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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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    Series expansion of the logarithm will be helpful, if we make use of some known facts on Riemann zeta function.2012-02-29

3 Answers 3

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$ \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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    @RagibZaman Oh, I get it! Thanks.2012-03-01
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$ 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.