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Question Number 145715 by Mrsof last updated on 07/Jul/21

Commented by Mrsof last updated on 07/Jul/21

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Answered by Dwaipayan Shikari last updated on 07/Jul/21

$$1){\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{x^4+4}dxx=\sqrt{2}u$$$$=\frac{1}{4\sqrt{2}}{\int }_0^{\mathrm{\infty }}\frac{1}{u^4+1}du=\frac{1}{4\sqrt{2}}.\frac{\pi }{sin\left(\frac{\pi }{4}\right)}=\frac{\pi }{4}$$

Answered by Dwaipayan Shikari last updated on 07/Jul/21

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{(x^2+1)}-\frac{1}{{(x^2+1)}^2}dx{x}^2=u$$$$=\pi -{\int }_0^{\mathrm{\infty }}\frac{u^{-1/2}}{{(u+1)}^2}du$$$$=\pi -B(\frac{1}{2},\frac{3}{2})=\pi -\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{3}{2}\right)}{\mathrm{\Gamma }\left(2\right)}=\pi -\frac{\pi }{2}=\frac{\pi }{2}$$

Answered by mathmax by abdo last updated on 07/Jul/21

$$2){\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{x}}^2}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{x}}^2+1-1}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^2+1}-{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^2+1)}^2}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^2+1}={\left[\mathrm{arctanx}\right]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}=\pi$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{(1+{\mathrm{x}}^2)}^2}{=}_{\mathrm{x}=\tan \theta }{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{(1+{\tan }^2\theta )}{{(1+{\tan }^2\theta )}^2}\mathrm{d}\theta ={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{\mathrm{d}\theta }{1+{\tan }^2\theta }$$$$=2{\int }_0^{\frac{\pi }{2}}{\cos }^2\theta \mathrm{d}\theta ={\int }_0^{\frac{\pi }{2}}(1+\cos \left(2\theta \right))\mathrm{d}\theta =\frac{\pi }{2}+{\left[\frac{1}{2}\sin \left(2\theta \right)\right]}_0^{\frac{\pi }{2}}=\frac{\pi }{2}$$$$\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{x}}^2}{{({\mathrm{x}}^2+1)}^2}=\pi -\frac{\pi }{2}=\frac{\pi }{2}$$

Answered by mathmax by abdo last updated on 07/Jul/21

$$1)\mathrm{{\rm Y}}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^4+4}\Rightarrow \mathrm{\Psi }{=}_{\mathrm{x}=\sqrt{2}\mathrm{t}}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\sqrt{2}\mathrm{dt}}{4(1+{\mathrm{t}}^4)}$$$$=\frac{\sqrt{2}}{4}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dt}}{{\mathrm{t}}^4+1}\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{1}{{\mathrm{z}}^4+1}\Rightarrow \phi \left(\mathrm{z}\right)=\frac{1}{({\mathrm{z}}^2-\mathrm{i})({\mathrm{z}}^2+\mathrm{i})}$$$$=\frac{1}{(\mathrm{z}-{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})(\mathrm{z}+{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})(\mathrm{z}-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})(\mathrm{z}+{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})}\mathrm{so}\mathrm{the}\mathrm{poles}\mathrm{are}\stackrel{-}{+}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}\mathrm{and}\stackrel{-}{+}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}$$$${\int }_{\mathrm{R}}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\mathrm{Res}(\phi ,{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})+\mathrm{Res}(\phi ,-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})\}$$$$\mathrm{Res}(\phi ,{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})=\frac{1}{{2\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}\left(2\mathrm{i}\right)}=\frac{1}{4\mathrm{i}}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}$$$$\mathrm{Res}(\phi ,-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})=\frac{1}{(-{2\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})(-2\mathrm{i})}=\frac{1}{4\mathrm{i}}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}\Rightarrow$$$${\int }_{\mathrm{R}}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\frac{1}{4\mathrm{i}}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}+\frac{1}{4\mathrm{i}}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}\}=\frac{\pi }{2}\left(2\cos \left(\frac{\pi }{4}\right)\right)=\frac{\pi }{\sqrt{2}}\Rightarrow$$$$\mathrm{\Psi }=\frac{\sqrt{2}}{4}\times \frac{\pi }{\sqrt{2}}\Rightarrow \mathrm{\Psi }=\frac{\pi }{4}$$

Commented by mathmax by abdo last updated on 07/Jul/21

$$\mathrm{another}\mathrm{way}\mathrm{\Psi }={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^4+4}\Rightarrow \mathrm{\Psi }{=}_{\mathrm{x}=\sqrt{2}\mathrm{t}}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\sqrt{2}\mathrm{dt}}{4(1+{\mathrm{t}}^4)}$$$$=\frac{\sqrt{2}}{4}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dt}}{1+{\mathrm{t}}^4}=\frac{1}{\sqrt{2}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{1+{\mathrm{t}}^4}{=}_{\mathrm{t}={\mathrm{z}}^{\frac{1}{4}}}\frac{1}{4\sqrt{2}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{z}}^{\frac{1}{4}-1}}{1+\mathrm{z}}\mathrm{dz}$$$$=\frac{1}{4\sqrt{2}}\times \frac{\pi }{\sin \left(\frac{\pi }{4}\right)}=\frac{\pi }{4\sqrt{2}\times \frac{1}{\sqrt{2}}}=\frac{\pi }{4}$$

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