Question Number 128707 by mnjuly1970 last updated on 09/Jan/21
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\mathrm{nice}\:\:\mathrm{calculus}… \\ $$$${prove}\:\:{that}::\: \\ $$$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\left(\mathrm{1}+\varphi^{\mathrm{2}} {x}^{\mathrm{2}} \right)}{\mathrm{1}+\pi^{\mathrm{2}} {x}^{\mathrm{2}} }\:{dx}={ln}\left(\frac{\pi+\varphi}{\pi}\right) \\ $$$$\varphi::=\:\:{golen}\:{ratio}… \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 09/Jan/21
$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{log}\left(\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)}{\mathrm{1}+\pi^{\mathrm{2}} {x}^{\mathrm{2}} }{dx}\:\Rightarrow{I}'\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{ax}^{\mathrm{2}} }{\left(\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)\left(\mathrm{1}+\pi^{\mathrm{2}} {x}^{\mathrm{2}} \right)}{dx} \\ $$$$=\frac{\mathrm{2}{a}}{{a}^{\mathrm{2}} −\pi^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+\pi^{\mathrm{2}} {x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} }{dx} \\ $$$$=\frac{\mathrm{2}{a}}{{a}^{\mathrm{2}} −\pi^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\pi}.\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{{a}}.\frac{\pi}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\pi+{a}} \\ $$$${I}\left({a}\right)={log}\left(\pi+{a}\right)+{C} \\ $$$${C}=−{log}\left(\pi\right)\:\:\:\:{when}\:{a}=\mathrm{0} \\ $$$${I}\left({a}\right)={log}\left(\mathrm{1}+\frac{{a}}{\pi}\right)\:\Rightarrow{I}\left(\varphi\right)={log}\left(\mathrm{1}+\frac{\varphi}{\pi}\right) \\ $$
Commented by mnjuly1970 last updated on 09/Jan/21
$$\:{short}\:{solution}\:{but} \\ $$$${very}\:{nice}.{thank}\:{you}… \\ $$
Commented by Dwaipayan Shikari last updated on 09/Jan/21
$${With}\:{pleasure}\:{sir}! \\ $$