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Question Number 36429 by prof Abdo imad last updated on 02/Jun/18

$$let\phi \left(\lambda \right)={\int }_{\frac{\lambda }{\pi }}^{\frac{\pi }{\lambda }}(1+\frac{1}{x^2})arctan\left(x\right)dxwith\lambda >0$$$$1)findasimpleformof\phi \left(\lambda \right)$$$$2)calculate{\phi }^{{}^{\prime }}\left(\lambda \right).$$

Commented by abdo mathsup 649 cc last updated on 03/Jun/18

$$letintegratebyparts{u}^{{}^{\prime }}=1+\frac{1}{x^2}and$$$$v^{{}^{\prime }}=arctan\left(x\right)$$$$\phi \left(\lambda \right)={\left[(1-\frac{1}{x})arctanx\right]}_{\frac{\lambda }{\pi }}^{\frac{\pi }{\lambda }}$$$$-{\int }_{\frac{\lambda }{\pi }}^{\frac{\pi }{\lambda }}(1-\frac{1}{x})\frac{dx}{1+x^2}$$$$=(1-\frac{\lambda }{\pi })arctan\left(\frac{\pi }{\lambda }\right)-(1-\frac{\pi }{\lambda })arctan\left(\frac{\lambda }{\pi }\right)$$$$-{\int }_{\frac{\lambda }{\pi }}^{\frac{\pi }{\lambda }}\frac{(x-1)dx}{x(1+x^2)}but$$$${\int }_{\frac{\lambda }{\pi }}^{\frac{\pi }{\lambda }}\frac{x-1}{x(1+x^2)}dx={\int }_{\frac{\lambda }{\pi }}^{\frac{\pi }{\lambda }}\frac{dx}{1+x^2}-{\int }_{\frac{\lambda }{\pi }}^{\frac{\pi }{\lambda }}\frac{dx}{x(1+x^2)}$$$$=arctan\left(\frac{\pi }{\lambda }\right)-arctan\left(\frac{\lambda }{\pi }\right)-{\int }_{\frac{\lambda }{\pi }}^{\frac{\pi }{\lambda }}\frac{dx}{x(1+x^2)}but$$$$F\left(x\right)=\frac{1}{x(1+x^2)}=\frac{a}{x}+\frac{bx+c}{x^2+1}$$$$a=1and{lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=a+b\Rightarrow b=-1$$$$c=0\Rightarrow {\int }_{\frac{\lambda }{\pi }}^{\frac{\pi }{\lambda }}\frac{dx}{x(1+x^2)}={\int }_{\frac{\lambda }{\pi }}^{\frac{\pi }{\lambda }}\{\frac{1}{x}-\frac{x}{1+x^2}\}dx$$$$={[ln\mid x\mid -\frac{1}{2}ln(1+x^2)]}_{\frac{\lambda }{\pi }}^{\frac{\pi }{\lambda }}$$$$={[ln\mid \frac{x}{\sqrt{1+x^2}}\mid ]}_{\frac{\lambda }{\pi }}^{\frac{\pi }{\lambda }}=ln\mid \frac{\pi }{\lambda \sqrt{1+{\left(\frac{\pi }{\lambda }\right)}^2}}\mid -ln\mid \frac{\lambda }{\pi \sqrt{1+{\left(\frac{\lambda }{\pi }\right)}^2}}$$$$\phi \left(\lambda \right)=(1-\frac{\lambda }{\pi })arctan\left(\frac{\pi }{\lambda }\right)-(1-\frac{\pi }{\lambda })arctan\left(\frac{\lambda }{\pi }\right)$$$$-ln\mid \frac{\pi }{\lambda \sqrt{1+{\left(\frac{\pi }{\lambda }\right)}^2}}\mid +ln\mid \frac{\lambda }{\pi \sqrt{1+{\left(\frac{\lambda }{\pi }\right)}^2}}\mid .$$

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jun/18

$${\int }_{\frac{\lambda }{\mathrm{\Pi }}}^{\frac{\mathrm{\Pi }}{\lambda }}\left(\frac{1+x^2}{x^2}\right){tan}^{-1}xdx$$$$I=\int (1+\frac{1}{x^2}){tan}^{-1}xdx$$$$={tan}^{-1}x(x+\frac{-1}{x})-\int \frac{1}{1+x^2}\times \left(\frac{x^2-1}{x}\right)dx$$$$=\left(\frac{x^2-1}{x}\right){tan}^{-1}x-\int \frac{x^2-1}{x(x^2+1)}$$$$=\left(\frac{x^2-1}{x}\right){tan}^{-1}x-\int \frac{x^2+1-2}{x(x^2+1)}dx$$$$=\left(\frac{x^2-1}{x}\right){tan}^{-1}x-\int \frac{dx}{x}+2\int \frac{xdx}{x^2(x^2+1)}$$$$=\left(\frac{x^2-1}{x}\right){tan}^{-1}x-lnx+\int \frac{dt}{t(t+1)}$$$$=\left(\frac{x^2-1}{x}\right){tan}^{-1}x-lnx+\int \frac{t+1-t}{(t+1)t}dt$$$$=\left(\frac{x^2-1}{x}\right){tan}^{-1}x-lnx+ln\left(\frac{t}{t+1}\right)$$$$=\left(\frac{x^2-1}{x}\right){tan}^{-1}x-lnx+ln\left(\frac{x^2}{x^2+1}\right)$$$$nowputupperandlowerlimit$$

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