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Evaluate the integral $$I=\int_0^{\infty}\frac{\log x}{x^2-1}dx$$

Attempt: $$I=-\int_{0}^{\infty}\log(x)\sum_{k=0}^{\infty}x^{2k}dx=-\sum_{k=0}^{\infty}\lim_{n\to\infty}I_n$$ where $I_n=\int_{0}^nx^{2k}\log(x)dx$.

For computing $I_n$, let $u=\log(x)$ and $dv=x^{2k}dx$. Then $du=(1/x)dx$ and $v=x^{2k+1}/(2k+1)^2$. So, \begin{equation} I_n=\frac{x^{2k+1}}{2k+1}log(x)-\frac{x^{2k+1}}{(2k+1)^2}\Bigg|_{x=0}^{x=n}=\frac{n^{2k+1}}{2k+1}log(n)-\frac{n^{2k+1}}{(2k+1)^2}-\lim_{x\to 0^{+}}\Big(\frac{x^{2k+1} }{2k+1}\log(x)\Big)\end{equation} On the other hand, $$\lim_{x\to 0^+}x^{2k+1}log(x)=\lim_{x\to 0^+}\frac{1/x}{1/x^{2k+1}}=\lim_{x\to 0^+}\frac{x^{2k+1}}{-2k-1}=0$$ So we have $$I_n=\frac{n^{2k+1}}{2k+1}\log(n)-\frac{n^{2k+1}}{(2k+1)^2}$$ Now we need to compute $\lim_{n\to\infty}I_n$. But I'm stuck here. How can we compute this limit? And can anyone check my attempt? Thanks!

Ergin Süer
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    See Marco Cantarini's answer there (different bounds for the integral) for the computation of one part of the integral. For the rest (or a more specific question): what have you tried, why did it fail, and what makes you "think that the method of integration by parts does not work"? – Clement C. May 28 '16 at 18:21
  • @ClementC. Thanks for your comments and brilliant answer. I refresh my question – Ergin Süer May 28 '16 at 21:25
  • @ErginSuer I am not sure you're on the right track here: if you read my answer, you'll see that I deal essentially only with the integral on $(0,1)$, and then show that the remaining part (integral on $(1,\infty)$) has the same value. The idea of decomposing $\frac{1}{1-x^2}$ as $\sum_k x^{2k}$ does not work for $x\in[1,\infty)$ (then the series diverges), and is only applicable for $\lvert x\rvert < 1$... hence the "trick" of writing $I=I_1+I_2$, showing than actually $I_2=I_1$ (via a change of variables), and only computing $I_1=\int_0^1 f(x) dx$ (where we can use this series). – Clement C. May 28 '16 at 21:41

2 Answers2

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Outline:

  • First, you actually should prove that your integral is indeed convergent. You may not want to, but that's always a good thing...

  • Rewrite your integral as $$ I = \underbrace{\int_0^1 dx \frac{\ln x}{x^2-1}}_{I_1} + \underbrace{\int_1^\infty dx \frac{\ln x}{x^2-1}}_{I_2} $$

  • Compute $I_1$ by writing $\frac{1}{x^2-1} = -\sum_{k=0}^\infty x^{2k}$, so that $$ I_1 = -\int_0^1 dx \sum_{k=0}^\infty x^{2k} \ln x = -\sum_{k=0}^\infty \int_0^1 dx\ x^{2k} \ln x = [...] $$ where you need to justify swapping sum and integral.$^{(\dagger)}$ See Marco Cantarini's answer there for more details on the computation of $I_1$.

  • Compute $I_2$ by reducing it to computing $I_1$... the change of variable $u=\frac{1}{x}$ will come in handy.$^{(\ddagger)}$

Spoilers.

$(\dagger)$ The words "absolute convergence" should have a role there.

$(\ddagger)$ More precisely, $$ I_2 = \int_1^\infty dx\frac{\ln x}{x^2-1} = \int_1^0 -\frac{du}{u^2}\frac{\ln \frac{1}{u}}{\frac{1}{u^2}-1} = \int_0^1 \frac{du}{u^2}\frac{-\ln u}{\frac{1}{u^2}-1} = \int_0^1 du\frac{-\ln u}{1-u^2} = I_1 $$

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Clement C.
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  • We need to dominate the functions $f_n(x):=\sum_{k=0}^{n}x^{2k}\ln(x)$. Right? But I couldn't do it. Could you give a hint? – Ergin Süer Jun 30 '16 at 11:47
  • It's been a while and I'm away from my laptop, but since all is negative, a fast way to argue would be Lebesgue's monotone convergence theorem. (To check.) – Clement C. Jun 30 '16 at 13:32
  • @ErginSuer (let me know if you have further questions) – Clement C. Jul 01 '16 at 08:49
  • @ErginSuer If the only issue you have is dcreasing instead of non-decreasing, consider studying convergence of $-f$ instead... the theorem, basically, applies to either decreasing or increasing, not only to increasing. – Clement C. Jul 01 '16 at 17:22
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Let $I$ be the integral given by

$$\begin{align} I&=\int_0^\infty \frac{\log(x)}{x^2-1}\,dx\\\\ &=\int_0^1 \frac{\log(x)}{x^2-1}\,dx+\int_1^\infty \frac{\log(x)}{x^2-1}\,dx\tag 1 \end{align}$$

Enforcing the substitution $x\to 1/x$ in the second integral on the right-hand side of $(1)$ reveals

$$\begin{align} I&=2\int_0^1 \frac{\log(x)}{x^2-1}\,dx\\\\ &=\int_0^1 \frac{\log(x)}{x-1}\,dx-\int_0^1 \frac{\log(x)}{x+1}\,dx\tag 2 \end{align}$$

For the first integral on the right-hand side of $(2)$, we enforce the substitution $x\to 1-x$ to obtain

$$\begin{align} \int_0^1 \frac{\log(x)}{x-1},dx&=-\int_0^1 \frac{\log(1-x)}{x}\,dx\\\\ &=\text{Li}_2(1)\\\\ &=\frac{\pi^2}{6} \end{align}$$

For the second integral on the right-hand side of $(2)$, we integrate by parts with $u=\log(x)$ and $v=\log(1+x)$ to obtain

$$\begin{align} \int_0^1 \frac{\log(x)}{x+1},dx&=-\int_0^1 \frac{\log(1+x)}{x}\,dx\\\\ &=-\int_0^{-1}\frac{\log(1-x)}{x}\,dx\\\\ &=\text{Li}_2(-1)\\\\ &=-\frac{\pi^2}{12} \end{align}$$

Hence, we find

$$\begin{align} I&=\frac{\pi^2}{6}+\frac{\pi^2}{12}\\\\ &=\frac{\pi^2}{4} \end{align}$$

Mark Viola
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