prove-that-0-ln-x-x-2-1-dx-0-

Question Number 140310 by liberty last updated on 06/May/21

prove that \int_{0}^{\infty} \frac{\ln x}{x^{2} + 1} dx = 0

Answered by qaz last updated on 06/May/21

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

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

Commented by TheSupreme last updated on 06/May/21

t = \frac{1}{x}

d x = - \frac{1}{t^{2}} d t

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

I = - I = 0

Answered by mathmax by abdo last updated on 06/May/21

\int_{0}^{\infty} \frac{lnx}{x^{2} + 1} dx = - \frac{1}{2} Re (Σ Res (f a_{i})) with f (z) = \frac{\log^{2} z}{z^{2} + 1}

f (z) = \frac{\log^{2} z}{(z - i) (z + i)} \Rightarrow Res (f, i) = \frac{\log^{2} i}{2 i} = \frac{{(\log (e^{\frac{i π}{2}}))}^{2}}{2 i} = \frac{{(\frac{i π}{2})}^{2}}{2 i} = \frac{- π^{2}}{8 i}

Res (f, - i) = \frac{\log^{2} (- i)}{2 i} = \frac{\log^{2} (e^{- \frac{i π}{2}})}{2 i} = - \frac{π^{2}}{8 i} \Rightarrow

Σ Res = \frac{i π^{2}}{8} + \frac{i π^{2}}{8} = \frac{i π^{2}}{4} \Rightarrow Re (Σ Res) = 0 \Rightarrow \int_{0}^{\infty} \frac{logx}{x^{2} + 1} dx = 0