# Thread: Proof that sum^infty of 1/n^2 = ...

1. No, I didn't, I thought you referenced it because it contained a proof--I didn't realize it had six. Thank you.

2. This is a quality thread! I now know 3 more proofs of this!

To simplependulum: did you think of that yourself?! If so, how long did that take you and what led you to it? Kudos!

3. Originally Posted by chiph588@
This is a quality thread! I now know 3 more proofs of this!

To simplependulum: did you think of that yourself?! If so, how long did that take you and what led you to it? Kudos!
At first i was thinking how to solve $\int_1^{\infty} \frac{\ln(x)}{x^2 - 1 }~dx$ without using the fact $\zeta(2) = \frac{\pi^2}{6}$ , in other words , I shall give another way to prove this fact starting at this point .

However , i failed , all my attempts eventually led to other troublesome integrals so I gave up , until I was asked to solve another integral $\int_0^1 \frac{ \tan^{-1}(x)}{\sqrt{1-x^2 }}~dx$ which brought some ideas to me in solving the integral I was struggling two weeks before .

4. Originally Posted by simplependulum
$\int_1^{\infty} \frac{\ln(x)}{ x^2 - 1 }~dx = \frac{3}{4} \zeta(2)$

$\vdots$

$= \frac{\pi^2}{8}$
Very creative! I'm impressed!

Two questions though:

First, how do we know $\displaystyle \int_1^\infty \frac{\ln(x)}{x^2-1}dx = \frac34 \zeta(2)?$

Also for $\displaystyle \int_0^1 \left[ \tan^{-1}\left(t\sqrt{1-y^2}\right) \right]_0^{\infty} \frac{dy}{\sqrt{1-y^2} }$, we have $\displaystyle y\in[0,1]$.

So $\displaystyle \lim_{t\to\infty} \tan^{-1}\left(t\sqrt{1-y^2}\right) = \begin{cases} \frac\pi2 ,\text{ if } y\neq1\\0,\text{ if } y=1 \end{cases}$.

Therefore, what do we do when $\displaystyle y=1?$

5. Here is my derivation of the first fact, although we need this definition of the zeta function

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So, in the initial integral, let $u = \ln(x)$. Then $dx = x du = e^{u} du$. $u(\infty) = \infty$ and $u(1) = 0$. This last bit gives you the new limits of integration.

[LaTeX ERROR: Compile failed]

Where on the RHS, I've let u = x so that i don't have to use u from now on. Now lets focus on the integrand.

$\displaystyle \frac{xe^x}{e^{2x}-1} = \frac{x(e^x+1-1)}{e^{2x}-1} = \frac{x}{e^{x}-1} - \frac{x}{e^{2x}-1}$.

The integral becomes

$\displaystyle \int_{0}^{\infty}\frac{x}{e^{x}-1} \,dx- \int_{0}^{\infty}\frac{x}{e^{2x}-1}\,dx
$

provided both converge. Now on the second piece, let $u = 2x$, $dx = du/2$ and $x = u/2$ to give that the piece is

$\displaystyle \frac{1}{4}\int_{0}^{\infty}\frac{u}{e^u-1}\,du$

Combining both pieces now we get

$\displaystyle \int_{0}^{\infty}\frac{x}{e^{x}-1} \,dx- \int_{0}^{\infty}\frac{x}{e^{2x}-1}\,dx = \frac{3}{4}\int_{0}^{\infty}\frac{x}{e^x-1}\,dx = \frac{3}{4}\Gamma(2)\zeta(2) = \frac{3}{4}\zeta(2)$

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