updated on 26/Sep/20

$$\mathrm{complex}\mathrm{method}2\mathrm{I}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^4+1}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{({\mathrm{x}}^2-\mathrm{i})({\mathrm{x}}^2+\mathrm{i})}$$$$=\frac{1}{2\mathrm{i}}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}(\frac{1}{{\mathrm{x}}^2-\mathrm{i}}-\frac{1}{{\mathrm{x}}^2+\mathrm{i}})=\frac{1}{2\mathrm{i}}({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^2-\mathrm{i}}-{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^2+\mathrm{i}})$$$$=\frac{1}{2\mathrm{i}}\left(2\mathrm{i}\mathrm{Im}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^2-\mathrm{i}}\right)\right)=\mathrm{Im}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^2-\mathrm{i}}\right)\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{1}{{\mathrm{z}}^2-\mathrm{i}}$$$$\mathrm{we}\mathrm{have}\phi \left(\mathrm{z}\right)=\frac{1}{(\mathrm{z}-{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})(\mathrm{z}+{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})}\mathrm{residus}\mathrm{theorem}\mathrm{give}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi -\mathrm{z})\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\phi ,{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})=2\mathrm{i}\pi .\frac{1}{{2\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}}=\mathrm{i}\pi {\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}$$$$=\mathrm{i}\pi \{\frac{1}{\sqrt{2}}-\frac{\mathrm{i}}{\sqrt{2}}\}=\frac{\mathrm{i}\pi }{\sqrt{2}}+\frac{\pi }{\sqrt{2}}\Rightarrow 2\mathrm{I}=\frac{\pi }{\sqrt{2}}\Rightarrow \mathrm{I}=\frac{\pi }{2\sqrt{2}}$$

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