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Question Number 143051 by mnjuly1970 last updated on 09/Jun/21

$${}_{\ast \ast \ast \ast \ast }::LobachevskyIntegral:{:}_{\ast \ast \ast \ast \ast }$$$$\mathit{\varphi }:={\int }_0^{\mathrm{\infty }}\frac{\mathrm{s}{in}^2\left(tan\left(x\right)\right)}{x^2}dx\stackrel{?}{=}\frac{\pi }{2}$$$$..........$$

Answered by Olaf_Thorendsen last updated on 09/Jun/21

$$\mathrm{Let}f\left(x\right)=\frac{{\sin }^2\left(\tan x\right)}{{\sin }^{2}x}$$$$\varphi ={\int }_0^{\mathrm{\infty }}\frac{{\sin }^2\left(\tan x\right)}{x^2}dx$$$$\varphi ={\int }_0^{\mathrm{\infty }}\frac{{\sin }^2\left(x\right)}{x^2}f\left(x\right)dx$$$$\mathrm{\forall }x⩾0f\mathrm{is}\mathrm{a}\mathrm{continuous}\mathrm{function}$$$$\mathrm{satisfying}\mathrm{the}\pi -\mathrm{periodic}\mathrm{assumption}$$$$f(x+\pi )=f\left(x\right),\mathrm{and}f(x-\pi )=f\left(x\right).$$$$$$$$\mathrm{We}\mathrm{can}\mathrm{apply}\mathrm{the}{\mathrm{Lobatchevsky}}^{\prime }\mathrm{s}$$$$\mathrm{Dirichlet}\mathrm{integral}\mathrm{formula}:$$$${\int }_0^{\mathrm{\infty }}\frac{{\sin }^{2}x}{x^2}f\left(x\right)dx={\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}f\left(x\right)dx={\int }_0^{\frac{\pi }{2}}f\left(x\right)dx$$$$\varphi ={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^2\left(\tan x\right)}{{\sin }^{2}x}dx$$$$\mathrm{Let}u=\tan x$$$$\varphi ={\int }_0^{\mathrm{\infty }}\frac{{\sin }^{2}u}{{\sin }^2\left(\arctan u\right)}.\frac{du}{1+u^2}$$$$\varphi ={\int }_0^{\mathrm{\infty }}\frac{{\sin }^{2}u}{1-{\cos }^2\left(\arctan u\right)}.\frac{du}{1+u^2}$$$$\varphi ={\int }_0^{\mathrm{\infty }}\frac{{\sin }^{2}u}{1-\frac{1}{1+{\tan }^2\left(\arctan u\right)}}.\frac{du}{1+u^2}$$$$\varphi ={\int }_0^{\mathrm{\infty }}\frac{{\sin }^{2}u}{1-\frac{1}{1+u^2}}.\frac{du}{1+u^2}$$$$\varphi ={\int }_0^{\mathrm{\infty }}\frac{{\sin }^{2}u}{u^2}du$$$$\mathrm{Let}g\left(u\right)=1\left(\mathrm{constant}\mathrm{function}\mathrm{unity}\right)$$$$\varphi ={\int }_0^{\mathrm{\infty }}\frac{{\sin }^{2}u}{u^2}g\left(u\right)du={\int }_0^{\mathrm{\infty }}\frac{\sin u}{u}g\left(u\right)du={\int }_0^{\frac{\pi }{2}}g\left(u\right)du$$$$\varphi ={\int }_0^{\frac{\pi }{2}}1.du=\frac{\pi }{2}$$

Commented by mnjuly1970 last updated on 09/Jun/21

$$thanksalotmrolaf....$$

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