Inverse Mellin transform of $\Delta(t):=\sum_{s=1}^\infty d(s)\sqrt{\frac{t}{s}}M_1(4\pi \sqrt{ts})$

Question

Define

$$ \Delta(t):=\sum_{s=1}^\infty d(s)\sqrt{\frac{t}{s}}M_1(4\pi \sqrt{ts}) $$

where $M_1(z)=-Y_1{(z)}-\frac{2}{\pi}K_1(z)$ and $d(s)$ is the divisor function.

What is the inverse Mellin transform of $\Delta$?

$\Delta(t)$ is conjectured to grow on the order of magnitude of $O(t^{1/4+\epsilon})$ as $t\to \infty$.

I expanded and distributed the sum to obtain

$$ \Delta(t)= -\sum_{s=1}^\infty d(s) \sqrt{\frac{t}{s}}Y_1{(4\pi\sqrt{ts})}-\frac{2}{\pi}\sum_{s=1}^\infty d(s)\sqrt{\frac{t}{s}}K_1(4\pi \sqrt{ts}) $$

I found that

$$\Delta(t)=-\Pi(t)-\frac{4}{\pi}\int_{(0,1)}\sum_{s=1}^\infty d(s)~e^{\frac{16\pi^2 t}{\log x}}~x^{s-1} ~dx $$

where the integrand is the inverse Mellin transform of the $K_1$ terms and where $\Pi(t)$ is the remainder of the $Y_1$ terms. I'm not sure how to express $\Pi(t)$ as a Mellin transform of another function.

Here are the definitions I am using

$$ Y_n(z):=\lim_{v \to n} Y_v(z) $$

for $n\in \Bbb Z.$

and

$$ K_n(z)=\lim_{v \to n} K_v(z) $$

for $n\in \Bbb Z$.

Where

$$ Y_v(z)= \frac{J_v(z)(v\pi)-J_{-v}(z)}{\sin(v\pi)} $$

for $z\in \Bbb C$ and $v \notin \Bbb Z$.

and finally

$$ K_v(z)= \frac{\pi}{2} \frac{I_{-v}(z)-I_{v}(z)}{\sin(v\pi)} $$

for $z\in \Bbb C$ and $v\notin \Bbb Z$.

Answer 1

For the divisor summatory function

$$D(x)=\sum\limits_{n=1}^x d(n)\tag{1}$$

one has $$\zeta(s)^2=s\, \mathcal{M}_x[D(x)](-s)=s \int\limits_0^\infty D(x)\, x^{-s-1}\, dx=\sum\limits_{n=1}^\infty \frac{d(n)}{n^s},\quad\Re(s)>1\tag{2}$$

where the integral is evaluated term-wise.

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Wikipedia indicates $\Delta(x)$ is given by the inverse Mellin transform

$$\Delta(x)=\mathcal{M}_s^{-1}\left[\frac{\zeta(s)^2}{s}\right]\left(\frac{1}{x}\right)=\frac{1}{2 \pi i} \int\limits_{c'-i \infty}^{c'+i \infty} \zeta(s)^2\, \frac{x^s}{s} \, ds,\quad 0<c'<1\tag{3}$$

where

$$D(x)=x \log(x)+(2 \gamma-1)\, x+\Delta(x)\tag{4}.$$

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Defining $\Delta(x)$ slightly differently as

$$\Delta(x)=\Delta_k(x)+\Delta_Y(x)\tag{5}$$

where

$$D(x)=x \log(x)+(2 \gamma-1)\, x+\frac{1}{4}+\Delta(x)\tag{6},$$

$$\Delta_K(x)=-\frac{2}{\pi} \sqrt{x} \sum\limits_{n=1}^\infty \frac{d(n)}{\sqrt{n}} K_1\left(4 \pi \sqrt{n x}\right)\tag{7},$$

and

$$\Delta_Y(x)=-\sqrt{x} \sum\limits_{n=1}^\infty \frac{d(n)}{\sqrt{n}} Y_1\left(4 \pi \sqrt{n x}\right)\tag{8}$$

evaluating term-wise Mathematica gives the Mellin Transform

$$s\, \mathcal{M}_x\left[\Delta_K(x)\right](-s)=s \int\limits_0^{\infty} \Delta_K(x)\, x^{-s-1} \, dx\\=2^{2 s-1}\, \pi^{2 s-2}\, \Gamma(1-s)^2 \sum\limits_{n=1}^{\infty} d(n)\, n^{s-1}=\frac{1}{2}\, \csc\left(\frac{\pi s}{2}\right)^2\, \zeta(s)^2,\quad\Re(s)<0\tag{9}$$

where the condition is inconsistent with the condition in formula (3) above.

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This would seem to imply

$$s\, \mathcal{M}_x\left[\Delta_Y(x)\right](-s)=s \int\limits_0^{\infty} \Delta_Y(x)\, x^{-s-1} \, dx\\=\left(1-\frac{1}{2}\, \csc\left(\frac{\pi s}{2}\right)^2\right)\, \zeta(s)^2\tag{10}$$

in some sense.

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Evaluating term-wise Mathematica indicates the inverse Mellin transform of $\Delta_K(x)$ is

$$\mathcal{M}_x^{-1}\left[\Delta_K(x)\right](y)=\frac{1}{2 \pi i} \int\limits_{c-i \infty}^{c+i \infty} \Delta_K(x)\, x^{-y} \, dx=\\-\frac{2}{\log^2(y)} \sum\limits_{n=1}^{\infty} d(n)\, e^{\frac{4 n \pi^2}{\log(y)}}=-\frac{2 \left(\psi_{e^{\frac{4 \pi^2}{\log(y)}}}^{(0)}(1)+\log\left(1-e^{\frac{4 \pi^2}{\log(y)}}\right)\right)}{\log\left(e^{\frac{4 \pi^2}{\log(y)}}\right) \log^2(y)},\quad 0<y<1\tag{11}$$

where $\psi _q^{(n)}(z)$ is the q-polygamma function.

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The following figures illustrate the magnitude of $\Delta_K(x)$ decreases rapidly when evaluated along vertical lines, whereas the magnitude of $\Delta_Y(x)$ increases rapidly when evaluated along vertical lines which makes me suspect the inverse Mellin transform of $\Delta_Y(x)$ doesn't converge.

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Figure (1): Illustration of $\left|\Delta_K(2+i t)\right|$ where formula (7) is evaluated at $N=100$

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Figure (2): Illustration of $\left|\Delta_Y(2+i t)\right|$ where formula (8) is evaluated at $N=100$