## 1.º Problema de 2010: um integral de Stieltjes :: 2010 Problem #1 – A Stieltjes Integral

Prove que/prove that

$\zeta \left( 2\right) =\dfrac{p}{q}\displaystyle\int_{-1}^{\sqrt{3}}\arctan (x)\,d\left( \arctan (x)\right)$,

where/em que  $(p,q)\in\mathbb{Z}^{2}$.

## Sobre Américo Tavares

eng. electrotécnico reformado / retired electrical engineer
Esta entrada foi publicada em Análise Matemática, Calculus, Cálculo, Integrais, Matemática, Math, Problemas com as tags , , , , . ligação permanente.

### 6 respostas a 1.º Problema de 2010: um integral de Stieltjes :: 2010 Problem #1 – A Stieltjes Integral

1. Wouldn’t there be a problem if $q = 0$?

• According to my evaluation $q\ne 0$

2. Also, I am having a little difficulty in understanding the right hand side expression. Isn’t there supposed to be some summation sign? I mean, $p/q$ should be summed over $\mathbb{Z}^2$, is that right (of course, not allowing $q = 0$)?

• What I mean is this: prove that the given integral can be expressed as a racional multiple of $\zeta(2)$, i. e. $\left( p,q\right) \in\mathbb{Z}^{2}$.
I edited the problem statement a little bit.

3. Ah, I see now. Thanks for the clarification. But, I must say that the integral in its present form is somewhat easy to calculate.

• Then you can move to the last one. That is an important intermediate result. One gets $\zeta(3)$ expressed as another series that converges faster to $\zeta(3)$ (see e.g. van der Poorten’s paper A proof Euler missed…, Apéry’s proof of the irrationality of $\zeta(3)$, An informal report). You find it online. Go to the page Consulta de publicações.
PS. I think that most of my readers doesn’t even know what a Stiltjes integral is.