# Evaluating improper integral <msubsup> &#x222B;<!-- ∫ --> 0 1 </msubsup>

Evaluating improper integral ${\int }_{0}^{1}\frac{\mathrm{log}\left(x\right)}{x+\alpha }\phantom{\rule{thickmathspace}{0ex}}dx$ for small positive $\alpha$
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alomjabpdl0
${\int }_{0}^{1}\frac{\mathrm{ln}\left(x\right)}{a+x}dx={\int }_{0}^{1/a}\frac{\mathrm{ln}\left(ay\right)}{1+y}dy$
$=\underset{0}{\underset{⏟}{\mathrm{ln}\left(1+y\right)\mathrm{ln}\left(ay\right){|}_{0}^{1/a}}}-{\int }_{0}^{1/a}\frac{\mathrm{ln}\left(1+y\right)}{y}dx={\text{Li}}_{2}\left(-y\right){|}_{0}^{1/a}={\text{Li}}_{2}\left(-1/a\right)$
and the result follows on using the dilogarithm inversion formula:
${\text{Li}}_{2}\left(-1/z\right)=-\frac{{\pi }^{2}}{6}-\frac{1}{2}{\mathrm{ln}}^{2}\left(z\right)-{\text{Li}}_{2}\left(-z\right).$
To prove the last identity, differentiate ${\text{Li}}_{2}\left(-1/z\right)$ then integrate back.
In geeneral we have
${\int }_{0}^{1}\frac{{\mathrm{ln}}^{p}\left(x\right)}{a+x}dx=\left(-1{\right)}^{p+1}\phantom{\rule{thinmathspace}{0ex}}p!\phantom{\rule{thinmathspace}{0ex}}{\text{Li}}_{p+1}\left(-1/a\right)$
which follows from the integral represenation of the polylogarithm function${\text{Li}}_{p+1}\left(z\right)=\frac{\left(-1{\right)}^{p}}{p!}{\int }_{0}^{1}\frac{z{\mathrm{ln}}^{p}\left(t\right)}{1-zt}dt$
upon replacing z by $-1/a$
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dikcijom2k
Changing the integration variable to $t:=x/\alpha$, we get
$I={\int }_{0}^{1/\alpha }\frac{\mathrm{log}\alpha +\mathrm{log}t}{1+t}\phantom{\rule{thickmathspace}{0ex}}dt.$
The integration of the $\mathrm{log}\alpha$ term is easy:
${\int }_{0}^{1/\alpha }\frac{\mathrm{log}\alpha }{1+t}\phantom{\rule{thickmathspace}{0ex}}dt=\mathrm{log}\alpha \mathrm{log}\frac{1+\alpha }{\alpha }.$
For the $\mathrm{log}t$ term, changing the integration variable to $u:=1+t$
$\begin{array}{rl}{\int }_{0}^{1/\alpha }\frac{\mathrm{log}t}{1+t}\phantom{\rule{thickmathspace}{0ex}}dt& ={\int }_{1}^{1+1/\alpha }\frac{\mathrm{log}\left(u-1\right)}{u}\phantom{\rule{thickmathspace}{0ex}}du\\ & =-\left[\mathrm{\Phi }\left(1+\frac{1}{\alpha }\right)-\mathrm{\Phi }\left(1\right)\right]\end{array}$
where $\mathrm{\Phi }$ is defined as in this Wikipedia article:
$\mathrm{\Phi }\left(x\right):=-{\int }_{0}^{x}\frac{|1-u|}{u}\phantom{\rule{thickmathspace}{0ex}}du=\left\{\begin{array}{ll}{\mathrm{Li}}_{2}\left(x\right)& \left(x\le 1\right)\\ \frac{{\pi }^{2}}{3}-{\mathrm{Li}}_{2}\left(\frac{1}{x}\right)-\frac{1}{2}\left(\mathrm{log}\left(x\right){\right)}^{2}& \left(x>1\right)\end{array}$
Combining the two and using ${\mathrm{Li}}_{2}\left(1\right)={\pi }^{2}/6$ yield
$I=-\frac{1}{2}\left(\mathrm{log}\alpha {\right)}^{2}-\frac{{\pi }^{2}}{6}+{\mathrm{Li}}_{2}\left(\frac{\alpha }{1+\alpha }\right)+\frac{1}{2}\left(\mathrm{log}\left(1+\alpha \right){\right)}^{2}.$
Finally, this identity:
${\mathrm{Li}}_{2}\left(z\right)+{\mathrm{Li}}_{2}\left(\frac{z}{z-1}\right)=-\frac{1}{2}\left(\mathrm{log}\left(1-z\right){\right)}^{2}$
simplifies the last two terms into $-{\mathrm{Li}}_{2}\left(-\alpha \right)$, giving the desired result.