Lobachevsky-Integral-0-sin-2-tan-x-x-2-dx-pi-2-

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

_{* * * * *} :: L o b a c h e v s k y I n t e g r a l : :_{* * * * *}

\boldsymbol ϕ := \int_{0}^{\infty} \frac{s {i n}^{2} (t a n (x))}{x^{2}} d x \overset{?}{=} \frac{π}{2}

\dots \dots \dots .

Answered by Olaf_Thorendsen last updated on 09/Jun/21

Let f (x) = \frac{\sin^{2} (\tan x)}{\sin^{2} x}

ϕ = \int_{0}^{\infty} \frac{\sin^{2} (\tan x)}{x^{2}} d x

ϕ = \int_{0}^{\infty} \frac{\sin^{2} (x)}{x^{2}} f (x) d x

\forall 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 :

\int_{0}^{\infty} \frac{\sin^{2} x}{x^{2}} f (x) d x = \int_{0}^{\infty} \frac{\sin x}{x} f (x) d x = \int_{0}^{\frac{π}{2}} f (x) d x

ϕ = \int_{0}^{\frac{π}{2}} \frac{\sin^{2} (\tan x)}{\sin^{2} x} d x

Let u = \tan x

ϕ = \int_{0}^{\infty} \frac{\sin^{2} u}{\sin^{2} (\arctan u)} . \frac{d u}{1 + u^{2}}

ϕ = \int_{0}^{\infty} \frac{\sin^{2} u}{1 - \cos^{2} (\arctan u)} . \frac{d u}{1 + u^{2}}

ϕ = \int_{0}^{\infty} \frac{\sin^{2} u}{1 - \frac{1}{1 + \tan^{2} (\arctan u)}} . \frac{d u}{1 + u^{2}}

ϕ = \int_{0}^{\infty} \frac{\sin^{2} u}{1 - \frac{1}{1 + u^{2}}} . \frac{d u}{1 + u^{2}}

ϕ = \int_{0}^{\infty} \frac{\sin^{2} u}{u^{2}} d u

Let g (u) = 1 (constant function unity)

ϕ = \int_{0}^{\infty} \frac{\sin^{2} u}{u^{2}} g (u) d u = \int_{0}^{\infty} \frac{\sin u}{u} g (u) d u = \int_{0}^{\frac{π}{2}} g (u) d u

ϕ = \int_{0}^{\frac{π}{2}} 1 . d u = \frac{π}{2}

Commented by mnjuly1970 last updated on 09/Jun/21

t h a n k s a l o t m r o l a f \dots .

t h a n k s a l o t m r o l a f \dots .