# On the convergence of $-\int_0^1\left(\int_1^\infty \left(\frac{1}{\zeta(x/y)}-1\right)dx\right)dy$

For fixed real numbers $1<x<\infty$ and $0<y<1$ one has that $$\frac{1}{\zeta\left(\frac{x}{y}\right)}-1=\sum_{k=2}^\infty \frac{\mu(k)}{k^{x/y}},$$ where $\mu(n)$ is the Möbius function and thus $\zeta(t)$ is the Riemann Zeta function for a real argument $t>1$.

I am asked myself if it is possible to get a good approximation, and thus show also the convergence of

$$I:=-\int_0^1\left(\int_1^\infty \left(\frac{1}{\zeta(x/y)}-1\right)dx\right)dy.$$

Notice that since I've created this exercise I believe that I can write by integration that $$I=-\sum_{k=2}^\infty\int_0^1\frac{\mu(k)yk^{-1/y}}{\log k}dy=\sum_{k=2}^\infty\frac{\mu(k)\left(k\log^2 k\left(\operatorname{Chi}(\log k)-\operatorname{Shi}(\log k)\right)-1+\log k\right)}{2k\log k}.$$

See if you need the codes from Wolfram Alpha online calculator to justify previous identity

integrate 1/k^(z/y) dz, from z=1 to z=infinite

integrate (y k^(-1/y))/(log(k)) dy, from y=0 to 1

Also as remark I believe that $$\sum_{k=2}^\infty\mu(k)\log (k)\left(\operatorname{Chi}(\log k)-\operatorname{Shi}(\log k)\right)$$ is convergent

sum mu(k)log(k)(Chi(log(k))-Shi(log(k))), from k=2 to 100

Question. Can you prove that $$-\int_0^1\left(\int_1^\infty \left(\frac{1}{\zeta(x/y)}-1\right)dx\right)dy$$ does converge? Can you provide us a good approximation of $I$? Thanks in advance.

• You can write $$I = \int_{1}^{\infty} \frac{1}{2}\left(1 - \frac{1}{x^2}\right)\left(1 - \frac{1}{\zeta(x)} \right) \, dx,$$ from which convergence easily follows. – Sangchul Lee May 9 '17 at 23:35
• Many thanks @SangchulLee feel free to add an answer (or what is the trick to set your integral). That you've a good day! – user243301 May 10 '17 at 5:34

I'll address convergence. Letting $x=yt$ gives

$$\int_0^1\int_1^\infty \left(1-\frac{1}{\zeta(x/y)}\right)dx\,dy = \int_0^1y\int_{1/y}^\infty \left(1-\frac{1}{\zeta(t)}\right)dt\,dy.$$

Since $1/y\ge 1,$ the inner integral on the right is bounded above by

$$\tag 1 \int_{1}^\infty \left(1-\frac{1}{\zeta(t)}\right)dt.$$

It's enough to show $(1)$ is finite. In fact, since the integrand in $(1)$ bounded, it's enough to prove the integral over $[2,\infty)$ is finite.

Note that

$$1-\frac{1}{\zeta (t)} = \frac{\zeta (t)-1}{\zeta (t)} \le \frac{\zeta (t)-1}{1} = \sum_{n=2}^{\infty}\frac{1}{n^t}.$$

For $t\ge 2,$ the last expression is no more than

$$\frac{1}{2^t}\sum_{n=2}^{\infty}\frac{2^t}{n^t}=\frac{1}{2^t}\sum_{n=2}^{\infty}\left (\frac{2}{n}\right)^t \le \frac{1}{2^t}\sum_{n=2}^{\infty}\left (\frac{2}{n}\right)^2 = \frac{1}{2^t}\cdot 4 \sum_{n=2}^{\infty}\left (\frac{1}{n}\right)^2 < \frac{1}{2^t}\frac{4\pi^2}{6}.$$

Since $2^{-t}$ is integrable over $[2,\infty),$ we have the desired convergence.

• It's very very clear, thank you very much. – user243301 May 10 '17 at 5:31