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Question Number 75793 by ~blr237~ last updated on 17/Dec/19

$$\mathrm{Prove}\mathrm{that}{\int }_0^{\mathrm{\infty }}{\left(\frac{\mathrm{arctanx}}{\mathrm{x}\sqrt{\log 2}}\right)}^2\mathrm{dx}=\pi$$

Commented by mathmax by abdo last updated on 18/Dec/19

$$letI={\int }_0^{\mathrm{\infty }}{\left(\frac{arctanx}{x\sqrt{ln\left(2\right)}}\right)}^{2}dx\Rightarrow ln\left(2\right)I={\int }_0^{\mathrm{\infty }}\frac{{arctan}^{2}x}{x^2}dxbyparts$$$$u^{{}^{\prime }}=\frac{1}{x^2}andv={arctan}^2\left(x\right)\Rightarrow ln\left(2\right)I={[-\frac{1}{x}{arctan}^2\left(x\right)]}_0^{+\mathrm{\infty }}$$$$-{\int }_0^{\mathrm{\infty }}(-\frac{1}{x})\frac{2arctan\left(x\right)}{1+x^2}dx=2{\int }_0^{\mathrm{\infty }}\frac{arcctan\left(x\right)}{x(1+x^2)}dx\Rightarrow$$$$I=\frac{2}{ln\left(2\right)}{\int }_0^{\mathrm{\infty }}\frac{arctan\left(x\right)}{x(1+x^2)}dx{=}_{arctanx=u}\frac{2}{ln\left(2\right)}{\int }_0^{\frac{\pi }{2}}\frac{u}{tan\left(u\right)(1+{tan}^{2}u)}(1+{tan}^{2}u)du$$$$=\frac{2}{ln\left(2\right)}{\int }_0^{\frac{\pi }{2}}\frac{u}{tanu}du=\frac{2}{ln\left(2\right)}{\int }_0^{\frac{\pi }{2}}u\frac{cosu}{sinu}dubypartsf=uand{g}^{{}^{\prime }}=\frac{cosu}{sinu}$$$$\frac{ln\left(2\right)}{2}I={\int }_0^{\frac{\pi }{2}}u\frac{cosu}{sinu}du={\left[uln\left(sinu\right)\right]}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}ln\left(sinu\right)du=-(-\frac{\pi }{2}ln\left(2\right))$$$$\frac{ln\left(2\right)}{2}I=\frac{\pi ln\left(2\right)}{2}\Rightarrow I=\pi$$

Answered by mind is power last updated on 17/Dec/19

$$=\left[\frac{-{\arctan }^2\left(\mathrm{x}\right)}{\mathrm{xlog}\left(2\right)}\right]+\frac{1}{\log \left(2\right)}{\int }_0^{+\mathrm{\infty }}\frac{2\arctan \left(\mathrm{x}\right)}{\mathrm{x}({\mathrm{x}}^2+1)}\mathrm{dx}$$$$=\frac{2}{\log \left(2\right)}{\int }_0^{+\mathrm{\infty }}\frac{\mathrm{arcran}\left(\mathrm{x}\right)}{\mathrm{x}({\mathrm{x}}^2+1)}\mathrm{dx}$$$$\mathrm{u}=\mathrm{arcran}\left(\mathrm{x}\right)\Rightarrow \mathrm{dx}=(1+{\mathrm{tg}}^2\left(\mathrm{u}\right))\mathrm{du}$$$$\iff \frac{2}{\log \left(2\right)}{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{u}}{\mathrm{tg}\left(\mathrm{u}\right)}\mathrm{du}=\frac{2}{\log \left(2\right)}{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{ucos}\left(\mathrm{u}\right)}{\sin \left(\mathrm{u}\right)}\mathrm{du}$$$$\mathrm{by}\mathrm{part}\iff$$$$\frac{2}{\log \left(2\right)}{\left[\mathrm{ulog}\left(\sin \left(\mathrm{u}\right)\right)\right]}_0^{\frac{\pi }{2}}-\frac{2}{\log \left(2\right)}{\int }_0^{\frac{\pi }{2}}\log \left(\sin \left(\mathrm{u}\right)\right)\mathrm{du}$$$$=-\frac{2}{\log \left(2\right)}{\int }_0^{\frac{\pi }{2}}\log \left(\sin \left(\mathrm{u}\right)\right)\mathrm{du}$$$${\int }_0^{\frac{\pi }{2}}\log \left(\sin \left(\mathrm{u}\right)\right)\mathrm{du}=-\log \left(2\right)\frac{\pi }{2}\mathrm{done}\mathrm{many}\mathrm{Times}$$$$\mathrm{so}\mathrm{We}\mathrm{get}-\frac{2}{\log \left(2\right)}.\frac{-\log \left(2\right)}{2}.\pi =\pi$$

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