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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?

Kraken
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3 Answers3

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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 $-\frac{\pi^2}{4}$
Henry
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  • Hmm..Could you please explain how you got $\frac{-\pi^2}{6}$ for $a=1$? – Kraken Oct 20 '23 at 10:40
  • @CeaealYT Lots of ways. One is to take the series expansion of $\log(1-x)/x$ and integrate term by term to get $-\zeta(2)$ – Henry Oct 20 '23 at 10:46
  • To expand on @Henry's comment. By plugging in $a=1$ into the series definition of the polylogarithm you get $-\mathrm{Li}2\left(\frac{1}{1}\right) =- \sum{k \ge 1} \frac{1}{k^2}$, and evaluating this series is the famous Basel problem of which there are indeed many solutions. Notice that the value $a=1$ is special because $\left( \frac{1}{a}\right)^k = \left( \frac{1}{1}\right)^k = 1$ is a constant, but this isn't true for any other random number. This is an example of why these evaluations are special and can't be generalized. – Robert Lee Oct 20 '23 at 10:53
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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 $u = \frac{y}{a}$.

Robert Lee
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  • Thanks for the explanation about the dilogarithm function. But for some given a, is there a way to evaluate it manually? – Kraken Oct 20 '23 at 10:31
  • What do you mean by "manually" exactly? Are you looking for a numerical approximation? If so, you can plug it into a calculator to get a value. Otherwise, nice closed-form solutions only exist for specific special values of $a$, as pointed out in @Henry's answer. – Robert Lee Oct 20 '23 at 10:35
  • Hmm...is it possible that if I understand how for specific values of a (like those in Henry's answers) have a nice closed-form solution, I can find a general method to evaluate it for others? – Kraken Oct 20 '23 at 10:40
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    A general method doesn't exist. If there were, then there wouldn't be a need to have invented the digamma function in the first place since we could just write the result in terms of simpler functions. Your integral is an example of a nonelementary integral, which means it can be written in terms of elementary results. The specific values with nice form usually exploit some symmetry or special property that doesn't generalize to any other number. – Robert Lee Oct 20 '23 at 10:47
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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, $$ I(1)=-\operatorname{Li}_2(1)=-\sum_{k=0}^{\infty} \frac{1}{k^2}=-\zeta(2)=-\frac{\pi^2}{6} $$ $$ I(2)=-\operatorname{Li}_2\left(\frac{1}{2}\right)=-\sum_{k=0}^{\infty} \frac{1}{k^2 2^k}=-\frac{\pi^2}{12}+\frac{\ln ^2 2}{2} $$

Lai
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