Question about Improper Integrals I am trying to find whether the following integral converges or diverges, but I am a bit uncertain about the steps I'm taking. I have to following integral.
$$\int_{-1}^1 \frac{e^x}{e^x-1} \,dx \\$$
Because I know it is discontinuous at x=0, I split the integral
$$\int_{-1}^0 \frac{e^x}{e^x-1} \,dx + \int_{0}^1 \frac{e^x}{e^x-1} \,dx \\$$
Now I set up the limit
$$\lim\limits_{t \to 0} \int_{-1}^t \frac{e^x}{e^x-1} \,dx + \lim\limits_{t \to 0} \int_{t}^1 \frac{e^x}{e^x-1} \,dx \\$$
After integrating using the following substitution:
$$u={e^x-1} \, \\$$
I get the following:
$$\lim\limits_{t \to 0} \int_{\frac1e-1}^{e^t-1} \frac{1}{u} \,du + \lim\limits_{t \to 0} \int_{e^t-1}^{\frac1e-1} \frac{1}{u} \,du \\$$
So finally, I get 
$$\lim\limits_{t \to 0} {(ln(e^t-1)-ln(\frac1e-1))}  + \lim\limits_{t \to 0} (ln(\frac1e-1)-ln(e^t-1))  \\$$
So my question is, from the last line, if t is going to 0, shouldn't e^t tend to 1 and thus I would get ln (0) which would make the integral divergent. This is not the answer in the book. Where have I gone wrong?
(Also, sorry for formatting, I'm still learning how to use all the symbols.)

Let us study $$I=\int_0^1\frac{e^x}{e^x-1}dx$$
with the change $t=e^x$,
then $I$ becomes:
$$\int_1^e\frac{dt}{t-1}$$ which diverges
since
$$\lim_{X\to 1^+}\int_X^2\frac{dt}{t-1}=\lim_{X\to1^+}(-\ln(|X-1|))=+\infty.$$
Your integral is therefore divergent.
