Can $\int_{2}^{\infty} [\zeta(x)-1] dx $ be evaluated?

I wonder whether an exact evaluation of $$\int_{2}^{\infty} [\zeta(x)-1] dx $$ can be obtained. Its sum analogue is already known: $$\sum_{n=2}^{\infty} [\zeta(n)-1] = 1. $$ Presumably, finding the indefinite integral is pretty difficult or even impossible, but I suspect the definite version could be found. So far, I have only found results on the mean values of the Riemann zeta function, but nothing on the integral mentioned above.

The two expressions are related via the Euler-Maclaurin summation formula, but thus far I have not been able to use it to calculate the integral.

This is not an answer \begin{align*}\begin{split}\int_2^{\infty}[\zeta(x)-1]\,dx &= \int_2^{\infty} \sum_{n=2}^{\infty} \frac{1}{n^x}\,dx=\sum_{n=2}^{\infty} \frac{1}{n^2\ln(n)}\\ &\sim \int_2^{\infty} \frac{1}{x^2\ln(x)}\,dx+\frac{0+\frac{1}{4\ln(2)}}{2}-\sum_{k=1}^{\infty} \frac{B_{2k}}{(2k)!}f^{(2k-1)}(2)\\ &= -\mathrm{Ei}(-\ln(2))+\frac{1}{\ln(2)}\bigg[\frac{1}{8}+\frac{1}{48}-\frac{1}{960}\bigg]\\ &+\frac{1}{\ln^2(2)}\bigg[\frac{1}{96}-\frac{13}{11520}\bigg]-\frac{1}{\ln^3(2)}\bigg[\frac{1}{1280}-\frac{29}{96768}\bigg]+\cdots\\ &\approx 0.60544739 \end{split}\end{align*}

A try, not a final answer yet, I will update and have to check and correct stuff.

We need to find: $$\sum_{c=1}^{\infty} \frac{1}{(c+1)^2\ln(c+1)}$$

We know from the product series of ln(c+1)/c Found similair to the product formula/refined stirling numbers found here How to find a power series representation for a divergent product?

$$\ln\left({\ln(1+c)\over c}\right)=\sum_{n=1}^{\infty}\int_{-1}^0 \frac{(n+x)!}{n! x! n}c^n (-1)^n dx$$

$$ \frac{1}{(c+1)\ln(c+1)} =1/c+\sum_{n=1}^{\infty}\int_{-1}^0 \frac{(n+x)!}{n! x!}c^{n-1} (-1)^n dx$$

$$ \frac{1}{(c+1)^2\ln(c+1)} =\frac{1}{c(c+1)}+\sum_{n=1}^{\infty}\int_{-1}^0 \frac{(n+x)!}{n! x!}\frac{c^{n-1}}{c+1} (-1)^n dx$$ $$=\frac{1}{c(c+1)}+\sum_{k=0}^{\infty}\sum_{n=1}^{\infty}\int_{-1}^0 \frac{(n+x)!}{n! x!}c^{n-1}c^{k} (-1)^{n+k} dx$$

Sum it diagonally under n+k=p

$$=\frac{1}{c(c+1)}+\sum_{p=1}^{\infty}\sum_{n=1}^{p}\int_{-1}^0 \frac{(n+x)!}{n! x!}c^{p-1} (-1)^{p} dx$$

$$=\frac{1}{c(c+1)}+\sum_{p=1}^{\infty}\int_{-1}^0 \bigg(\frac{(p+x+1)!}{p! (x+1)!}-1\bigg)c^{p-1} (-1)^{p} dx$$

$$=\frac{1}{c(c+1)}+\sum_{p=1}^{\infty}\int_{0}^1 \bigg(\frac{(p+x)!}{p! (x)!}-1\bigg)c^{p-1} (-1)^{p} dx$$

We can now extend all it's powers over all p, in order to regularize it over c. $$=1+\sum_{p \in \mathbb{Z}}^{\infty}\int_{0}^1 \bigg(\frac{(p+x)!}{p! (x)!}-1\bigg) \zeta(1-p) (-1)^{p} dx$$

If p goes to limit 0, it's $-\gamma$ If p<0 it's 0.

Now we have now over all positive integers, which we need to regularize.

$$=1-\gamma+\sum_{p \in \mathbb{N}}^{\infty}\int_{0}^1 \bigg(\frac{(p+x)!}{p! (x)!}-1\bigg) \zeta(1-p) (-1)^{p} dx$$

Unfortunally I don't know the answer yet. I've to do some rewriting.

To check Because regularised sum over the negative zeta's (e.g. postive numbers) will be roughly equal to it's first few terms, before it diverges, we can actually see it moves around the value to be found.

We know numerically the answer is ~0.6055. Thuse

0.6055~=+1-0,577+1/4-11/144+1181/720/120-~0.009= between 0.61 and 0.60 so it's somewhat okay, especially when refined a little by taking the second order average of the alternating value's.

Alternative derivation:

$$\sum_{n=1}^{\infty} \frac{1}{(n+1)^2\ln(n+1)}=\sum_{n=1}^{\infty}\frac{1}{(n+1)n\ln(n+1)}-\frac{1}{(n+1)^2n\ln(n+1)}=$$$$\sum_{n=1}^{\infty}\sum_{j=1}^{\infty} \frac{(-ln(n+1))^{j-1}}{j!n(n+1)} $$ $$1+\sum_{n=1}^{\infty}\sum_{j=2}^{\infty} \frac{(-ln(n+1))^{j-1}}{(j)!n(n+1)} $$

$\frac{(-ln(n+1))^{j-1}}{j!n(n+1)}$ when extended over n writen as polynomal are the stirling numbers of the first $-S(n,j)/j/(n-1)!$ So you get

$$\sum_{j=2}^{\infty}\sum_{n=1}^{\infty} \frac{-S(n,j) \zeta(j-n)}{j*(n-1)!} $$

Or if use them unisgned

$$\sum_{j=2}^{\infty}\sum_{n=1}^{\infty} )\frac{(-1)^{n+1}S(n,j) \zeta(j-n)}{j*(n-1)!} $$

Which is equal to the integral/factorial above when writen out. However the difference is with this formula you need to extend it beyond n<1.

Basically this:

$$\sum_{p =1}^{\infty}\int_{0}^1 \bigg(\frac{(p+x)!}{p! (x)!}-1\bigg) \zeta(1-p) (-1)^{p}/c^p dx=\sum_{j=2}^{\infty}\sum_{n=1}^{\infty} \frac{-S_1(n,j) \zeta(n-j)}{j*(n-1)!} $$ $$ \sum_{n=1}^{\infty}\sum_{j=2}^{\infty} \frac{(-ln(n+1))^{j-1}}{(j)!n(n+1)}=\sum_{j=2}^{\infty}\sum_{n\in \mathbb{Z}}^{\infty} \frac{-S_1(n,j) \zeta(n-j)}{j*(n-1)!} =$$

Proof for the negative side here $$\sum_{n=-\infty}^{0} \frac{-S_1(n,j) \zeta(n-j)}{j*(n-1)!} =(-1)^{j+1}\zeta(j)/j$$ $$\sum_{j=2}^{\infty}(-1)^{j+1}\zeta(j)/j=-\gamma$$

Continuation/some more rewriting: You can also rewrite the previous found integral, as those coefficients turned out to be common:

$$=1-\gamma+\sum_{p=1}^{\infty}\int_{0}^1 \bigg(\frac{(p+x)!}{p! (x)!}-1\bigg) \zeta(1-p) (-1)^{p} dx$$

If W_i(n) are the bernoulli polynomals of the second kind then: $$\int_{0}^1 \bigg(\frac{(p+x)!}{p! (x)!}\bigg)=W_p(p)$$ Turns out there are Gregory $G_i$ coefficients which are related to them.

$$W_p(p)-1=\sum_{i=1}^{p*}\frac{p!}{(p-i)!(i)!} G_{i}$$

So if we rewrite our to calculate sum part: $$\sum_{p=1}^{\infty} \sum_{i=1}^{n*}\frac{p!}{(p-i)!(i)!} \zeta(1-p) (-1)^{p} G_{i} =$$

Remove the first term because I think it might help $$-\zeta(0)(G_1)+\sum_{p=2}^{\infty} \sum_{i=1}^{n*}\frac{p!}{(p-i)!(i)!} \zeta(1-p) G_{i} =$$ $$1-\gamma+1/4+\sum_{p=2}^{\infty} \sum_{i=1}^{n*}\frac{p!}{(p-i)!(i)!} \zeta(1-p) G_{i} =$$

We know that: $$n!=\bigg(\frac{n}{e}\bigg)^{n}\big(\sqrt{2 n\pi}\big) \exp\bigg(-\sum_{j=1}^{\infty}\frac{\zeta(-j)}{j(n)^{j}}\bigg)$$

$$\frac{d}{dn} \ln(n!)=\ln(n)+\frac{1}{2n}+\bigg(\sum_{j=1}^{\infty}\frac{\zeta(-j)}{(n)^{j+1}}\bigg)$$

I've haven't often used the polgamma yet, but it's nice to introduce it: $\frac{d^s}{dn}\ln(n!)=\psi_{s-1}(n+1)$

$$\psi_{0}(n+1)=\ln(n)+\frac{1}{2n}+\bigg(\sum_{j=2}^{\infty}\frac{\zeta(1-j)}{(n)^{j}}\bigg)$$

m=1/n $$\frac{d^s}{d^sn}\bigg(\sum_{j=2}^{\infty}\zeta(1-j)(n)^{j}\bigg)= \bigg(\sum_{j=2}^{\infty} \frac{j!}{(j-s)!}\zeta(1-j)(m)^{j-s}\bigg)$$

$$\frac{d^s}{d^sn}\bigg(\psi_{0}(1+1/m)+\ln(m)-m/2 \bigg) = \bigg(\sum_{j=2}^{\infty} \frac{j!}{(j-s)!}\zeta(1-j)(m)^{j-s}\bigg)$$

$$\frac{d^s}{d^sn}\psi_{0}(1+1/m)=(-1)^s\sum_{k=0}^{s-1}\frac{\frac{s!}{(s-k)!k!}\frac{(s-1)!}{(s-k-1)!}\psi_{s-k}(1+1/m)}{m^{2s-k}}$$

With m goes to 1

$$\bigg(\sum_{j=2}^{\infty} \frac{j!}{(j-s)!}\frac{\zeta(1-j)}{}\bigg)=\bigg((-1)^s\sum_{k=0}^{*s-1}\frac{s!}{(s-k)!k!}\frac{(s-1)!}{(s-k-1)!}\psi_{s-k}(2)\bigg)+\bigg|\frac{d^s}{d^sn} ln(m)-m/2 \bigg|_{m=1}$$

$$\bigg|\frac{d^s}{d^sn} ln(m)\bigg|_{m=1}=(-1)^{s+1}(s-1)!$$

the derivative of -m/2 is only relevant if s=1, which gives -1/2. The other coefficents are $\gamma$ and $-1/4$

$$1-\gamma+1/4+\sum_{p=2}^{\infty} \sum_{s=1}^{p*}\frac{p!}{(p-s)!(s)!} \zeta(1-p) G_{s} =$$

$$1+ \sum_{s=1}^{\infty}\sum_{k=0}^{*s-1}(-1)^sG_{s}\bigg(\frac{1}{(s-k)!k!}\frac{(s-1)!}{(s-k-1)!}\psi_{s-k}(2)\bigg) =$$

$$1+ \sum_{s=1}^{\infty}\sum_{k=0}^{*s-1}(-1)^sG_{s}\bigg(\frac{1}{(s-k)!k!}\frac{(s-1)!}{(s-k-1)!} (s-k)!((-1)^{1+s-k}\zeta(1+s-k)-1)\bigg) =$$

$$1+ \sum_{s=1}^{\infty}\sum_{k=0}^{*s-1}(-1)^{1+k}G_{s}\bigg(\frac{1}{k!}\frac{(s-1)!}{(s-k-1)!} (\zeta(1+s-k)-1)\bigg) =$$

$$1+ \sum_{s=1}^{\infty}\sum_{j=1}^{s}(-1)^{1+s-j}G_{s}\bigg(\frac{1}{(s-j)!}\frac{(s-1)!}{(j-1)!} (\zeta(1+j)-1)\bigg) =$$

The -1 part is 0 when summe over j, unless s=1 and at s=1, $G_1=1/2$

$$ \sum_{s=1}^{\infty}\sum_{j=1}^{s}(-1)^{1+s-j}G_{s}\bigg(\frac{1}{(s-j)!}\frac{(s-1)!}{(j-1)!} (-1)\bigg) =1/2$$

So now we have rewriten it to: $$3/2+ \sum_{s=1}^{\infty}\sum_{j=1}^{s}(-1)^{1+s-j}G_{s}\bigg(\frac{1}{(s-j)!}\frac{(s-1)!}{(j-1)!} \zeta(1+j)\bigg)$$

Probably could been derived much easier but I thought it was cool. Obviously there must be some rule for how and when you can rewrite a positive power function into an negative one. I think it got something to do with asymtotics, but maybe someone else could explain it, as i am just starting to figure it.