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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: enter image description here

The integrand for $t>w$ has been replaced by $\frac{1}{16t^2}$, resulting in the term $\frac{1}{16w}$.

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    $\begingroup$ @JackD'Aurizio it was simple curiosity. I think this is enough reason to consider an integral. $\endgroup$ – Nemo Jul 8 '17 at 18:27
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    $\begingroup$ @JackD'Aurizio I do not remember making such a claim. Is it so hard to read carefully what other people are saying? Maybe it is when you have such an inflated ego. $\endgroup$ – Nemo Jul 9 '17 at 9:25
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    $\begingroup$ Be Nice. $\endgroup$ – Glorfindel Jul 9 '17 at 9:45
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    $\begingroup$ Now I understand when people are saying what a hostile place MSE is becoming for asking questions... $\endgroup$ – Nemo Jul 9 '17 at 10:53
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    $\begingroup$ By numerical computation one can expect that $\int_0^1 \frac{t\ln(2\sinh t)}{\left(3t^2+\ln^2(2\sinh t)\right)^2}~dt=-\int_1^\infty \frac{t\ln(2\sinh t)}{\left(3t^2+\ln^2(2\sinh t)\right)^2}~dt$ $\endgroup$ – FDP Jul 9 '17 at 22:12
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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.

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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 $\displaystyle \int_0^{1} g(t)\,dt=-\int_{1}^{\infty} g(t)\,dt$

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