# Definite integral problem

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?

• 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. Commented Apr 24, 2014 at 2:01
• Hint: $\ln(1+t) = -\sum\limits_{k=1}^\infty \dfrac{(-1)^k t^k} k = t -\dfrac{t^2}{2}+\dfrac{t^3}{3}-\ldots , \forall t: |t|<1$ Commented Apr 24, 2014 at 2:06
• @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? Commented Apr 24, 2014 at 9:16
• @Cheeku Here there are tons of methods listed to compute that identity.
– S L
Commented Apr 24, 2014 at 10:31

To calculate that sum, let us assume that the value of the following series $$S_n = 1 + \frac 1 {2^2} + \frac 1 {3^2} + \dots = \frac{\pi^2}{6}$$
To get the value of our series, we take $S_n - 2 S_{2n}$.