I need to determine whether the improper integral converges or not:
$\int_{0}^{\infty}\left(\frac{1}{2x}-\frac{1}{e^{x}-e^{-x}}\right)dx$
any ideas how to start?
Edit: What I've done:
$$\int_{0}^{\infty}\frac{1}{2x}-\frac{1}{e^{x}-e^{-x}} = \int_{0}^{\infty}\frac{1}{2x} - \int_{0}^{\infty}\frac{1}{e^{x}-e^{-x}} = \frac{1}{2}\int_{0}^{\infty}\frac{1}{x} - \int_{0}^{\infty}\frac{1}{e^{x}-e^{-x}} = $$
It seems to me that the left one diverges but the integral actually converges
Thanks!
Note that for $x>1$:
$$\frac1{2x}-\frac1{e^x-e^{-x}}>\frac1{2x}-\frac1{e^x-\frac12e^x}=\frac1{2x}-2e^{-x}$$
But the integral
$$\int_1^\infty\frac1{2x}-2e^{-x}~\mathrm dx$$
diverges to $+\infty$, hence by the comparison test for divergence, the given integral diverges as well.
Note that the integrand has singularities at $x=0$ and $x\to \infty$. The singularity at $x=0$ is removable since we have
$$\begin{align} \frac1{2x}-\frac{1}{e^x-e^{-x}}&=\frac1{2x}-\frac{1}{2x+O(x^3)}\\\\ &=O(x)\,\,\text{as}\,\,x\to 0 \end{align}$$
However, we note that as $x\to \infty$, we have
$$\begin{align} \frac1{2x}-\frac{1}{e^x-e^{-x}}=\frac1{2x}+o\left(\frac1{x}\right) \end{align}$$
Therefore, the integral of interest diverges logarithmically.
