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Question Number 126493 by mathmax by abdo last updated on 21/Dec/20

$$\mathrm{calculate}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^4+1)}^2}$$

Answered by Dwaipayan Shikari last updated on 21/Dec/20

$${\int }_0^{\mathrm{\infty }}\frac{dx}{{(x^4+1)}^2}{x}^4=t\Rightarrow 4x^3=\frac{dt}{dx}$$$$=\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{t^{-\frac{3}{4}}}{{(t+1)}^2}dt\frac{t}{t+1}=u$$$$=\frac{1}{4}{\int }_0^1{\left(\frac{u}{1-u}\right)}^{-\frac{3}{4}}du=\frac{1}{4}\beta (\frac{1}{4},\frac{7}{4})=\frac{3\mathrm{\Gamma }\left(\frac{1}{4}\right)\mathrm{\Gamma }\left(\frac{3}{4}\right)}{16}=\frac{3\pi }{8\sqrt{2}}$$

Answered by mathmax by abdo last updated on 21/Dec/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^4+{\mathrm{a}}^4}\mathrm{with}\mathrm{a}>0\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-{\int }_0^{\mathrm{\infty }}\frac{{4\mathrm{a}}^3}{{({\mathrm{x}}^4+{\mathrm{a}}^4)}^2}\mathrm{dx}\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(1\right)=-4{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^4+1)}^2}\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^4+1)}^2}=-\frac{1}{4}{\mathrm{f}}^{{}^{\prime }}\left(1\right)\mathrm{let}\mathrm{explicit}\mathrm{f}\left(\mathrm{a}\right)$$$$\mathrm{f}\left(\mathrm{a}\right){=}_{\mathrm{x}=\mathrm{at}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{adt}}{{\mathrm{a}}^4({\mathrm{t}}^4+1)}=\frac{1}{{\mathrm{a}}^3}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{1+{\mathrm{t}}^4}{=}_{\mathrm{t}={\mathrm{z}}^{\frac{1}{4}}}$$$$=\frac{1}{{4\mathrm{a}}^3}{\int }_0^{\mathrm{\infty }}\frac{1}{1+\mathrm{z}}{\mathrm{z}}^{\frac{1}{4}-1}\mathrm{dz}=\frac{1}{{4\mathrm{a}}^3}\times \frac{\pi }{\sin \left(\frac{\pi }{4}\right)}=\frac{\pi }{{4\mathrm{a}}^3\times \frac{\sqrt{2}}{2}}=\frac{\pi }{2\sqrt{2}{\mathrm{a}}^3}\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{\pi }{2\sqrt{2}}{\mathrm{a}}^{-3}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=\frac{\pi }{2\sqrt{2}}(-3){\mathrm{a}}^{-4}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(1\right)=\frac{-3\pi }{2\sqrt{2}}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^4+1)}^2}=-\frac{1}{4}\times \frac{-3\pi }{2\sqrt{2}}=\frac{3\pi }{8\sqrt{2}}=\frac{3\pi \sqrt{2}}{16}$$

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