Integral related to $\psi(x)$

Question

I'm trying to solve this integral:

$$I(x)=\int_{0}^{\infty}\frac{\ln(t)\sin(2\pi x)\sinh(t)}{\left(\cosh(t)-\cos(2\pi x)\right)^2}dt,$$ where $0<x<1$.

But I can't find intelligent ways to approach it (with elementary substitutions it gets much more complicated).

The only thing I know is the solution to a very similar integral:

$$\int_{0}^{\infty}\frac{\ln(t)\left(1-\cos(2\pi x)\cosh(t)\right)}{\left(\cosh(t)-\cos(2\pi x)\right)^2}dt=\ln(2\pi)+\frac{\psi(x)+\psi(1-x)}{2},$$ where $\psi(x)$ is the digamma function.

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Graphing the first integral $I(x)$ saw that $I(1/2)=0$, so I imagine that the solution is something like $f(x)-f(1-x)$.

Since $2\cosh(x)=e^x+e^{-x}$ and $2 \sinh(x)=e^x-e^{-x}$ structurally I think it might make sense, but I'm not sure.

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Motivation of the question

I was trying to find a formula similar to what is asked in this question: I used the formula the OP posted on his IG page

Answer 1

Not an answer.

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Let, $$I(z)=\sin(z)\int_{0}^{\infty}\frac{\ln(t)\sinh(t)}{\left(\cosh t-\cos z \right)^2}dt$$

Substitute $t\mapsto-\ln t$, $$I(z)=2\int_0^1\ln(-\ln t)\left[\frac{(1-t^2)\sin z}{(t^2-2t\cos z+1)^2}\right]dt$$

Differentiate the following with respect to $t$, $$\sum_{r=1}^{\infty}t^{r}\sin rz=\frac{t\sin z}{t^{2}-2t\cos z+1}$$

$$\sum_{r=1}^{\infty}rt^{r-1}\sin(rz)=\frac{(1-t^2)\sin z}{(t^2-2t\cos z+1)^2}$$

Also,

$$I(z)=2\frac{d}{ds}\int_0^1(-\ln t)^{s-1}\left[...\right]dt\bigg|_{s=1}$$

Using these we get the following:

$$\boxed{\int_0^{\infty}t^{s-1}\frac{\ln(t)\sin(z)\sinh (t)}{(\cosh t-\cos z)^2}dt=2\frac{d}{ds}\Gamma(s)\sum_{r=1}^{\infty}\frac{\sin rx}{r^{s-1}}}$$

But the problem is it doesn't work for $s=1$.

I don't know how to proceed further.