Definite integral of quotients of exponentials I am trying to solve the following integral:
$$
   I = \int_{0}^{+\infty} \dfrac{e^{-at}}{(e^{bt} + e^{-bt})^2} dt
$$
Now by reason, at $t=0$, the integral evaluates to $\frac{1}{4}$, and as $t \rightarrow +\infty$, the whole expression tends to $0$.
I don't think that there are any remarkable rules to be employed here, and integral by parts seems to make things worse (due to the denominator). I thought about substitution but the different $a$ and $b$ multiplier doesn't seem to make it easy.
A lazy check with WolframAlpha did not give anything better. It did perform a limited Taylor expansion around $t = 0$, which I guess only evaluates best when $t$ is in the neighbourhood of $0$.
Many checks across the web seems to give no useful hints, so I'm starting to think that maybe another approach has to be taken. Can anyone guide me if such an approach exists?
Thanks,
 A: $$
\frac{e^{-at}}{(e^{bt}+e^{-bt})^2} = \frac{e^{-(a+2b)t}}{(1+e^{-2bt})^2}
$$
now using for $|x| < 1$
$$
\frac{1}{(1+x)^2} = \sum_{k=1}^{\infty}(-1)^{k-1}k x^{k-1}
$$
then
$$
\frac{e^{-at}}{(e^{bt}+e^{-bt})^2} = e^{-(a+2b)t}\sum_{k=1}^{\infty}(-1)^{k-1}k e^{-2b(k-1)t}
$$
and thus
$$
\int_0^{\infty}\frac{e^{-at}}{(e^{bt}+e^{-bt})^2}dt = \int_{0}^{\infty}e^{-(a+2b)t}\sum_{k=1}^{\infty}(-1)^{k-1}k e^{-2b(k-1)t}dt
$$
which gives a series etc.
NOTE
$$
c_k = \int_0^{\infty}(-1)^{k-1} k e^{-(a+2b+2b(k-1))t}dt = \frac{(-1)^{k-1} k}{a+2kb}
$$
or
$$
\int_0^{\infty}(-1)^{k-1} k e^{-(a+2b+2b(k-1))t}dt = \sum_{k=1}^{\infty}c_k
$$
This series representation is well behaved for the integral
$$
\int_{\epsilon}^{\infty}(-1)^{k-1} k e^{-(a+2b+2b(k-1))t}dt \ \ \mbox{with} \ \ \epsilon > 0
$$
or
$$
\int_{\epsilon}^{\infty}(-1)^{k-1} k e^{-(a+2b+2b(k-1))t}dt  = \sum_{k=1}^{\infty}\frac{(-1)^{k-1} k e^{-\epsilon  (a+2 b k)}}{a+2 b k}
$$
for instance for $a=1, b= 1, \epsilon = 0.01$ and for $N = 500$ we have
$$
\int_{0.01}^{\infty}(-1)^{k-1} k e^{-(a+2b+2b(k-1))t}dt  \approx 0.140212\\
\sum_{k=1}^{500}\frac{(-1)^{k-1} k e^{-\epsilon  (a+2 b k)}}{a+2 b k}\approx 0.140201
$$
