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Question Number 135099 by 0731619177 last updated on 10/Mar/21

Answered by Dwaipayan Shikari last updated on 10/Mar/21

$$I\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{f\left(ax\right)-f\left(bx\right)}{x}dx$$$$I^{\prime }\left(a\right)={\int }_0^{\mathrm{\infty }}{f}^{\prime }\left(ax\right)dx=\frac{1}{a}\underset{z\to \mathrm{\infty }}{\lim }(f\left(z\right)-f\left(0\right))$$$$I\left(a\right)=\underset{z\to \mathrm{\infty }}{\lim }(f\left(z\right)-f\left(0\right))log\left(a\right)+C$$$$I\left(b\right)=\underset{z\to \mathrm{\infty }}{\lim }(f\left(z\right)-f\left(0\right))log\left(b\right)+C=0\Rightarrow C=-log\left(b\right)(f\left(z\right)-f\left(0\right))$$$$I\left(a\right)=(f\left(z\right)-f\left(0\right))log\left(\frac{a}{b}\right)$$$$f\left(x\right)={tan}^{-1}x$$$${\int }_0^{\mathrm{\infty }}\frac{{tan}^{-1}ax-{tan}^{-1}bx}{x}dx=\underset{z\to \mathrm{\infty }}{\lim }({tan}^{-1}z-{tan}^{-1}0)log\left(\frac{a}{b}\right)$$$$=\frac{\pi }{2}log\left(\frac{a}{b}\right)$$

Answered by Ñï= last updated on 10/Mar/21

$${\int }_0^{\mathrm{\infty }}\frac{{\tan }^{-1}ax-{\tan }^{-1}bx}{x}dx={\int }_0^{\mathrm{\infty }}{\int }_b^a\frac{1}{1+y^2{x}^2}dydx$$$$={\int }_b^a{\int }_0^{\mathrm{\infty }}\frac{1}{1+y^2{x}^2}dxdy={\int }_b^a\frac{1}{y}{tan}^{-1}yx{\mid }_0^{\mathrm{\infty }}dy=\frac{\pi }{2}{\int }_b^a\frac{1}{y}dy$$$$=\frac{\pi }{2}ln\frac{a}{b}$$

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