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[tex] \int_{0}^{\infty}x\frac{d}{dx}\left(\frac{\zeta(x)}{x}+\frac{1}{2x}-\frac{1}{x-1} \right )dx[/tex]

Where [itex]\zeta(x)[/itex] is the Riemann zeta function. Of course, we can integrate by parts to obtain :

[tex]\frac{1}{2}-\int_{0}^{\infty}\left(\frac{\zeta(x)}{x}+\frac{1}{2x}-\frac{1}{x-1}\right) dx [/tex]

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

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# Does this integral converge ?

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