Evaluate $\int_0^1 {\ln(1+x)\over x}\,dx$.

Question

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.

Looks like you have something in the form of this: https://en.wikipedia.org/wiki/Polylogarithm
According to Wolfram, you should get pi^2/12. — Joe, Feb 29 '12 at 02:47
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 02:51
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 02:52

score 12 · Answer 1 · answered Feb 29 '12 at 02:50

12

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

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

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

answered Feb 29 '12 at 02:50

Ragib Zaman

35,127

Looks good! Char limit. – Joe Feb 29 '12 at 02:59
Where did the 2 coefficient right after the S come from? – badreferences Feb 29 '12 at 20:31
1

@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 01 '12 at 12:30
@RagibZaman Oh, I get it! Thanks. – badreferences Mar 01 '12 at 20:19

score 1 · Answer 2 · answered Feb 29 '12 at 02:55

$$ I = \int_0^1 \frac{\ln(1+x)}{x}\,dx = \int_0^\infty \ln(1+e^{-t})\,dt\,, $$ 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 $$ \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}\,? $$

score 1 · Answer 3 · answered Feb 29 '12 at 02:59

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.

Evaluate $\int_0^1 {\ln(1+x)\over x}\,dx$.

3 Answers3

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