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Question Number 52418 by Abdo msup. last updated on 07/Jan/19

$$findthevalueof{\int }_0^{\mathrm{\infty }}\frac{arctan\left(x\right)}{1+x^4}dx$$

Commented by maxmathsup by imad last updated on 08/Jan/19

$$letf\left(t\right)={\int }_0^{\mathrm{\infty }}\frac{arctan\left(tx\right)}{1+x^4}dx\Rightarrow f^{{}^{\prime }}\left(t\right)={\int }_0^{\mathrm{\infty }}\frac{x}{(1+x^2{t}^2)(1+x^4)}dx$$$${=}_{xt=u}{\int }_0^{\mathrm{\infty }}\frac{u}{t(1+u^2)(1+\frac{u^4}{t^4})}\frac{du}{t}={\int }_0^{\mathrm{\infty }}\frac{u}{(1+u^2)(t^2+\frac{u^4}{t^2})}du$$$$=t^2{\int }_0^{\mathrm{\infty }}\frac{u}{(u^2+1)(u^4+t^4)}duletdecomposeF\left(u\right)=\frac{u}{(u^2+1)(u^4+t^4)}$$$$F\left(u\right)=\frac{u}{(u^2+1)({(u^2+t^2)}^2-2u^2{t}^2)}=\frac{u}{(u^2+1)(u^2+t^2-\sqrt{2}tu)(u^2+t^2+\sqrt{2}tu)}$$$$=\frac{au+b}{u^2+1}+\frac{cu+d}{u^2-\sqrt{2}tu+t^2}+\frac{eu+f}{u^2+\sqrt{2}tu+t^2}$$$$F(-u)=-F\left(u\right)\Rightarrow \frac{-au+b}{u^2+1}+\frac{-cu+d}{u^2+\sqrt{2}tu+t^2}+\frac{-eu+f}{u^2-\sqrt{2}tu+t^2}$$$$=\frac{-au-b}{u^2+1}+\frac{-cu-d}{u^2-\sqrt{2}tu+t^2}+\frac{-eu-f}{u^2+\sqrt{2}tu+t^2}\Rightarrow$$$$b=0andc=eandf=-d\Rightarrow$$$$F\left(u\right)=\frac{au}{u^2+1}+\frac{cu+d}{u^2-\sqrt{2}tu+t^2}+\frac{cu-d}{u^2+\sqrt{2}tu+t^2}$$$${lim}_{u\to +\mathrm{\infty }}uF\left(u\right)=0=a+2c\Rightarrow c=-\frac{a}{2}\Rightarrow$$$$F\left(u\right)=\frac{au}{u^2+1}-\frac{1}{2}\frac{au-2d}{u^2-\sqrt{2}tu+t^2}-\frac{1}{2}\frac{au+2d}{u^2+\sqrt{2}tu+t^2}$$$$F\left(1\right)=\frac{1}{2(1+t^4)}=\frac{a}{2}-\frac{1}{2}\frac{a-2d}{1+t^2-\sqrt{2}t}-\frac{1}{2}\frac{a+2d}{1+t^2+\sqrt{2}t}$$$$F(-1)=\frac{-1}{2(1+t^4)}-\frac{1}{2}\frac{a-2d}{1+t^2+\sqrt{2}t}-\frac{1}{2}\frac{-a+2d}{1+t^2-\sqrt{2}t}$$$$....becontinued...$$

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