nice-integral-T-0-arctan-x-x-ln-x-1-dx-pi-pi-4-

Question Number 143454 by mnjuly1970 last updated on 14/Jun/21

\dots \dots . . n i c e \dots \dots . i n t e g r a l \dots \dots .

T := \int_{0}^{\infty} \frac{a r c t a n (x)}{x^{l n (x) + 1}} d x \overset{?}{=} \frac{π \sqrt{π}}{4}

Answered by mindispower last updated on 14/Jun/21

x \to \frac{1}{x} \Rightarrow

A = \int_{0}^{\infty} \frac{\frac{π}{2} - a r c t a n (x)}{{(\frac{1}{x})}^{- l n (x) + 1}} . \frac{d x}{x^{2}} = \frac{π}{2} \int_{0}^{\infty} \frac{d x}{x^{l n (x) + 1}} - \int_{0}^{\infty} \frac{a r c t a n (x) d x}{x^{l n (x) + 1}}

2 A = \frac{π}{2} \int_{0}^{\infty} \frac{d x}{x^{l n (x) + 1}}

l n (x) = u

A = \frac{π}{4} \int_{- \infty}^{\infty} \frac{d u}{e^{u^{2}}} = \frac{π}{2} \int_{0}^{\infty} e^{- u^{2}} d u = \frac{π}{4} \int_{0}^{\infty} e^{- t} . t^{- \frac{1}{2}} d t

= \frac{π}{4} Γ (\frac{1}{2}) = \frac{π}{4} \sqrt{π}

\int_{0}^{\infty} \frac{a r c t a n (x)}{x^{l n (x) + 1}} d x = \frac{π \sqrt{π}}{4}

Commented by mnjuly1970 last updated on 14/Jun/21

t h a n k d a l o t . .

Commented by mindispower last updated on 14/Jun/21

p l e a s u r