Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
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. –  André Nicolas Apr 24 at 2:01
2  
Hint: $\ln(1+t) = -\sum\limits_{k=1}^\infty \dfrac{(-1)^k t^k} k = t -\dfrac{t^2}{2}+\dfrac{t^3}{3}-\ldots , \forall t: |t|<1$ –  Graham Kemp Apr 24 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? –  Cheeku Apr 24 at 9:16
1  
@Cheeku Here there are tons of methods listed to compute that identity. –  Santosh Linkha Apr 24 at 10:31

1 Answer 1

up vote 8 down vote accepted

\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 $S_n - 2 S_{2n}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.