Can the integral be evaluated without Mathematica?

[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 !

 

[mfb]

Looks like the first part is fine
But for large x, your function approaches 1/(2x)+o(1/x^2)...

 

[mmzaj]

Looks like the first part is fine
But for large x, your function approaches 1/(2x)+o(1/x^2)...

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}