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Question Number 128707 by mnjuly1970 last updated on 09/Jan/21

$$...\mathrm{nice}\mathrm{calculus}...$$$$provethat::$$$${\int }_0^{\mathrm{\infty }}\frac{ln(1+{\phi }^2{x}^2)}{1+{\pi }^2{x}^2}dx=ln\left(\frac{\pi +\phi }{\pi }\right)$$$$\phi ::=golenratio...$$$$$$

Answered by Dwaipayan Shikari last updated on 09/Jan/21

$$I\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{log(1+a^2{x}^2)}{1+{\pi }^2{x}^2}dx\Rightarrow I^{\prime }\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{2{ax}^2}{(1+a^2{x}^2)(1+{\pi }^2{x}^2)}dx$$$$=\frac{2a}{a^2-{\pi }^2}{\int }_0^{\mathrm{\infty }}\frac{1}{1+{\pi }^2{x}^2}-\frac{1}{1+a^2{x}^2}dx$$$$=\frac{2a}{a^2-{\pi }^2}(\frac{1}{\pi }.\frac{\pi }{2}-\frac{1}{a}.\frac{\pi }{2})=\frac{1}{\pi +a}$$$$I\left(a\right)=log(\pi +a)+C$$$$C=-log\left(\pi \right)whena=0$$$$I\left(a\right)=log(1+\frac{a}{\pi })\Rightarrow I\left(\phi \right)=log(1+\frac{\phi }{\pi })$$

Commented by mnjuly1970 last updated on 09/Jan/21

$$shortsolutionbut$$$$verynice.thankyou...$$

Commented by Dwaipayan Shikari last updated on 09/Jan/21

$$Withpleasuresir!$$

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