# Alternating series combining harmonic number and zeta values

While evaluating the following fractional part integral, I get stuck on an almost euler sum as highlighted in red colour. Could someone evaluate the red series in terms of well-known constants ? $$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\bigg\{\frac{1}{1-x\,y\,z}\bigg\}dx\,dy\,dz$$ $$=\color{red}{\sum_{n\geq1}(-1)^{n+1}\frac{\text{H}_{n}\zeta(1+n)}{n(1+n)}}-\frac{1}{2}\zeta(2)-\frac{5}{4}\zeta(3)$$

• It might be worth noting that $$\sum_{n\geq1} (-1)^{n+1} \frac{1}{n(1+n)} = \log(4)-1.$$ – Klangen May 27 at 10:47

Like Yuriy S., I think the constant $$S$$ is unlikely to be solved in terms of well-known constants. Below is a one-integral representation: $$S = \sum_{n=1}^\infty (-1)^{n+1} \frac{H_n}{n} \frac{\zeta(n+1)}{n+1} = -\int_0^1 \frac{dx}{x^2} \ \log{(1-x)}\big( \log{\Gamma(1+x)} +\gamma \ x \big) .$$

To derive it, use $$\sum_{n=1}^\infty (-x)^{n+1} \frac{\zeta(n+1)}{n+1} = \log{\Gamma(1+x)} +\gamma \ x$$ and $$\quad (*) \quad -\int_0^1 \frac{dx}{x} \ x^n \log{(1-x)} = \frac{H_n}{n}.$$

Obviously, $$S$$ can be constructed by inserting (*) into it's definition, interchanging $$\sum$$ and $$\int,$$ and recognizing the resulting series as that for the log of Euler's gamma function. It has been checked numerically.

• The (*) equation can't be right: the left hand side doesn't depend on $n$. Is there a typo? – Yuriy S May 31 at 7:39
• @Yuriy S. You are correct. I forgot the $x^n.$ I'll edit the answer. – skbmoore May 31 at 14:41
• Great answer in any case, +1 – Yuriy S May 31 at 14:56

I highly suspect that the series in question can't be evaluated in terms of known constants, so the following is not an answer, but rather an extended comment.

I'll try to find an integral expression for the series, without using the OP's original integral.

$$S=\sum_{n=1}^\infty (-1)^{n+1}\frac{\text{H}_{n}\zeta(n+1)}{n(n+1)}$$

Let's use the usual integral representations for the Harmonic numbers and the Zeta function:

$$H_n=\int_0^1 \frac{1-x^n}{1-x}dx$$

$$\zeta(n+1)=\frac{1}{n!} \int_0^\infty \frac{y^n dy}{e^y -1}$$

Also let's represent:

$$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$

We need to sum two supplementary series:

$$f(t)=\sum_{n=1}^\infty (-1)^{n+1} \frac{t^n}{n! n}=\int \frac{1-e^{-t}}{t} dt =\ln t-\operatorname{Ei}(-t)$$

$$g(t)=\sum_{n=1}^\infty (-1)^{n+1} \frac{t^n}{(n+1)!}=\frac{e^{-t}-1}{t}+1$$

So we have:

$$S=\int_0^1 \frac{dx}{1-x} \int_0^\infty \frac{dy}{e^y -1} \left(f(y)-g(y)-f(x y)+g(x y) \right)$$

Let's simplify the expression:

$$f(y)-f(x y)= -\ln x-\operatorname{Ei}(-y)+\operatorname{Ei}(-x y)$$

Note that for $$t \to 0$$ the following holds:

$$\operatorname{Ei}(-t)=\ln t+\gamma-t + \frac{t^2}{4}+O(t^3)$$

It's easy to check that for $$x y \to 0$$ the resulting expression is proportional to $$(1-x) y$$ which makes both the integrals convergent.

Can't really think of anything better for now than this ugly double integral:

$$S=\int_0^1 \frac{dx}{1-x} \int_0^\infty \frac{dy}{e^y -1} \left(\operatorname{Ei}(-x y)-\operatorname{Ei}(-y)-\ln x-\frac{e^{-y}-1}{y}+\frac{e^{-x y}-1}{x y} \right)$$

I will update with the numerical value from Mathematica. The series' convergence is not very fast.