Hi all, I am trying to solve an equation and in the way, I have come across the following.
I am stuck here since the whole day. I will really appreciate if someone can hint me any thing on the followin
$$\int_{0}^{\infty} e^{-ne^{-at}}dt$$
Hi all, I am trying to solve an equation and in the way, I have come across the following.
I am stuck here since the whole day. I will really appreciate if someone can hint me any thing on the followin
$$\int_{0}^{\infty} e^{-ne^{-at}}dt$$
$\begin{align*}\displaystyle \int_0^\infty e^{-ne^{-at}}dt & = \int_0^\infty \sum_{k\ge 0} \dfrac{(-n)^ke^{-akt}}{k!}dt \\ & = \displaystyle \sum_{k\ge 0} \int_0^\infty \dfrac{(-n)^ke^{-akt}}{k!}dt \\ & = \displaystyle \sum_{k\ge 0} \dfrac{(-n)^k}{ak(k!)}\end{align*}$
This might not work... The first term is undefined. Well, that was my guess :-P.
Hmm, let me go back to my idea one more time. That will give us:
$\begin{align*}\displaystyle \int_0^\infty \sum_{k\ge 0}\dfrac{(-n)^ke^{-akt}}{k!}dt & = \lim_{p \to \infty} \int_0^p \sum_{k\ge 0}\dfrac{(-n)^ke^{-akt}}{k!}dt \\ & = \displaystyle \lim_{p \to \infty} \sum_{k\ge 0} \int_0^p \dfrac{(-n)^ke^{-akt}}{k!}dt \\ & = \displaystyle \lim_{p \to \infty} \left[\int_0^p dt + \sum_{k\ge 1} \int_0^p \dfrac{(-n)^k e^{-akt}}{k!}dt\right] \\ & = \displaystyle \lim_{p \to \infty} \left[p+\sum_{k\ge 1} \dfrac{(-n)^k - (-n)^ke^{-akp}}{ak\cdot k!}\right]\end{align*}$
Perhaps this can be evaluated to something simpler.
The integration by parts suggested by HallsofIvy and chiro would both be infinite. There you have the added challenge that every term will be undefined (because you will be integrating from 1 to 0 with respect to y).
I forgot to mention, I was a bit cavalier with swapping the summation with the integral. In reality, you need to be careful. I technically should have taken the limit as $q\to \infty$ and taken the summation from 0 to $q$. Then, I would have needed to take care to make sure I could swap the two limits. I think it should be ok in this problem, but I really did not check it out enough.