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w$ has been replaced by $\frac{1}{16t^2}$, resulting in the term $\frac{1}{16w}$.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. $\endgroup$ – user82588 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. $\endgroup$ – user82588 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... $\endgroup$ – user82588 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

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:


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:


Or immediately:


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.


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, $\displaystyle \alpha=\ln\left(\frac{1+\sqrt{5}}{2}\right)$

$\begin{align}\int_0^{\alpha} g(t)\,dt+\int_{\alpha}^{\infty} g(t)\,dt&= \left(\int_0^{1}g(t)\,dt-\int_{\alpha}^1 g(t)\,dt\right)+\left(\int_1^{+\infty}g(t)\,dt+\int_{\alpha}^1 g(t)\,dt\right)\\ &=\int_0^{1} g(t)\,dt+\int_{1}^{\infty} g(t)\,dt \end{align}$


$\displaystyle \int_0^{\infty}g(t)\,dt=0$ is equivalent to $\displaystyle \int_0^{1} g(t)\,dt=-\int_{1}^{\infty} g(t)\,dt$


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