prove-0-ln-1-x-2-x-2-1-x-2-dx-pi-ln-e-2-

Question Number 151230 by mnjuly1970 last updated on 19/Aug/21

p r o v e :

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

Commented by Lordose last updated on 19/Aug/21

Commented by Lordose last updated on 19/Aug/21

Answered by Lordose last updated on 19/Aug/21

Ω = \int_{0}^{1} \frac{\ln (1 + x^{2})}{x^{2} (1 + x^{2})} dx \overset{P . F}{=} \int_{0}^{1} \frac{\ln (1 + x^{2})}{x^{2}} dx - \int_{0}^{1} \frac{\ln (1 + x^{2})}{1 + x^{2}} dx

Ω = Φ - Ψ

Φ \overset{IBP}{=} - \frac{\ln (1 + x^{2})}{x} ∣_{0^{+}}^{1} + \int_{0}^{1} \frac{2}{1 + x^{2}} dx

Φ = - \log (2) + \frac{π}{2}

Ψ = \int_{0}^{1} \frac{\ln (1 + x^{2})}{1 + x^{2}} dx \overset{x = \tan (y)}{=} - 2 \int_{0}^{\frac{π}{4}} \ln (\cos (y)) dy

Ψ = 2 \int_{0}^{\frac{π}{4}} (\underset{k = 1}{\sum^{\infty}} \frac{{(- 1)}^{k}}{k} \cos (2 ky) + \log (2)) dy = 2 \underset{k = 1}{\sum^{\infty}} \frac{{(- 1)}^{k}}{k} \int_{0}^{\frac{π}{4}} \cos (2 ky) dy + 2 \log (2) \int_{0}^{\frac{π}{4}} 1 dy

Ψ = \underset{k = 1}{\sum^{\infty}} \frac{{(- 1)}^{k} \sin (\frac{π k}{2})}{k^{2}} + \frac{π \log (2)}{2}

Ψ = \frac{π}{2} \log (2) - G

Ω = \frac{π}{2} - \log (2) - \frac{π}{2} \log (2) + G

Ω = \frac{π}{2} - \log (2) (1 + \frac{π}{2}) + G

G = {Catalan}^{'} s Constant = \underset{k = 1}{\sum^{\infty}} \frac{{(- 1)}^{k} \sin (\frac{π k}{2})}{k^{2}}

Commented by mnjuly1970 last updated on 19/Aug/21

t h a n k y o u s o m u c h m r l o r d o s e

m y m i s t a k e Ω = \int_{0}^{\infty} \frac{l n (1 + x^{2})}{x^{2} (1 + x^{2})} d x

g r a t e f u l \dots

Answered by qaz last updated on 19/Aug/21

\int_{0}^{1} \frac{\ln (1 + x^{2})}{x^{2} (1 + x^{2})} dx

= \int_{0}^{1} \frac{\ln (1 + x^{2})}{x^{2}} dx - \int_{0}^{1} \frac{\ln (1 + x^{2})}{1 + x^{2}} dx

= - \frac{\ln (1 + x^{2})}{x} ∣_{0}^{1} + \int_{0}^{1} \frac{2}{1 + x^{2}} dx + 2 \int_{0}^{π / 4} lncos xdx

= - \ln 2 + \frac{π}{2} + \int_{0}^{π / 2} lncos \frac{x}{2} dx

= - \ln 2 + \frac{π}{2} + \frac{1}{2} \int_{0}^{π / 2} \ln \frac{1 + \cos x}{2} dx

= - \ln 2 + \frac{π}{2} + \frac{1}{2} \int_{0}^{π / 2} \ln \frac{\sin x}{\tan \frac{x}{2}} dx - \frac{1}{2} \int_{0}^{π / 2} \ln 2 dx

= - \ln 2 + \frac{π}{2} - \frac{π}{2} \ln 2 - \int_{0}^{π / 4} lntan xdx

= - \ln 2 + \frac{π}{2} - \frac{π}{2} \ln 2 - \int_{0}^{1} \frac{lnx}{1 + x^{2}} dx

= - \ln 2 + \frac{π}{2} - \frac{π}{2} \ln 2 - \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \int_{0}^{1} x^{2 n} lnxdx

= - \ln 2 + \frac{π}{2} - \frac{π}{2} \ln 2 - \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{{(2 n + 1)}^{2}}

= - \ln 2 + \frac{π}{2} - \frac{π}{2} \ln 2 + G

Answered by Lordose last updated on 19/Aug/21

Ω = \int_{0}^{\infty} \frac{\ln (1 + x^{2})}{x^{2} (1 + x^{2})} dx \overset{x = \tan (y)}{=} - 2 \int_{0}^{\frac{π}{2}} \cot^{2} yln (\cos (y)) dy

Ω = - 2 \frac{\partial}{\partial a} ∣_{a = 2} \int_{0}^{\frac{π}{2}} \cos^{a} (y) \sin^{- 2} (y) dy

Ω = - \frac{\partial}{\partial a} ∣_{a = 2} (B (\frac{a + 1}{2}, - \frac{1}{2})) = - \frac{1}{2} (ψ^{(0)} (\frac{a + 1}{2}) - ψ^{(0)} (\frac{a}{2})) B (\frac{a + 1}{2}, - \frac{1}{2})

Ω = \frac{1}{2} π (γ + ψ^{(0)} (\frac{3}{2}))

Answered by qaz last updated on 19/Aug/21

\int_{0}^{\infty} \frac{\ln (1 + x^{2})}{x^{2} (1 + x^{2})} dx

= \int_{0}^{π / 2} \frac{lnsec^{2} x}{\tan^{2} x} dx

= - 2 \int_{0}^{π / 2} \cot^{2} xlncos xdx

= 2 \int_{0}^{π / 2} (1 - \csc^{2} x) lncos xdx

= - π \ln 2 - 2 \int_{0}^{π / 2} \csc^{2} xlncos xdx

= - π \ln 2 + 2 \cot xlncos x ∣_{0}^{π / 2} + 2 \int_{0}^{π / 2} dx

= - π \ln 2 + π

Commented by mnjuly1970 last updated on 19/Aug/21

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

Commented by peter frank last updated on 19/Aug/21

thank you