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How would one evaluate $\int_0^1 {\ln(1+x)\over x}\,dx$?

I'd like to do this without approximations. Not quite sure where to start. What really bothers me is that I came across this while reviewing my old intro to calculus book... but I'm fairly certain I've exhausted all the basic methods they teach in that text.

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Looks like you have something in the form of this: According to Wolfram, you should get pi^2/12. – Joe Feb 29 '12 at 2:47
The section of the text it's in suggests definitive evaluation methods. Which is what gets me, otherwise I could just apply Simpson's. – badreferences Feb 29 '12 at 2:51
Series expansion of the logarithm will be helpful, if we make use of some known facts on Riemann zeta function. – Sangchul Lee Feb 29 '12 at 2:52

$$ \int^1_0 \frac{ \log (1+x) }{x} dx = \int^1_0 \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^{n-1}}{n} dx$$

$$ =\sum_{n=1}^{\infty} (-1)^{n-1} \int^1_0 \frac{x^{n-1} }{n} dx = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^2}. $$

Denote $\displaystyle S = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}.$ Then $$\sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^2} = S - 2\left( \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \cdots \right) = S - \frac{S}{2} = \frac{\pi^2}{12}.$$

Thus $$\int^1_0 \frac{ \log (1+x) }{x} dx = \frac{\pi^2}{12}.$$

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Looks good! Char limit. – Joe Feb 29 '12 at 2:59
Where did the 2 coefficient right after the S come from? – badreferences Feb 29 '12 at 20:31
@badreferences the alternating sum is $S$ but with every second term negative, not positive. So to get back from $S$ to the alternating sum, we subtract $2$ times those terms. Subtracting only once makes those terms disappear, subtracting twice makes the terms have a negative sign, and that's what we want. – Ragib Zaman Mar 1 '12 at 12:30
@RagibZaman Oh, I get it! Thanks. – badreferences Mar 1 '12 at 20:19

$$ I = \int_0^1 \frac{\ln(1+x)}{x}\,dx = \int_0^\infty \ln(1+e^{-t})\,dt\,, $$ where $x = e^{-t}$. Then expand $$ \ln(1 + e^{-t}) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\,e^{-tn}\,, $$ So we find $$ I = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\,\int_0^\infty e^{-tn}\,dt = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2} = \frac{\pi^2}{12}\,. $$ Is the last sum familiar to you? What about the closely related and easier sum $$ \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}\,? $$

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The Taylor series for $\frac{\ln(1+x)}{x}$ is $\sum\limits_{n=1}^\infty (-1)^{n-1}\frac{x^{n-1}}{n}$, and this converges absolutely in $(0,1)$ thus we can use it for our integral. This means $$\int_0^1\frac{\ln(1+x)}{x}=\int_0^1\sum\limits_{n=1}^\infty (-1)^{n-1}\frac{x^{n-1}}{n}=\sum\limits_{n=1}^\infty (-1)^{n-1}\int_0^1\frac{x^{n-1}}{n}=\sum\limits_{n=1}^\infty (-1)^{n-1}\frac{1}{n^2}$$ and this series is equal to $\pi^2/12$ according to Wolfram|Alpha.

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