Evaluating an integral from 0 to 1 with a parameter, (and a dilogarithm)

Question

So I need to evaluate the following integral (in terms of a): $$\int_{0}^{1} \frac{\ln{|1-\frac{y}{a}|}}{y} dy$$

Till now I have tried u-sub ($u = \ln{|1-\frac{y}{a}|}$, $u=\frac{y}{a}$) and integration by parts. Wolfram Alpha gives me the evaluation in terms of a dilogarithm ($-$Li$_2(\frac{y}{a}) + C$), but the problem with this solution is I don't know how to evaluate a dilogarithm. Any ideas on how one might evaluate it?

Answer 1

There may not be a simplification in general since your definite integral is $-\textrm{Li}_2(\frac1a)$ when $a\ge 1$, and that usually does not have a simplification.

There will be special cases:

• when $a=1$ it is $-\frac{\pi^2}{6}$; this is also $\int_{0}^{a} \frac{\ln{|1-\frac{y}{a}|}}{y} dy$.
• when $a=2$ it is $\frac12 (\log_e 2)^2 - \frac{\pi^2}{12}$
• when $a=\frac12$ it is $-\frac{\pi^2}{4}$

Answer 2

The dilogarithm is a special case of the polylogarithm. It has the following series definition: $$ \mathrm{Li}_{s}(z) = \sum_{k\ge 1} \frac{z^k}{k^s} $$ the dilogarithm is the special case of $s=2$. From here we can notice the following: $$ \int_{0}^{z}\frac{\mathrm{Li}_{s}(t)}{t}\mathrm{d}t = \int_{0}^{z}\frac{1}{t}\sum_{k\ge 1} \frac{t^k}{k^s}\mathrm{d}t = \sum_{k\ge 1}\frac{1}{k^s}\int_{0}^{z} t^{k-1}\mathrm{d}t = \sum_{k\ge 1}\frac{1}{k^s}\frac{z^{k}}{k} = \sum_{k\ge 1} \frac{z^k}{k^{s+1}} = \mathrm{Li}_{s+1}(z) $$ This gives us a recursive definition for the polylogarithms. Finally, recalling the geometric series formula $\sum_{n\ge 0} t^n = \frac{1}{1-t}$ we get $$ \mathrm{Li}_{1}(z) = \sum_{k\ge 1} \frac{z^k}{k} = \sum_{k\ge 1} \int_{0}^{z}t^{k-1}\mathrm{d}t= \int_{0}^{z}\sum_{k\ge 1} t^{k-1}\mathrm{d}t \overset{\color{purple}{n=k-1}}{=} \int_{0}^{z}\sum_{n\ge \color{purple}{0}} t^{\color{purple}{n}}\mathrm{d}t = \int_{0}^{z}\frac{1}{1-t}\mathrm{d}t =-\ln(1-z) $$ So combining this with the recursion proven before we get $$ \mathrm{Li}_{2}(z) = \int_{0}^{z} \frac{\mathrm{Li}_{1}(u)}{u}\mathrm{d}u = -\int_{0}^{z}\frac{\ln(1-u)}{u} \mathrm{d}u $$ which is of the form of your integral after the $u$-sub $u = \frac{y}{a}$.

Answer 3

Putting $x=\frac ya$, we have $$I(a)=\int_{0}^{1} \frac{\ln{|1-\frac{y}{a}|}}{y} dy= \int_{0}^{\frac1a} \frac{\ln{|1-x|}}{x} dy $$ Using the expansion for $|x|<1$, $$\frac{1}{1-x}=\sum_{k=0}^{\infty} x^k \\ \Leftrightarrow -\ln (1-x)=\sum_{k=0}^{\infty} \frac{x^{k+1}}{k+1}=\sum_{k=1}^{\infty} \frac{x^k}{k} $$ Back to our integral, $$ \begin{aligned} I(a) & =-\int_0^1 \frac{1}{a} \sum_{k=1}^{\infty} \frac{x^{k-1}}{k} d x \\ & =-\sum_{k=1}^{\infty} \int_0^{\frac{1}{a}} \frac{x^{k-1}}{k} d x \\ & =-\sum_{k=1}^{\infty} \frac{\left(\frac{1}{a}\right)^k}{k^2} \\ & =-\operatorname{Li}_2\left(\frac{1}{a}\right) \end{aligned} $$ For examples, $$ I(1)=-\operatorname{Li}_2(1)=-\sum_{k=0}^{\infty} \frac{1}{k^2}=-\zeta(2)=-\frac{\pi^2}{6} $$ $$ I(2)=-\operatorname{Li}_2\left(\frac{1}{2}\right)=-\sum_{k=0}^{\infty} \frac{1}{k^2 2^k}=-\frac{\pi^2}{12}+\frac{\ln ^2 2}{2} $$