# Does this integral converge ?

1. Feb 1, 2013

### mmzaj

Does this integral converge !?

$$\int_{0}^{\infty}x\frac{d}{dx}\left(\frac{\zeta(x)}{x}+\frac{1}{2x}-\frac{1}{x-1} \right )dx$$
Where $\zeta(x)$ is the Riemann zeta function. Of course, we can integrate by parts to obtain :

$$\frac{1}{2}-\int_{0}^{\infty}\left(\frac{\zeta(x)}{x}+\frac{1}{2x}-\frac{1}{x-1}\right) dx$$

Unfortunately, i don't know how to do this integral, and i don't have mathematica, hence the thread !!

2. Feb 1, 2013

3. Feb 1, 2013

### mmzaj

Re: Does this integral converge !?

And it doesn't converge . i kinda knew it , thanks . i was trying to evaluate a limiting case of the integral :
$$\int_{0}^{\infty}\frac{d}{dx}\left(\frac{\zeta(x)}{x}+\frac{1}{2x}-\frac{1}{x-1} \right )\ln\left(-e^{-zx}\right)dx$$
For $z\in \mathbb{C}$