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

_(∗∗∗∗∗) :: Lobachevsky Integral ::_(∗∗∗∗∗) 𝛗:=∫_0 ^( ∞) ((sin^2 ( tan(x)))/x^( 2) )dx=^? (π/2) ..........

$$\:\:\:\:\:\:\:_{\ast\ast\ast\ast\ast} ::\:\:{Lobachevsky}\:{Integral}\:::_{\ast\ast\ast\ast\ast} \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}:=\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{s}{in}^{\mathrm{2}} \left(\:{tan}\left({x}\right)\right)}{{x}^{\:\mathrm{2}} }{dx}\overset{?} {=}\frac{\pi}{\mathrm{2}} \\ $$$$\:\:\:\:.......... \\ $$

Answered by Olaf_Thorendsen last updated on 09/Jun/21

Let f(x) = ((sin^2 (tanx))/(sin^2 x)) φ = ∫_0 ^∞ ((sin^2 (tanx))/x^2 ) dx φ = ∫_0 ^∞ ((sin^2 (x))/x^2 )f(x) dx ∀ x≥0 f is a continuous function satisfying the π−periodic assumption f(x+π) = f(x), and f(x−π) = f(x). We can apply the Lobatchevsky′s Dirichlet integral formula : ∫_0 ^∞ ((sin^2 x)/x^2 )f(x)dx = ∫_0 ^∞ ((sinx)/x)f(x)dx = ∫_0 ^(π/2) f(x)dx φ = ∫_0 ^(π/2) ((sin^2 (tanx))/(sin^2 x))dx Let u = tanx φ = ∫_0 ^∞ ((sin^2 u)/(sin^2 (arctanu))).(du/(1+u^2 )) φ = ∫_0 ^∞ ((sin^2 u)/(1−cos^2 (arctanu))).(du/(1+u^2 )) φ = ∫_0 ^∞ ((sin^2 u)/(1−(1/(1+tan^2 (arctanu))))).(du/(1+u^2 )) φ = ∫_0 ^∞ ((sin^2 u)/(1−(1/(1+u^2 )))).(du/(1+u^2 )) φ = ∫_0 ^∞ ((sin^2 u)/u^2 ) du Let g(u) = 1 (constant function unity) φ = ∫_0 ^∞ ((sin^2 u)/u^2 )g(u)du = ∫_0 ^∞ ((sinu)/u)g(u)du = ∫_0 ^(π/2) g(u)du φ = ∫_0 ^(π/2) 1.du = (π/2)

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

Commented by mnjuly1970 last updated on 09/Jun/21

thanks alot mr olaf....

$$\:\:{thanks}\:{alot}\:{mr}\:{olaf}.... \\ $$

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