Evaluating improper integral ${\int}_{0}^{1}\frac{\mathrm{log}(x)}{x+\alpha}\phantom{\rule{thickmathspace}{0ex}}dx$ for small positive $\alpha $

Patatiniuh
2022-07-15
Answered

Evaluating improper integral ${\int}_{0}^{1}\frac{\mathrm{log}(x)}{x+\alpha}\phantom{\rule{thickmathspace}{0ex}}dx$ for small positive $\alpha $

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alomjabpdl0

Answered 2022-07-16
Author has **12** answers

${\int}_{0}^{1}\frac{\mathrm{ln}(x)}{a+x}dx={\int}_{0}^{1/a}\frac{\mathrm{ln}(ay)}{1+y}dy$

$=\underset{0}{\underset{\u23df}{\mathrm{ln}(1+y)\mathrm{ln}(ay){|}_{0}^{1/a}}}-{\int}_{0}^{1/a}\frac{\mathrm{ln}(1+y)}{y}dx={\text{Li}}_{2}(-y){|}_{0}^{1/a}={\text{Li}}_{2}(-1/a)$

and the result follows on using the dilogarithm inversion formula:

${\text{Li}}_{2}(-1/z)=-\frac{{\pi}^{2}}{6}-\frac{1}{2}{\mathrm{ln}}^{2}(z)-{\text{Li}}_{2}(-z).$

To prove the last identity, differentiate ${\text{Li}}_{2}(-1/z)$ then integrate back.

In geeneral we have

${\int}_{0}^{1}\frac{{\mathrm{ln}}^{p}(x)}{a+x}dx=(-1{)}^{p+1}\phantom{\rule{thinmathspace}{0ex}}p!\phantom{\rule{thinmathspace}{0ex}}{\text{Li}}_{p+1}(-1/a)$

which follows from the integral represenation of the polylogarithm function${\text{Li}}_{p+1}(z)=\frac{(-1{)}^{p}}{p!}{\int}_{0}^{1}\frac{z{\mathrm{ln}}^{p}(t)}{1-zt}dt$

upon replacing z by $-1/a$

$=\underset{0}{\underset{\u23df}{\mathrm{ln}(1+y)\mathrm{ln}(ay){|}_{0}^{1/a}}}-{\int}_{0}^{1/a}\frac{\mathrm{ln}(1+y)}{y}dx={\text{Li}}_{2}(-y){|}_{0}^{1/a}={\text{Li}}_{2}(-1/a)$

and the result follows on using the dilogarithm inversion formula:

${\text{Li}}_{2}(-1/z)=-\frac{{\pi}^{2}}{6}-\frac{1}{2}{\mathrm{ln}}^{2}(z)-{\text{Li}}_{2}(-z).$

To prove the last identity, differentiate ${\text{Li}}_{2}(-1/z)$ then integrate back.

In geeneral we have

${\int}_{0}^{1}\frac{{\mathrm{ln}}^{p}(x)}{a+x}dx=(-1{)}^{p+1}\phantom{\rule{thinmathspace}{0ex}}p!\phantom{\rule{thinmathspace}{0ex}}{\text{Li}}_{p+1}(-1/a)$

which follows from the integral represenation of the polylogarithm function${\text{Li}}_{p+1}(z)=\frac{(-1{)}^{p}}{p!}{\int}_{0}^{1}\frac{z{\mathrm{ln}}^{p}(t)}{1-zt}dt$

upon replacing z by $-1/a$

dikcijom2k

Answered 2022-07-17
Author has **6** answers

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 $u:=1+t$

$\begin{array}{rl}{\int}_{0}^{1/\alpha}\frac{\mathrm{log}t}{1+t}\phantom{\rule{thickmathspace}{0ex}}dt& ={\int}_{1}^{1+1/\alpha}\frac{\mathrm{log}(u-1)}{u}\phantom{\rule{thickmathspace}{0ex}}du\\ & =-[\mathrm{\Phi}(1+\frac{1}{\alpha})-\mathrm{\Phi}(1)]\end{array}$

where $\mathrm{\Phi}$ is defined as in this Wikipedia article:

$\mathrm{\Phi}(x):=-{\int}_{0}^{x}\frac{|1-u|}{u}\phantom{\rule{thickmathspace}{0ex}}du=\{\begin{array}{ll}{\mathrm{Li}}_{2}(x)& (x\le 1)\\ \frac{{\pi}^{2}}{3}-{\mathrm{Li}}_{2}\left(\frac{1}{x}\right)-\frac{1}{2}(\mathrm{log}(x){)}^{2}& (x>1)\end{array}$

Combining the two and using ${\mathrm{Li}}_{2}(1)={\pi}^{2}/6$ yield

$I=-\frac{1}{2}(\mathrm{log}\alpha {)}^{2}-\frac{{\pi}^{2}}{6}+{\mathrm{Li}}_{2}\left(\frac{\alpha}{1+\alpha}\right)+\frac{1}{2}(\mathrm{log}(1+\alpha ){)}^{2}.$

Finally, this identity:

${\mathrm{Li}}_{2}(z)+{\mathrm{Li}}_{2}\left(\frac{z}{z-1}\right)=-\frac{1}{2}(\mathrm{log}(1-z){)}^{2}$

simplifies the last two terms into $-{\mathrm{Li}}_{2}(-\alpha )$, giving the desired result.

$I={\int}_{0}^{1/\alpha}\frac{\mathrm{log}\alpha +\mathrm{log}t}{1+t}\phantom{\rule{thickmathspace}{0ex}}dt.$

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 $u:=1+t$

$\begin{array}{rl}{\int}_{0}^{1/\alpha}\frac{\mathrm{log}t}{1+t}\phantom{\rule{thickmathspace}{0ex}}dt& ={\int}_{1}^{1+1/\alpha}\frac{\mathrm{log}(u-1)}{u}\phantom{\rule{thickmathspace}{0ex}}du\\ & =-[\mathrm{\Phi}(1+\frac{1}{\alpha})-\mathrm{\Phi}(1)]\end{array}$

where $\mathrm{\Phi}$ is defined as in this Wikipedia article:

$\mathrm{\Phi}(x):=-{\int}_{0}^{x}\frac{|1-u|}{u}\phantom{\rule{thickmathspace}{0ex}}du=\{\begin{array}{ll}{\mathrm{Li}}_{2}(x)& (x\le 1)\\ \frac{{\pi}^{2}}{3}-{\mathrm{Li}}_{2}\left(\frac{1}{x}\right)-\frac{1}{2}(\mathrm{log}(x){)}^{2}& (x>1)\end{array}$

Combining the two and using ${\mathrm{Li}}_{2}(1)={\pi}^{2}/6$ yield

$I=-\frac{1}{2}(\mathrm{log}\alpha {)}^{2}-\frac{{\pi}^{2}}{6}+{\mathrm{Li}}_{2}\left(\frac{\alpha}{1+\alpha}\right)+\frac{1}{2}(\mathrm{log}(1+\alpha ){)}^{2}.$

Finally, this identity:

${\mathrm{Li}}_{2}(z)+{\mathrm{Li}}_{2}\left(\frac{z}{z-1}\right)=-\frac{1}{2}(\mathrm{log}(1-z){)}^{2}$

simplifies the last two terms into $-{\mathrm{Li}}_{2}(-\alpha )$, giving the desired result.

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