nice-calculus-prove-that-i-dx-1-x-e-x-2-pi-2-2-3-ii-0-sin-tan-x-x-dx-pi-2-1-1-e-

Question Number 136868 by mnjuly1970 last updated on 27/Mar/21

\dots \dots n i c e c a l c u l u s \dots .

p r o v e t h a t ::

i : \int_{- \infty}^{\infty} \frac{d x}{{(1 + x + e^{x})}^{2} + π^{2}} = \frac{2}{3}

i i : \int_{0}^{\infty} \frac{s i n (t a n (x))}{x} d x = \frac{π}{2} (1 - \frac{1}{e})

Answered by Ñï= last updated on 27/Mar/21

I = \int_{0}^{\infty} \frac{\sin (\tan x)}{x} d x = \underset{n = 0}{\sum^{\infty}} \int_{n \frac{π}{2}}^{(n + 1) \frac{π}{2}} \frac{\sin (\tan x)}{x} d x ? ? ? ?

= \underset{n = 0}{\sum^{\infty}} \int_{n π}^{(2 n + 1) \frac{π}{2}} \frac{\sin (\tan x)}{x} d x + \underset{n = 0}{\sum^{\infty}} \int_{(2 n - 1) \frac{π}{2}}^{n π} \frac{\sin (\tan x)}{x} d x ? ? ? ?

= \underset{n = 0}{\sum^{\infty}} \int_{0}^{π / 2} \frac{\sin (\tan x)}{n π + x} d x - \underset{n = 0}{\sum^{\infty}} \int_{0}^{π / 2} \frac{\sin (\tan x)}{n π - x} d x

= \int_{0}^{π / 2} \underset{n = 0}{\sum^{\infty}} \frac{2 x}{x^{2} - n^{2} π^{2}} \sin (\tan x) d x

= \int_{0}^{π / 2} \cot x \sin (\tan x) d x

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

= \int_{0}^{1} d a \int_{0}^{\infty} \frac{\cos (a u)}{1 + u^{2}} d u

= \frac{π}{2} \int_{0}^{1} e^{- a} d a

= \frac{π}{2} (1 - e^{- 1})

Commented by mnjuly1970 last updated on 27/Mar/21

v e r y n i c e . .

t h a n k y o u s o m u c h \dots