# 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?

-
I know I have to split the limits so going from -infinity to 0 plus 0 to infinity. –  EhBabay Feb 20 '13 at 21:55
Possibly you want the Principal Value, since in the ordinary sense of convergence there is trouble at $x=0$. –  André Nicolas Feb 20 '13 at 22:00
I've returned some hours after posting my answer and finished it, including an explanation of why you do need a Cauchy principal value for this. –  Michael Hardy Feb 21 '13 at 1:31
What do you mean by "the first substitution"? If you are doing this integral by substitution, no one can tell you how to adjust the limits of integration without knowing what substitution you propose to use. –  Gerry Myerson Feb 21 '13 at 1:53

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 $a\to0$.