What is the result of the complex integral with Riemann zeta function?

[eljose]

let be the integral..where $$\zeta(s)$$ is the Riemann zeta function.

$$\int_{c-i\infty}^{c+i\infty}ds\zeta(s)(x^{s}/s)$$

then what would be the result?..there would be two singularities at the points s=0 and s=1 the problem is if there would be any other singularitiy on the integral

 

[shmoe]

This is a straightforward application of Perron's if c>1 and x>0. You surely know where the poles of zeta are by now, no?

 

[eljose]

well i see two poles inside the integral..so i would get the result:

$$A+Bx$$ (poles at s=0 and s=1 with A and B real constants) but it seems a very easy integral, i would expected a sum over the zeros of Riemann Zeta or something similar..uummm..perhaps i have made something wrong.

 

[shmoe]

well i see two poles inside the integral..so i would get the result:

$$A+Bx$$ (poles at s=0 and s=1 with A and B real constants)

How did you get this? What contour did you try to apply the residue theorem to? If you want to say something about an unbounded contour you can't apply the residue theorem directly, you have to look at bounded contours and look at limits.

i would expected a sum over the zeros of Riemann Zeta or something similar..uummm..perhaps i have made something wrong.

You're seen perron's formula, look at it again closely. There's no way you should expect a sum over the zeros of zeta here.