fixing upper and lower limits $$\int_{-\infty}^\infty \frac{e^{-x} \, dx}{1-e^{-2x}}$$
Not really sure how to fix my upper and lower limits when I get through the first substitution. Anybody know what I should do to to fix my new upper and lower limits?
 A: This is $0$ because it's the integral of an odd function over an interval symmetric about $0$, at least if there are no convergence problems (I posted this thinking about what happens as $x\to\pm\infty$ and thought there are no convergence problems, but André Nicolas points out above that one must also consider $x=0$).
To see that it is odd, first notice that when you multiply the numerator and denominator by $e^x$ you get
$$
\frac{e^{-x}}{1-e^{-2x}} = \frac{1}{e^x - e^{-x}}
$$
and then notice what happens when you plug in $-x$ in place of $x$.
OK, just to get really concrete: if $0<a<b$ then
$$
\int_{-b}^{-a} \cdots + \int_a^b \cdots = 0
$$
because it's an odd function.  Letting $b\to\infty$ and $a\to0$, you see that the principal value is $0$.  But do we need a principal value, or is everything (at least among values of integrals) in sight finite?  That's the next thing to think about.
Later edit: OK, I'm not as rushed now.  Look at
$$
e^x - e^{-x} = (1+ x + \text{higher-degree terms}) - (1 - x + \text{higher-degree terms})
$$
$$
= 2x + \text{higher-degree terms}.
$$
So
$$
\frac{1}{e^x - e^{-x}} = \frac{1}{2x + (\text{h.d.t.})}.
$$
So the integral of this from $0$ to a positive number blows up, and so does the integral from $0$ to a negative number, and they have opposite signs.  The positive and negative parts are both infinite.  That's just the situation where Cauchy principal values are the thing to look for.  I said $a\to0$ above.  If one of the integrals were from $-\infty$ to $-2a$, and the other from $a$ to $\infty$ (without the "$2$"), then we'd actually get something other than $0$, because the positive part would be growing faster than the negative parts as $a\to0$.
