What would be the integral of the zeta function or $ \sum\limits_{n=1}^{\infty} \frac {1}{n^x} $? The zeta function is defined as:
$$ \zeta (x) = \sum\limits_{n=1}^{\infty} \frac {1}{n^x}  $$
Does an integral of this function exist? If it does then what would it be?
More information about zeta function can be found here .
 A: [Rough Calculation] You may take it as an answer. I have been calculating the integration in the usual way, assuming $x$ to be real. $C$ is arbitrary constant.
$\displaystyle \int \zeta(x)dx=x-(\frac{1}{2^x\log 2}+\frac{1}{3^x\log 3}+\dots)+C$
Now, I claim that the infinite sum converges.
Since, for all $n>1$ and $x>1$ we have $\displaystyle\frac{1}{n^x}>\frac{1}{n^x\log n}$ summing over $n=2$ to $\infty$ we get,
$$\zeta(x)-1>\sum_{n=2}^\infty\frac{1}{n^x\log n}$$ [I have not used weak inequality as I have not worked on the fact that when they will be equal]
The convergence is followed by the comparison test. As an overview we can say, $$\int\zeta(x)dx>x-1+\zeta(x)$$
Hope this works.
A: Okay I think I may have a solution, though it isn't pretty.
We can write $ \zeta (x) $ as:
$$ \sum_{n=1}^{\infty}{\frac {1}{n^x}}= \sum_{n=1}^{\infty}{e^{-x \ln (n)}}  $$
The expansion of $e^x$ is:
$$ e^x=\sum_{k=0}^\infty \frac {x^k}{k!}$$ 
Similarly the expansion of ${e^{-x \ln (n)}}$ is:
$$ \sum_{n=1}^{\infty}{\frac {1}{n^x}}=\sum_{n=1}^{\infty}{e^{-x \ln (n)}}=\sum_{n=1}^{\infty} \left(\sum_{k=0}^\infty \left( \frac {(-x)^k(\ln (n))^k}{k!} \right) \right) $$
Integrating the expression:
$$\int \left(\sum_{n=1}^{\infty} \left(\sum_{k=0}^\infty \left( \frac {(-x)^k(\ln (n))^k}{k!} \right) \right)\right)dx= \sum_{n=1}^{\infty} \left(\sum_{k=0}^\infty \left( \frac {(-1)^k(x)^{k+1}(\ln (n))^k}{(k+1)k!} \right) \right)$$

$$\therefore~ \int\zeta (x)dx=\int \left (\sum_{n=1}^{\infty}{\frac {1}{n^x}} \right) dx=  \sum_{n=1}^{\infty} \left(\sum_{k=0}^\infty \left( \frac {(-1)^k(x)^{k+1}(\ln (n))^k}{(k+1)k!} \right) \right)$$

