∫_0^∞ #160; log x1 + x^2dx = ^ ln( x )/( x i)( x + i) 1mudx = ^ ( iln( x )2( x + i) iln( x )2( x i))dx = ^ ln( iu + 1)/u 1mudu + ln( ...

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Integration Using Substitution

∫ 0 ∞log x 1+...

Question

$\int_{0}^{\infty} \frac{log x}{1 + x^{2}} d x$

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Solution

$\int_{0}^{\infty} \frac{log x}{1 + x^{2}} d x$
$= \int^{\frac{ln (x)}{(x - i) (x + i)}} d x$
$= \int^{⎛ ⎜ ⎝ \frac{i ln (x)}{2 (x + i)} - \frac{i ln (x)}{2 (x - i)} ⎞ ⎟ ⎠} d x$
$= \int^{\frac{ln (i u + 1)}{u}} d u + ln (- i) \int^{\frac{1}{u}} d u$
$= - \int^{-} \frac{ln (1 - v)}{v} d v$
$\int^{\frac{ln (i u + 1)}{u}} d u + ln (- i) \int^{\frac{1}{u}} d u$
$= \int^{⎛ ⎜ ⎝ \frac{ln (1 - i u)}{u} + \frac{ln (i)}{u} ⎞ ⎟ ⎠} d u$
$\frac{i}{2} \int^{\frac{ln (x)}{x + i}} d x - \frac{i}{2} \int^{\frac{ln (x)}{x - i}} d x$
$= \frac{i ln (- i) ln (| x + i |)}{2} - \frac{i ln (i) ln (| x - i |)}{2} - \frac{i L i_{2} (- i (x + i))}{2} + \frac{i L i_{2} (i (x - i))}{2} + C$
= $0$

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