Show this $\int_0^\infty \frac{t\ln(2\sinh t)}{\left(3t^2+\ln^2(2\sinh t)\right)^2}~dt=0$ While evaluating the integral
$$
I_1=\int_{0}^\infty\frac{\sin\pi x~dx}{x\prod\limits_{k=1}^\infty\left(1-\frac{x^3}{k^3}\right)},\tag{1}
$$
I came to this integral of elementary function
$$
I_2=\int_0^\infty \frac{dt}{\left(i t\sqrt{3}+\ln(2\sinh t)\right)^2}.\tag{2}
$$
In fact $I_2$ is real and
$$
I_1=-2\pi I_2.
$$
These formulas imply the closed form
$$
\int_{0}^{\infty}\frac{t\ln\left(\,2\sinh\left(\,t\,\right)\,\right)}{\left[\,3t^{2} + \ln^{2}\left(\,2\sinh\left(\,t\,\right)\,\right)\right]^{\,2}}\,{d}t = 0,\tag{3}
$$
or alternatively
$$
\text{Im}\int_0^\infty \frac{dt}{\left(i t\sqrt{3}+\ln(2\sinh t)\right)^2}=0.
$$
Brief outline of proof is as follows. Write the infinite product in terms of Gamma functions, apply reflection formula for Gamma function to get rid of $\sin\pi x$, then use integral representation for Beta function and change the order of integration. Then one can integrate over $x$ to obtain the desired formula.
It seems that this should have a simple proof, but I don't see it. 

Q: Can anybody provide a direct proof ?. 

Such a direct proof may shed light on possible routes to calculation or simplification of $(2)$.
Here is a numerical demonstration using Mathematica that the integral under consideration is $0$ up to at least $100$ digits:

The integrand for $t>w$ has been replaced by $\frac{1}{16t^2}$, resulting in the term $\frac{1}{16w}$.
 A: This answer directly proves that:
$$\text{Im}\int_0^\infty \frac{dt}{\left(i t\sqrt{3}+\ln(2\sinh t)\right)^2}=0$$
First, we make a change of variable:
$$x=e^{-2t}$$
Which transforms the identity to:
$$\text{Im} \int_0^1 \frac{dx}{x \left(\ln(1-x)-e^{\pi i/3} \ln x \right)^2}=0$$
Finding the imaginary part explicitly, we now need to prove:
$$ \int_0^1 \frac{\ln x \ln(1-x)-\frac{1}{2} \ln^2 x}{x \left(\ln^2 x+\ln^2(1-x)-\ln x \ln (1-x)\right)^2}dx=0$$

Let's introduce a function:

$$f(x)=f(1-x)=\ln^2 x+\ln^2(1-x)-\ln x \ln (1-x)$$

As we have a difference of two positive definite functions under the integral, the identity is equivalent to:

$$\int_0^1 \frac{\ln x \ln(1-x)}{x f(x)^2}dx=\frac{1}{2}\int_0^1 \frac{ \ln^2 x}{x f(x)^2}dx$$

Let's denote the integrals $J_1$ and $J_2$. We need to prove that $J_1=J_2$.
Using the substitution $x \to 1-x$ we can prove the following identities:
$$J_1=\int_0^1 \frac{\ln x \ln(1-x)}{(1-x) f(x)^2}dx=\frac{1}{2} \int_0^1 \frac{\ln x \ln(1-x)}{x (1-x) f(x)^2}dx$$
$$J_2=\frac{1}{2}\int_0^1 \frac{ \ln^2 (1-x)}{(1-x) f(x)^2}dx=\frac{1}{4}\int_0^1 \frac{ (1-x)\ln^2 x+x\ln^2 (1-x)}{x(1-x) f(x)^2}dx$$
Subtracting the two forms of $J_2$ gives us another set of identities:
$$J_3=\int_0^1 \frac{ \ln^2 x}{x (1-x)f(x)^2}dx=\int_0^1 \frac{ \ln^2 (1-x)}{x (1-x)f(x)^2}dx=\int_0^1 \frac{ \ln^2 x+\ln^2 (1-x)}{x f(x)^2}dx$$

From the above follows a relation:
$$J_3-J_1=\int_0^1 \frac{ 1}{x f(x)}dx$$
Now we use integration by parts with:
$$u(x)=\frac{ 1}{f(x)}, \qquad v(x)= \ln x$$
The limits for $u(x)v(x)$ at $0$ and $1$ are both equal to zero. After simplifications, we can write:
$$J_3-J_1=\int_0^1 \frac{(2-x) \ln^2 x-(1+x) \ln x \ln (1-x)}{x(1-x)f(x)^2}dx$$
Making a substitution $x \to 1-x$ and adding the two results, we obtain a symmetric form of the integral:
$$J_3-J_1=\frac{1}{2} \int_0^1 \frac{(2-x) \ln^2 x+(1+x) \ln^2 (1-x)-3 \ln x \ln (1-x)}{x(1-x)f(x)^2}dx$$
From the other identities above, it can be finally seen that:

$$J_3-J_1=J_3+2J_2-3J_1$$

Or immediately:
$$J_1=J_2$$
The proof is finished.

Remark. This doesn't use the other identity shown by  FDP in the comments. To me it looks very difficult to prove.
A: What is following is too lengthy for a comment.
Define $g(t)=\displaystyle \frac{t\ln(2\sinh t)}{\left(3t^2+\ln^2(2\sinh  t)\right)^2}$ on $[0;+\infty[$.
Observe that, for $\displaystyle t>\ln\left(\frac{1+\sqrt{5}}
{2}\right),g(t)>0$
and,
$\displaystyle t\leq\ln\left(\frac{1+\sqrt{5}}
{2}\right),g(t)\leq 0$
let, $\displaystyle \alpha=\ln\left(\frac{1+\sqrt{5}}{2}\right)$
$\begin{align}\int_0^{\alpha} g(t)\,dt+\int_{\alpha}^{\infty} g(t)\,dt&=
\left(\int_0^{1}g(t)\,dt-\int_{\alpha}^1 g(t)\,dt\right)+\left(\int_1^{+\infty}g(t)\,dt+\int_{\alpha}^1 g(t)\,dt\right)\\
&=\int_0^{1} g(t)\,dt+\int_{1}^{\infty} g(t)\,dt
\end{align}$
Therefore,
$\displaystyle \int_0^{\infty}g(t)\,dt=0$ is equivalent to $\displaystyle \int_0^{1} g(t)\,dt=-\int_{1}^{\infty} g(t)\,dt$
