I happen to watch the video here, which gives a solution to the definite integral below using the power series approach. Then answer is $\frac{\pi^2}{6}$, given by:

$$\int_0^1 \frac{\ln x}{x-1}dx=\int_{-1}^0 \frac{\ln(1+u)}{u}du=\sum_{n=0}^{\infty}\frac{1}{(n+1)^2}=\frac{\pi^2}{6},$$

where the power seires expansion of the function $\ln(1+u)$ is used.

I tried for some time, but could not find another approach. Does anyone know any alternative methods to evaluate above definite integral without using the infinite series expansion?

Any comments, or ideas, are really appreciated.

  • 2
    $\begingroup$ This might answer your question. $\endgroup$
    – r9m
    Oct 22 '19 at 6:02
  • 1
    $\begingroup$ @r9m, Thank you very much! $\endgroup$
    – student
    Oct 22 '19 at 6:46
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    $\begingroup$ Your integral is a special case of $~\displaystyle\frac{1}{n!}\int\frac{(\ln(x+z))^n}{x-a}\,dx =\sum\limits_{k=0}^n\frac{(\ln(x+z))^k}{k!}(-1)^{n-k+1}\text{Li}_{n-k+1}\left(\frac{x+z}{a+z}\right) + C~$ ; $~(n,a,z):=(1,1,0)~$ . $\endgroup$
    – user90369
    Oct 22 '19 at 7:21

Here is to integrate without resorting to power series. Note \begin{align} \int_0^1\frac{\ln x}{1-x}dx& =\frac43\int_0^1 \frac{\ln x }{1-x}dx -\frac13\int_0^1 { \frac{\ln x }{1-x} } \overset{x\to x^2}{dx} \\ &= \frac43\int_0^1 \frac{\ln x}{1-x^2}dx = \frac23\int_0^\infty \frac{\ln x}{1-x^2}dx=\frac23J(1) \end{align}

where $ J(\alpha) =-\frac 12 \int_0^\infty \frac{\ln (1-\alpha^2 + \alpha^2 x^2)}{x^2-1}dx $

$$ J'(\alpha) =-\int_0^\infty \frac{\alpha dx}{1-\alpha^2 + \alpha^2 x^2} = -\frac{\pi/2}{\sqrt{1-\alpha^2}}$$


$$ \int_0^1\frac{\ln x}{1-x}dx = \frac23\int_0^1 J'(\alpha) d\alpha =-\frac{\pi}{3}\int_0^1 \frac{d\alpha}{\sqrt{1-\alpha^2}}= -\frac{\pi^2}{6}$$

  • 2
    $\begingroup$ Where does this function $J$ come from? It doesn't look like one could guess that just from the computation up that point. $\endgroup$
    – quarague
    Oct 22 '19 at 13:20
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    $\begingroup$ @quarague - $I = \int_0^1 J'(a)da$ is a known technique in definite integration to 'bypass' the original integral which is usually hard to do directly. The actual form of $J(a)$ depends on the integrand of $I$; it requires some search or guess, though. $\endgroup$
    – Quanto
    Oct 22 '19 at 13:31
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    $\begingroup$ @Quanto, thanks very much for this nice answer! $\endgroup$
    – student
    Oct 22 '19 at 16:42
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    $\begingroup$ @Quanto "$I = \int_0^1 \,J'(a)da$ is a known technique...". Really? My calculus experience is limited to Volume 1 of "Calculus 2nd Edition" by Apostol. It probably doesn't surprise you that I didn't know about this technique. Can you suggest Calculus/Real-Analysis books (or pdf's) devoted to an organized presentation of various techniques for definite &/or indefinite integration? $\endgroup$ Jul 28 '20 at 23:57
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    $\begingroup$ @user2661923 The technique in question is often referred to as Feynman's trick, so googling that may help you find the most impressive examples of this inventive but often thorny technique. $\endgroup$
    – J.G.
    Aug 19 '20 at 20:31

Here is another way of showing that $\int_0^1 \frac{\log(x)}{1-x}\,dx=\pi^2/6$. First, enforce the substitution $x\mapsto 1-x$ to obtain

$$I=-\int_0^1 \frac{\log(1-x)}{x}\,dx$$

Then, noting that $\int_0^1 \frac{1}{1-xy}\,dy=-\frac{\log(1-x)}{x}$ we write $I$ as

$$I=\int_0^1 \int_0^1 \frac{1}{1-xy}\,dx\,dy\tag1$$

In THIS ANSWER, I used the transformation $x=s+t$, $y=s-t$ to write the double integral in $(1)$ as

$$\begin{align} \int_0^1\int_0^1 \frac{1}{1-xy}\,dx\,dy&=\int_0^{1/2}\int_{-s}^{s}\frac{2}{(1-s^2)+t^2}\,dt\,ds+\int_{1/2}^{1}\int_{s-1}^{1-s}\frac{2}{(1-s^2)+t^2}\,dt\,ds\\\\ &=\int_0^{1/2}\frac{4}{\sqrt{1-s^2}}\arctan\left(\frac{s}{\sqrt{1-s^2}}\right)\,ds\\\\&+\int_{1/2}^{1}\frac{4}{\sqrt{1-s^2}}\arctan\left(\sqrt{\frac{1-s}{1+s}}\right)\,ds\\\\ &=4\int_0^{1/2}\frac{\arcsin(s)}{\sqrt{1-s^2}}\,ds+4\int_{1/2}^1\frac{\arccos(s)}{2\sqrt{1-s^2}}\,ds\\\\ &=2\arcsin^2(1/2)+\arccos^2(1/2)\\\\ &=2\left(\frac{\pi}{6}\right)^2+\left(\frac{\pi}{3}\right)^2\\\\ &=\frac{\pi^2}{6} \end{align}$$

And we are done!

  • $\begingroup$ thank you very much for this nice answer! I have already voted this answer. Because Quanto's answer was posed earlier, so I accept that one. Wish you stay strong and safe too ! $\endgroup$
    – student
    Mar 3 '21 at 15:07
  • $\begingroup$ You're welcome. My pleasure. Pleased to see that this was useful. $\endgroup$
    – Mark Viola
    Mar 3 '21 at 17:34

$$\begin{align*} & \color{blue}{I = \int_0^1 \frac {\ln x}{x-1}dx}\\ \end{align*}$$

Now, $$\color{red}{\psi_0 (z) = -\gamma + \int_0^{1} \frac {x^{z-1}-1}{x-1}dx}$$ We'll differentiate and that is trigamma function $\psi_1(z)$

$$\implies \frac {\partial\psi_0}{\partial z}= \psi_1 (z) = \int_0^1 \frac {x^{z-1}\ln x}{x-1}dx$$ $$\color{green}{I = \left[\psi_1(z)\right]_{z=1} = \frac {\pi^2}{6}}$$

I have copied this answer from my old answer


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