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Computing the integration.

$\int_0^1\frac{\ln x}{1-x}d x$

I have no idea ... how to do that?

... The answer is $-\frac{\pi ^2}{6}$ I get it by software.

integer
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  • Hint: what is the region of convergence for the Maclaurin series of $\ln(1-t)$? What modes of convergence could be considered on subsets of that region? – Jonathan Y. Oct 15 '13 at 06:43
  • An elementary method https://math.stackexchange.com/a/3404200/686284 – Quanto Mar 25 '23 at 14:10

1 Answers1

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Let $u = (1-x)$, so that this integral is the same as $\displaystyle \int_0^1 \frac{\ln(1-u)}{u} \mathrm{d}u$.

Recall the Taylor expansion for $\ln(1-u)$:

$$ \ln(1-u) = -\sum_{n = 1}^\infty \frac{u^n}{n}.$$

Then our problem becomes

$$-\int_0^1 \sum_{n = 1}^\infty \frac{u^{n-1}}{n} \mathrm{d}u= -\sum_{n = 1}^\infty \int_0^1 \frac{u^{n-1}}{n} \mathrm{d}u = -\sum_{n=1}^\infty \frac{1}{n^2} = -\frac{\pi^2}{6},$$

where I take it for granted that we know how to evaluate the last sum. (For a reference, it's called the Basel Problem. It has been asked and answered in varying flavors on MSE before.)

davidlowryduda
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  • Hi, Do you think this is the unique simple method? – integer Oct 15 '13 at 07:04
  • @integer: When you say the unique simple method, do you mean do I think this is the easiest way to get the answer? If that's the case, then yes, I think this is the easiest way. – davidlowryduda Oct 15 '13 at 07:29
  • I do not think that you will find simpler methods except if you want a solution which is PolyLog[2,1-x] (thgis is the antiderivative of Log[x] / (1-x). – Claude Leibovici Oct 15 '13 at 09:31