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Question Number 44002 by maxmathsup by imad last updated on 19/Sep/18

$$calculateI={\int }_0^{\frac{\pi }{4}}\frac{arctanx}{1+x}andJ={\int }_0^{\frac{\pi }{4}}\frac{arctanx}{1-x}dx$$

Commented by maxmathsup by imad last updated on 24/Sep/18

$$letf\left(\alpha \right)={\int }_0^{\frac{\pi }{4}}\frac{arctan\left(\alpha x\right)}{1+x}dx\Rightarrow I=f\left(1\right)wehave$$$$f^{{}^{\prime }}\left(\alpha \right)={\int }_0^{\frac{\pi }{4}}\frac{xdx}{(1+x)(1+{\alpha }^2{x}^2)}{=}_{\alpha x=t}{\int }_0^{\frac{\alpha \pi }{4}}\frac{tdt}{\alpha (1+\frac{t}{\alpha })(1+t^2)}$$$$={\int }_0^{\frac{\alpha \pi }{4}}\frac{tdt}{(t+\alpha )(t^2+1)}letdecomposeF\left(t\right)=\frac{t}{(t+\alpha )(t^2+1)}$$$$F\left(t\right)=\frac{a}{t+\alpha }+\frac{bt+c}{t^2+1}wehavea={lim}_{t\to -\alpha }(t+\alpha )F\left(t\right)=\frac{-\alpha }{1+{\alpha }^2}$$$${lim}_{t\to +\mathrm{\infty }}tF\left(t\right)=0=a+b\Rightarrow b=\frac{\alpha }{1+{\alpha }^2}\Rightarrow F\left(t\right)=-\frac{\alpha }{(1+{\alpha }^2)(t+\alpha )}+\frac{\frac{\alpha }{1+{\alpha }^2}t+c}{1+t^2}$$$$F\left(0\right)=0=-\frac{\alpha }{(1+{\alpha }^2)\alpha }+c\Rightarrow c=\frac{1}{1+{\alpha }^2}\Rightarrow$$$$F\left(t\right)=-\frac{\alpha }{(1+{\alpha }^2)(t+\alpha )}+\frac{1}{1+{\alpha }^2}\frac{\alpha t+1}{t^2+1}\Rightarrow$$$$f^{{}^{\prime }}\left(\alpha \right)=\frac{-\alpha }{1+{\alpha }^2}{\int }_0^{\frac{\alpha \pi }{4}}\frac{dt}{t+\alpha }+\frac{1}{1+{\alpha }^2}{\int }_0^{\frac{\alpha \pi }{4}}\frac{\alpha t+1}{t^2+1}dtbut$$$${\int }_0^{\frac{\alpha \pi }{4}}\frac{dt}{t+\alpha }={[ln\mid t+\alpha \mid ]}_0^{\frac{\alpha \pi }{4}}=ln\mid \frac{\alpha \pi }{4}+\alpha \mid -ln\mid \alpha \mid =ln\mid 1+\frac{\pi }{4}\mid$$$${\int }_0^{\frac{\alpha \pi }{4}}\frac{\alpha t+1}{t^2+1}dt=\frac{\alpha }{2}{\left[ln(t^2+1)\right]}_0^{\frac{\alpha \pi }{4}}+arctan\left(\frac{\alpha \pi }{4}\right)$$$$=\frac{\alpha }{2}ln(1+\frac{{\alpha }^2{\pi }^2}{16})+arctan\left(\frac{\alpha \pi }{4}\right)\Rightarrow$$$$f^{{}^{\prime }}\left(\alpha \right)=\frac{-\alpha }{1+{\alpha }^2}ln(1+\frac{\pi }{4})+\frac{\alpha }{2(1+{\alpha }^2)}ln(1+\frac{{\alpha }^2{\pi }^2}{16})+\frac{1}{1+{\alpha }^2}arctan\left(\frac{\alpha \pi }{4}\right)\Rightarrow$$$$f\left(\alpha \right)=-ln(1+\frac{\pi }{4}){\int }_0^{\alpha }\frac{x}{1+x^2}dx+\frac{1}{2}{\int }_0^{\alpha }\frac{x}{1+x^2}ln(1+\frac{x^2{\pi }^2}{16})dx+{\int }_0^{\alpha }\frac{1}{1+x^2}arctan\left(\frac{\pi x}{4}\right)dx$$$$+c(c=f\left(0\right)=0)\Rightarrow$$$$f\left(1\right)=-ln(1+\frac{\pi }{4}){\int }_0^1\frac{x}{1+x^2}dx+\frac{1}{2}{\int }_0^1\frac{x}{1+x^2}ln(1+\frac{x^2{\pi }^2}{16})dx$$$$+{\int }_0^1\frac{arctan\left(\frac{\pi x}{4}\right)}{1+x^2}dx....becontinued...$$$$$$

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