Finding if functions converge or diverge

Question

$$\int_{0}^{\infty} \frac{1}{e^x-e^{-x}}dx$$

So I managed to integrate this getting something like $\tanh^{-1} (e^x) + C$ (or $\coth^{-1}$). The problem is I don't think my teacher wants us using hyperbolic trig so I'm wondering if there is a way to prove or integrate this without the use of hyperbolic trig. Comparison test maybe?

$0$ and $\infty$ are both improper. For comparison test it is $0 \leq f(x) \leq g(x)$: So could I let my $g(x) = \int_{1}^{\infty} \frac{1}{e^x}dx$ ?

$$\int_{1}^{\infty} \frac{e^x}{e^{2x}}dx = \int_{1}^{\infty} \frac{1}{e^x} dx$$ I am not sure what to do about the zero or how to explain this.

Answer 1

Change the limits after the substitution:$$\int_0^\infty\frac{e^x}{e^{2x}-1}dx=\int_1^\infty\frac{du}{u^2-1}=\frac12\int_1^\infty\left(\frac1{u-1}-\frac1{u+1}\right)du=$$

$$=\left.\frac12\log\frac{u+1}{u-1}\right|_1^\infty=\lim_{b\to\infty}\frac12\log\frac{b+1}{b-1}-\lim_{c\to0^+}\frac12\log\frac{c+1}{c-1}...$$

and this clearly diverges, because of the lower limit

Answer 2

Use the substitution $u=e^x$:

$$\int \frac{e^x}{e^{2x} - 1} \, \mathrm{d}x = \int \frac{u}{u^2 - 1} \frac{\mathrm{d}u}{u} = \int \frac{\mathrm{d}u}{u^2 - 1}$$

Then $\frac{1}{u^2 - 1} = \frac{1}{(u-1)(u+1)} =\frac{1}{2}\left(\frac{1}{u-1} - \frac{1}{u+1}\right)$ so

$$\frac{1}{2} \int \frac{1}{u-1} - \frac{1}{u+1} \, \mathrm{d}u = \frac{1}{2}\bigg[\log (u-1) - \log (u+1)\bigg] = \frac{1}{2} \log \left(\frac{e^x-1}{e^x+1}\right)$$

You can see this diverges.