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Question Number 31969 by abdo imad last updated on 17/Mar/18

$$findthevalueof{\int }_0^{\mathrm{\infty }}\left(\frac{1+t^2}{1+t^4}\right)arctantdt.$$

Commented by abdo imad last updated on 19/Mar/18

$$letputI={\int }_0^{\mathrm{\infty }}\frac{arctant}{1+t^4}dt.ch.x=\frac{1}{t}give$$$$I={\int }_0^{\mathrm{\infty }}\frac{\frac{\pi }{2}-arctant}{1+\frac{1}{t^4}}\frac{dt}{t^2}={\int }_0^{\mathrm{\infty }}\frac{\frac{\pi }{2}-arctant}{t^2+\frac{1}{t^2}}dt$$$$={\int }_0^{\mathrm{\infty }}\frac{t^2(\frac{\pi }{2}-arctant)}{1+t^4}dt=\frac{\pi }{2}{\int }_0^{\mathrm{\infty }}\frac{t^2}{1+t^4}dt-{\int }_0^{\mathrm{\infty }}\frac{t^{2}arctant}{1+t^2}dt\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\left(\frac{1+t^2}{1+t^4}\right)arctantdt=\frac{\pi }{2}{\int }_0^{\mathrm{\infty }}\frac{t^2}{1+t^4}dt.ch.t^4=ugive$$$${\int }_0^{\mathrm{\infty }}\frac{t^2}{1+t^4}dt={\int }_0^{\mathrm{\infty }}\frac{{\left(u^{\frac{1}{4}}\right)}^2}{1+u}\frac{1}{4}{u}^{\frac{1}{4}-1}du$$$$=\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{1}{2}+\frac{1}{4}-1}}{1+u}du=\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{3}{4}-1}}{1+u}du$$$$=\frac{1}{4}\frac{\pi }{sin\left(\frac{3\pi }{4}\right)}=\frac{\pi }{4}\frac{1}{\frac{1}{\sqrt{2}}}=\frac{\pi \sqrt{2}}{4}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\left(\frac{1+t^2}{1+t^4}\right)arctantdt=\frac{\pi }{2}\frac{\pi \sqrt{2}}{4}=\frac{{\pi }^2}{8}\sqrt{2}.$$$$Ihaveusedtheresult{\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t}dt=\frac{\pi }{sin\left(\pi a\right)}with0<a<1$$

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