I-0-pi-1-pi-2-x-2-1-1-x-2-lnx-dx-put-pix-tanA-x-tanB-I-0-pi-2-ln-tanA-lnpi-dA-0-pi-2-ln-tanB-dB-I-pi-2-lnpi-

Question Number 113766 by Riteshgoyal last updated on 15/Sep/20

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

p u t π x = t a n A, x = t a n B

I = \int_{0}^{\frac{π}{2}} (l n (t a n A) - l n π) d A - \int_{0}^{π / 2} l n (t a n B) d B

I = \frac{- π}{2} l n π

Answered by mathmax by abdo last updated on 16/Sep/20

I = \int_{0}^{\infty} (\frac{π}{1 + π^{2} x^{2}} - \frac{1}{1 + x^{2}}) lnx dx = π \int_{0}^{\infty} \frac{lnx}{1 + π^{2} x^{2}} - \int_{0}^{\infty} \frac{lnx}{1 + x^{2}} dx

\int_{0}^{\infty} \frac{lnx}{1 + x^{2}} dx = 0 (put x = \frac{1}{u})

π \int_{0}^{\infty} \frac{\ln (x)}{1 + {(π x)}^{2}} dx =_{π x = t} π \int_{0}^{\infty} \frac{\ln (\frac{t}{π})}{1 + t^{2}} \times \frac{dt}{π}

= \int_{0}^{\infty} \frac{lnt - \ln π}{1 + t^{2}} dt = \int_{0}^{\infty} \frac{lnt}{1 + t^{2}} dt - \ln (π) \times \frac{π}{2} = 0 - \frac{π}{2} \ln (π) \Rightarrow

I = - \frac{π}{2} \ln (π)