Absolute Convergence of Improper Integral of sinx/(e^x-e^(-x))

Question

I want to prove absolute convergence of improper integral ;

\begin{equation} \int_0^{\infty} \dfrac{\text{sin}x}{e^x-e^{-x}} dx. \end{equation}

But I couldn't prove this. (I don't have to calcurate the value. I only have to prove absolute convergence.)

My attempt is as follows.

$ \left| \dfrac{\text{sin}x}{e^x-e^{-x}} \right| \leqq \dfrac{1}{e^x-e^{-x}} \\ \int_{\epsilon}^N \dfrac{1}{e^x-e^{-x}} dx =_{(t=e^x)} \int_{e^{\epsilon}}^{e^N} \dfrac{1}{t^2-1} dt = \dfrac{1}{2} \left[ \text{ln} \left| \dfrac{t-1}{t+1} \right| \right]^{e^{N}}_{e^{\epsilon}} $

I want to let $\epsilon \rightarrow 0$ and $N \rightarrow \infty.$

However, if I let $\epsilon \rightarrow 0$, $ \text{ln} \left| \dfrac{e^{\epsilon}-1}{e^{\epsilon}+1} \right| \rightarrow - \infty $

I would like you to give me any ideas or hints to prove absolute convergence of

\begin{equation} \int_0^{\infty} \dfrac{\text{sin}x}{e^x-e^{-x}} dx. \end{equation}

Answer 1

The integrand is $$\frac 12 \sin (x) \text{csch}(x)$$ Composing Taylor series, you have $$\frac 12 \sin (x) \text{csch}(x)=\frac{1}{2}-\frac{x^2}{6}+O\left(x^4\right)$$ so, no problem around $x=0$

Answer 2

For $x>1$ we have $$\left|\frac{\sin x}{e^x-e^{-x}}\right|\leq \frac{1}{e^{x/2}}$$

because for $x>1$ $$e^x-e^{-x}>e^{x/2}$$

thus $$\int_1^{\infty} \frac{\sin x}{e^x-e^{-x}} \,dx\le \int_1^{\infty} \frac{dx}{e^{x/2}} $$ as the last integral converges, the given integral converges at $+\infty$.

It converges at $x=0$ as proved above by Mr.Claude Leibovici:

as $$\frac{\sin (x)}{e^x-e^{-x}}=\frac{1}{2}-\frac{x^2}{6}+O\left(x^3\right)$$ the integral converges at $x=0$ either.

Its value is $\frac{1}{4} \pi \tanh \left(\frac{\pi }{2}\right)$

Thanks to contributor Gary for the useful collaboration.