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I was solving a definite integral problem which was reduced to : $$\int^{1}_{0} \frac{\ln(1+t)}{t} dt$$

I couldn't solve it and when I saw the solution, the answer was simply given as $\frac{\pi^2}{12}$, and claimed that this is an identity.

Can anybody give me a proof of this identity?

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    $\begingroup$ This is a close relative of the famous result of Euler that $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots=\frac{\pi^2}{6}$. The proof of the result is mildly complicated. The simplest way is a Fourier series manipulation. $\endgroup$ Commented Apr 24, 2014 at 2:01
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    $\begingroup$ Hint: $\ln(1+t) = -\sum\limits_{k=1}^\infty \dfrac{(-1)^k t^k} k = t -\dfrac{t^2}{2}+\dfrac{t^3}{3}-\ldots , \forall t: |t|<1$ $\endgroup$ Commented Apr 24, 2014 at 2:06
  • $\begingroup$ @AndréNicolas I could solve the integral. Was exactly asking a proof for this identity...My bad to not specify it. :( Can you give me a name for this identity, or a link to its proof? $\endgroup$
    – Cheeku
    Commented Apr 24, 2014 at 9:16
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    $\begingroup$ @Cheeku Here there are tons of methods listed to compute that identity. $\endgroup$
    – S L
    Commented Apr 24, 2014 at 10:31

1 Answer 1

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\begin{align*} \int_0^1 \frac{\log(x+1)}{x} \, dx &= \int_0^1 \sum_{n=1}^\infty (-1)^{n+1} \frac{x^{n-1}}{n}dx\\ &= \sum_{n=1}^\infty (-1)^{n+1}\int_0^1 \frac{x^{n-1}}{n}dx\\ &= \sum_{n=1}^\infty (-1)^{n+1} \frac{1}{n^2}\\ &= \frac{\pi^2}{12}\\ \end{align*}

To calculate that sum, let us assume that the value of the following series $$S_n = 1 + \frac 1 {2^2} + \frac 1 {3^2} + \dots = \frac{\pi^2}{6}$$

Now if we consider only the even values,

\begin{align*} S_{2n} &= \frac 1 {2^2} + \frac 1 {4^2} + \frac 1 {6^2} + \dots \\ &= \frac{1}{2^2 \cdot 1} + \frac 1 {2^2\cdot 2^2} + \frac{1}{2^2 \cdot 3^2} + \dots \\ &= \frac{1}{4} S_n\\ &= \frac{\pi^2}{24}\\ \end{align*}

To get the value of our series, we take $S_n - 2 S_{2n}$.

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