# 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?

• If you are happy with a sum you can put $\frac{1}{n^x} = e^{-ln(n)x}$ and integrate – Paul Feb 12 at 12:48
• This is my mathematica code for the integrated Euler Maclaurin formula for the Riemann zeta function: pastebin.com/9wSSSmR4 You can skip the plot and go directly to the second part for faster execution. – Mats Granvik Feb 12 at 15:26
• What do you mean precisely by exist ? – Yves Daoust Feb 13 at 8:44

[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.

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

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

• Why to expand $e^{-x \log n}$? You can compute the integral directly according to Paul's comment: $$\int e^{-x \log n} \,dx = -\dfrac{e^{-x \log n}}{\log n}+c,$$ for $n > 1$. – Alex Silva Feb 13 at 9:31