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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.

Quanto
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student
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3 Answers3

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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 $

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

Thus

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

Quanto
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    Where does this function $J$ come from? It doesn't look like one could guess that just from the computation up that point. – quarague Oct 22 '19 at 13:20
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    @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. – Quanto Oct 22 '19 at 13:31
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    @Quanto, thanks very much for this nice answer! – student Oct 22 '19 at 16:42
  • I'm having a tough time following the early steps. Where did $t$ come from and how is it defined? – Polygon Jul 14 '20 at 16:02
  • @Polygon - there was a typo, which is fixed now – Quanto Jul 14 '20 at 16:13
  • See https://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-limits-k-1-infty-frac1k2-basel-pro – FDP Jul 14 '20 at 17:46
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    @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?

     – user2661923 Jul 28 '20 at 23:57
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    @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. – J.G. Aug 19 '20 at 20:31
  • @J.G. nice response, thanks. – user2661923 Aug 19 '20 at 20:43
  • @user2661923 - check out https://math.stackexchange.com/questions/3453227/when-to-differentiate-under-the-integral-sign/3453393#3453393 – Quanto Feb 11 '21 at 16:53
  • Nice solution. I've posted a complementary solution that you might enjoy. ;-) – Mark Viola Feb 26 '21 at 17:51
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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!

Mark Viola
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  • 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 ! – student Mar 03 '21 at 15:07
  • You're welcome. My pleasure. Pleased to see that this was useful. – Mark Viola Mar 03 '21 at 17:34
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$$\begin{align*} & \color{blue}{I = \int_0^1 \frac {\ln x}{x-1}dx}\\ \end{align*}$$

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