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Question Number 31098 by abdo imad last updated on 02/Mar/18

$$findthevalueof{\int }_1^{\mathrm{\infty }}\frac{arctan(x+1)-arctanx}{x^2}dx.$$

Commented by abdo imad last updated on 04/Mar/18

$$letputI\left(\xi \right)={\int }_1^{\xi }\frac{arctan(x+1)-arctanx}{x^2}dx$$$$wehaveI={lim}_{\xi \to +\mathrm{\infty }}I\left(\xi \right)$$$$I\left(\xi \right)={\int }_2^{\xi +1}\frac{arctant}{{(t-1)}^2}dt-{\int }_1^{\xi }\frac{arctanx}{x^2}dtletintegratebypsrts$$$${\int }_2^{1+\xi }\frac{arctant}{{(t-1)}^2}dt={\left[\frac{-1}{t-1}arctant\right]}_2^{1+\xi }-{\int }_2^{1+\xi }\frac{-1}{t-1}\frac{dt}{1+t^2}$$$$=arctan2-\frac{1}{\xi }arctan(1+\xi )+{\int }_2^{1+\xi }\frac{dt}{(t-1)(1+t^2)}$$$$\to arctan2+{\int }_2^{+\mathrm{\infty }}\frac{dt}{(t-1)(1+t^2)}but$$$${\int }_2^{+\mathrm{\infty }}\frac{dt}{(t-1)(1+t^2)}={\int }_1^{+\mathrm{\infty }}\frac{dx}{x(1+{(x+1)}^2)}={\int }_1^{+\mathrm{\infty }}\frac{dx}{x(x^2+2x+2)}$$$$\frac{1}{x(x^2+2x+2)}=\frac{a}{x}+\frac{bx+c}{x^2+2x+2}=f\left(x\right)$$$$a={lim}_{x\to 0}xf\left(x\right)=\frac{1}{2},{lim}_{x\to \mathrm{\infty }}xf\left(x\right)=0=a+b\Rightarrow b=-\frac{1}{2}so$$$$f\left(x\right)=\frac{1}{2x}+\frac{\frac{-1}{2}x+c}{x^2+2x+2}wehavef\left(1\right)=\frac{1}{5}=\frac{1}{2}+\frac{1}{5}(-\frac{1}{2}+c)$$$$\Rightarrow 1=\frac{5}{2}-\frac{1}{2}+c\Rightarrow 1=2+c\Rightarrow c=-1and$$$${\int }_1^{+\mathrm{\infty }}\frac{dx}{x(x^2+2x+2)}={\int }_1^{+\mathrm{\infty }}(\frac{1}{2x}-\frac{1}{2}\frac{x+2}{x^2+2x+2})$$$$={\int }_1^{+\mathrm{\infty }}\frac{dx}{2x}-\frac{1}{4}{\int }_1^{+\mathrm{\infty }}\frac{2x+4}{x^2+2x+2}dx$$$$={\int }_1^{+\mathrm{\infty }}\frac{dx}{2x}-\frac{1}{4}{\int }_1^{+\mathrm{\infty }}\frac{2x+2}{x^2+2x+2}dx-\frac{1}{2}{\int }_1^{+\mathrm{\infty }}\frac{dx}{x^2+2x+2}$$$$={[\frac{1}{2}ln\mid x\mid -\frac{1}{4}ln\mid x^2+2x+2\mid ]}_1^{+\mathrm{\infty }}-\frac{1}{2}{\int }_1^{+\mathrm{\infty }}\frac{dx}{{(x+1)}^2+1}$$$$=[ln(\frac{\sqrt{\mid x\mid }}{\sqrt{x^2+2x+2}}4$$

Commented by abdo imad last updated on 04/Mar/18

$$=-ln\left({\left({}^4\sqrt{5}\right)}^{-1}\right)-\frac{1}{2}{\int }_2^{+\mathrm{\infty }}\frac{dt}{1+t^2}$$$$=\frac{1}{4}ln\left(5\right)-\frac{1}{2}{\left[arctant\right]}_2^{+\mathrm{\infty }}=\frac{1}{4}ln\left(5\right)-\frac{1}{2}(\frac{\pi }{2}-arctan2)$$$$=\frac{1}{4}ln\left(5\right)-\frac{\pi }{4}+\frac{1}{2}arctan2$$$${\int }_1^{\xi }\frac{arctanx}{x^2}={\left[\frac{-1}{x}arctanx\right]}_1^{\xi }-{\int }_1^{\xi }\frac{-1}{x}\frac{dx}{1+x^2}\to$$$$\frac{\pi }{4}+{\int }_1^{+\mathrm{\infty }}\frac{dx}{x(1+x^2)}=\frac{\pi }{4}+{\int }_1^{+\mathrm{\infty }}(\frac{1}{x}-\frac{x}{1+x^2})dx$$$$=\frac{\pi }{4}+{[ln\mid \frac{x}{\sqrt{1+x^2}}\mid ]}_1^{+\mathrm{\infty }}=\frac{\pi }{4}-ln\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi }{4}+ln\left(\sqrt{2}\right)\Rightarrow$$$$I=\frac{1}{4}ln\left(5\right)-\frac{\pi }{2}+\frac{1}{2}arctan2-ln\left(\sqrt{2}\right)$$

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