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How to prove $$\int _0^1 {\ln(x)\over{1-x^2}}={-\pi^{2}\over 8}$$ My solution: If we can prove$\int _0^1 {\ln(x)\over{1-x^2}}= \lim_{n\to \infty} \int _0^1\ln(x)(1+x^2+x^4+......+x^{2n})$,then I think we can prove the equality above. because the right hand =$\lim_{n\to \infty} $$(-1-{1\over{3^2}}-{1\over 5^2}......-{1\over{2n+1}^2})={-\pi^{2}\over 8}$.

Can someone help me prove why$\int _0^1 {\ln(x)\over{1-x^2}}= \lim_{n\to \infty} \int _0^1\ln(x)(1+x^2+x^4+......+x^{2n})$? I don't know how to prove it? Or, can someone use other methods to solve the equality above?

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hint: find $\int_{0}^1 x^k\cdot lnxdx$ – DeepSea Jun 8 '14 at 0:56
The most important thing is that lnx/(1-x^2) does not equal lnx(1+x^2+x^4+......+x^2n) when x=0, So we cannot say that they are always equal in [0,1],so I think we need to use knowledge of uniform convergence to prove the integrals of two functions are the same. But I can't prove it. – python3 Jun 8 '14 at 1:00
I'm not sure why this was downvoted but this is a good question. I'll give a short answer for this. – Cameron Williams Jun 8 '14 at 1:05
In general: $$\int_0^1 \frac{(\log{x})^n}{1-x^2}\, dx = (-1)^n \left(1-\frac{1}{2^{n+1}}\right) n!\, \zeta(n+1)$$ – gar Jun 8 '14 at 11:56

This is an improper integral so it should be viewed as a limiting procedure of the form

$$\lim_{b\rightarrow 1^-}\lim_{a\rightarrow 0^+} \int_a^b \frac{\ln x}{1-x^2}\,dx.$$

The series for $\dfrac{1}{1-x^2}$ converges uniformly (which is the key) over such intervals so you can interchange series and integral without trouble. This is a standard theorem in real analysis.

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For your approach:

Define the sequence $(f_n)_{n=1}^{\infty}$ by $f_n:(0,1) \to \mathbb{R}$ s.t. $f_n(x) = \left(\displaystyle\sum_{k=1}^n x^{2k}\right) \ln x $; and define $f:(0,1) \to \mathbb{R}$ s.t. $f(x) = \dfrac{\ln x}{1-x^2}$.

Show that $f_n \to f$ uniformly.

Then $\displaystyle\int_0^1 f_n \to \int_0^1 f$, so that $\displaystyle\int_0^1 f(x) dx = \displaystyle\lim_{n\to \infty} \displaystyle\int_0^1 f_n(x) dx$

(as explained by Cameron)

Alternative approach:

Choose the branch of the logarithm that corresponds to taking a cut along $\mathbb{R}_{\geq 0}$ and consider $I = \displaystyle\oint _C \dfrac{(\log z)^2}{1-z^2}dz$ along the contour $C$ defined as $\gamma_+ \cup\gamma_R \cup\gamma_- \cup\gamma_{\epsilon}$ (traversed anti-clockwise) where:

$ \gamma_+ = \{z\in \mathbb{C} : z=x, \ \epsilon \leq x \leq R \} \ \ \ \ \ \ $ $\gamma _R = \{z\in \mathbb{C} : z=Re^{i\theta},\ 0<\theta<2\pi\}$

$\gamma_- = \{z\in \mathbb{C}: z=te^{2\pi i}, \ R\geq t\geq \epsilon \} \ \ \ \ \ \ $ $\gamma_{\epsilon} = \{z\in \mathbb{C} : z=\epsilon e^{i\theta}, \ 2\pi >\epsilon > 0\}$

By the residue theorem, $I = 2\pi i \cdot \dfrac{-\pi^2}{-2e^{i\pi}} = \dfrac{i\pi^3}{e^{i\pi}}$.

Taking $R\to \infty, \epsilon \to 0$, noting that the contributions from integrating along $\gamma_{\epsilon}$ and $\gamma_R$ are vanishingly small in the limit, we also have:

$\begin{eqnarray*} I&=& \displaystyle\int_0^{\infty} \dfrac{(\log x)^2}{1-x^2} dx - \displaystyle\int_0^{\infty} \dfrac{(\log x - 2\pi i)^2}{1-x^2} dx \\ &=& 4\pi i \displaystyle\int_0^{\infty} \dfrac{\log x}{1-x^2} dx + 4\pi ^2 \cdot \underbrace{\text{PV} \left(\displaystyle\int_0^{\infty} \dfrac{1}{1-x^2} dx\right)}_{=\ 0}\\ &=&4\pi i \left(\displaystyle\int _0^1 \dfrac{\log x}{1-x^2}dx +\displaystyle\int _1^{\infty} \dfrac{\log x}{1-x^2}dx \right) \\ &=& 4\pi i \left(\displaystyle\int _0^1 \dfrac{\log x}{1-x^2}dx +\displaystyle\int _1^0 \dfrac{\log \frac{1}{x}}{1-\frac{1}{x^2}}\cdot -\dfrac{1}{x^2} dx \right) \\ &=& 8\pi i \displaystyle\int _0^1 \dfrac{\log x}{1-x^2}dx \end{eqnarray*}$

Hence $\displaystyle\int _0^1 \dfrac{\log x}{1-x^2}dx = \dfrac{-i\pi ^3}{8\pi i} = -\dfrac{\pi ^2}{8}$

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As suggested by 8 pi r $$\int x^{2k} \log(x)~ dx=\frac{x^{2 k+1} (2 k \log (x)+\log (x)-1)}{(2 k+1)^2}$$ so $$\int_{0}^1 x^{2k} \log(x)~ dx=-\frac{1}{(2 k+1)^2}$$ and then $$\int _0^1 {\ln(x)\over{1-x^2}}=-\sum_{k=0}^\infty\frac{1}{(2 k+1)^2}=-{\pi^{2}\over 8}$$

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Perhaps you could think of it slightly differently. For values of $x$ in $(0,1)$, the geometric expansion for $\frac{1}{1-x^2}$ converges absolutely and uniformly and there is no problem.

So instead consider

$$ \lim_{a \nearrow 1} \int_0^a \frac{\ln x}{1-x^2} \mathrm{d}x,$$

which exists and is equal to the expansion you've written down.

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Remember that $\frac{1}{1-x^{2}}=1+x^{2}+\ldots$ for $-1<x<1$ so $1+x^{2}+\ldots+x^{2n}$ converge pointwise to $\frac{1}{1-x^{2}}$ as $n$ increase. Thus $\ln(x)(1+x^{2}+\ldots+x^{2n})$ converge pointwise to $\frac{ln(x)}{1-x^{2}}$ for $0<x<1$.

Since $\ln(x)(1+x^{2}+\ldots+x^{2n})$ are negative and decrease monotonically and converge pointwise almost everywhere (only missing $0$ and $1$) to $\frac{\ln(x)}{1-x^{2}}$, using Lebesgue monotone convergence theorem we immediately have $\int\limits_{0}^{1}\frac{\ln(x)}{1-x^{2}}dx=\lim\limits_{n\rightarrow\infty}\int\limits_{0}^{1}\ln(x)(1+x^{2}+\ldots+x^{2n})dx$.

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