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Question Number 133433 by greg_ed last updated on 22/Feb/21

$$\mathbf{hi},\mathbf{everybody}!$$$$\mathbf{with}\mathbf{I}={\int }_0^{\mathrm{\infty }}\frac{1}{{\mathit{x}}^4+1}\mathit{d}\mathit{x},$$$$\mathbf{prove}\mathbf{that}:2\mathbf{I}={\int }_0^{\mathrm{\infty }}\frac{{\mathit{x}}^2+1}{{\mathit{x}}^4+1}\mathit{d}\mathit{x}.$$

Answered by Dwaipayan Shikari last updated on 22/Feb/21

$$I={\int }_0^{\mathrm{\infty }}\frac{1}{x^4+1}dx=\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{1}{4}-1}}{{(u+1)}^{\frac{3}{4}+\frac{1}{4}}}du{x}^4=u$$$$=\frac{1}{4}\mathrm{\Gamma }\left(\frac{3}{4}\right)\mathrm{\Gamma }\left(\frac{1}{4}\right)=\frac{\pi }{2\sqrt{2}}$$$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{x^2+1}{x^4+1}dx={\int }_0^{\mathrm{\infty }}\frac{x^2}{x^4+1}+I$$$$=\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{3}{4}-1}}{{(1+u)}^{\frac{3}{4}+\frac{1}{4}}}du+I$$$$=\frac{1}{4}\mathrm{\Gamma }\left(\frac{3}{4}\right)\mathrm{\Gamma }\left(\frac{1}{4}\right)+I=\frac{\pi }{2\sqrt{2}}+I=I+I=2I$$

Answered by Dwaipayan Shikari last updated on 22/Feb/21

$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{1}{(x^4+1)}dxx=\frac{1}{t}$$$$={\int }_0^{\mathrm{\infty }}\frac{t^2}{t^4+1}dt=\mathrm{\Phi }$$$$2\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{t^2}{t^4+1}+\frac{1}{1+t^4}dt$$

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