Prove-that-0-arctanx-x-log2-2-dx-pi-

Question Number 75793 by ~blr237~ last updated on 17/Dec/19

Prove that \int_{0}^{\infty} {(\frac{arctanx}{x \sqrt{\log 2}})}^{2} dx = π

Commented by mathmax by abdo last updated on 18/Dec/19

l e t I = \int_{0}^{\infty} {(\frac{a r c t a n x}{x \sqrt{l n (2)}})}^{2} d x \Rightarrow l n (2) I = \int_{0}^{\infty} \frac{{a r c t a n}^{2} x}{x^{2}} d x b y p a r t s

u^{^{'}} = \frac{1}{x^{2}} a n d v = {a r c t a n}^{2} (x) \Rightarrow l n (2) I = {[- \frac{1}{x} {a r c t a n}^{2} (x)]}_{0}^{+ \infty}

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

I = \frac{2}{l n (2)} \int_{0}^{\infty} \frac{a r c t a n (x)}{x (1 + x^{2})} d x =_{a r c t a n x = u} \frac{2}{l n (2)} \int_{0}^{\frac{π}{2}} \frac{u}{t a n (u) (1 + {t a n}^{2} u)} (1 + {t a n}^{2} u) d u

= \frac{2}{l n (2)} \int_{0}^{\frac{π}{2}} \frac{u}{t a n u} d u = \frac{2}{l n (2)} \int_{0}^{\frac{π}{2}} u \frac{c o s u}{s i n u} d u b y p a r t s f = u a n d g^{^{'}} = \frac{c o s u}{s i n u}

\frac{l n (2)}{2} I = \int_{0}^{\frac{π}{2}} u \frac{c o s u}{s i n u} d u = {[u l n (s i n u)]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} l n (s i n u) d u = - (- \frac{π}{2} l n (2))

\frac{l n (2)}{2} I = \frac{π l n (2)}{2} \Rightarrow I = π

Answered by mind is power last updated on 17/Dec/19

= [\frac{- \arctan^{2} (x)}{xlog (2)}] + \frac{1}{\log (2)} \int_{0}^{+ \infty} \frac{2 \arctan (x)}{x (x^{2} + 1)} dx

= \frac{2}{\log (2)} \int_{0}^{+ \infty} \frac{arcran (x)}{x (x^{2} + 1)} dx

u = arcran (x) \Rightarrow dx = (1 + {tg}^{2} (u)) du

\Leftrightarrow \frac{2}{\log (2)} \int_{0}^{\frac{π}{2}} \frac{u}{tg (u)} du = \frac{2}{\log (2)} \int_{0}^{\frac{π}{2}} \frac{ucos (u)}{\sin (u)} du

by part \Leftrightarrow

\frac{2}{\log (2)} {[ulog (\sin (u))]}_{0}^{\frac{π}{2}} - \frac{2}{\log (2)} \int_{0}^{\frac{π}{2}} \log (\sin (u)) du

= - \frac{2}{\log (2)} \int_{0}^{\frac{π}{2}} \log (\sin (u)) du

\int_{0}^{\frac{π}{2}} \log (\sin (u)) du = - \log (2) \frac{π}{2} done many Times

so We get - \frac{2}{\log (2)} . \frac{- \log (2)}{2} . π = π