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 .

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    $\begingroup$ If you are happy with a sum you can put $\frac{1}{n^x} = e^{-ln(n)x}$ and integrate $\endgroup$ – Paul Feb 12 at 12:48
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    $\begingroup$ 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. $\endgroup$ – Mats Granvik Feb 12 at 15:26
  • $\begingroup$ What do you mean precisely by exist ? $\endgroup$ – 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: $$\int \left(\sum_{n=1}^{\infty} \left(\sum_{k=0}^\infty \left( \frac {(-x)^k(\ln (n))^k}{k!} \right) \right)\right)dx= \sum_{n=1}^{\infty} \left(\sum_{k=0}^\infty \left( \frac {(-1)^k(x)^{k+1}(\ln (n))^k}{(k+1)k!} \right) \right)$$

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

  • $\begingroup$ 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 $n > 1$. $\endgroup$ – Alex Silva Feb 13 at 9:31

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