"Nice" Solution to repeated integral I have a problem wherein I have defined a function $I_r(t) = \int e^{(2r-1)at} \int e^{(2r-3)at} \cdots \int e^{at} dt\cdots dt$, and $I_r(0) = 0$, for $r = 1,2,3,\ldots$.
I find that $e^{-ar^2t} I_r(t) = \left(1-e^{-at}\right)^r q(t)$, where $q(exp(-at))$ is a polynomial of $e^{-at}$. Is there a general technique for evaluating repeated integrals of this type that allows me to write $q$ in a nice clean way?
If I took $I^*_r(t) = \int e^{at} \int e^{at}\cdots \int e^{at} dt\cdots dt$ with $I^*_r(0) = 0$, and multiplied by $e^{-art}$, I would get $e^{-art}I^*_r(t) = (1-e^{-at})^r$. I am looking for a nice closed-form solution where I have a quadratic in $r$.
This is related to the derivation of a discrete probability distribution where the transition rate function is quadratic w/r.t. the number of events per cell.
 A: I don't think you'll get a closed form but you get get an expression that involves products and sums that isn't too horrible and is easy to evaluate for particular cases. It'll take me forever to type it up here with nice formatting but I'll describe the approach in a way that's easy to reproduce.
We're starting with the number 1 and repeatedly performing these two operations: integration and multiplication by $e^{nat}$ for various $n$. Switch to working with Laplace transforms. In Laplace transform space, integration (starting from zero) is simply multiplication by $1/s$ and multiplication by the exponential takes $f(s)$ to $f(s-na)$.
So the Laplace transform of $I_1$ is $1/(s(s-a))$, the Laplace transform of $I_2$ is $1/(s(s-3a)(s-4a))$ and so on. Notice how when we keep going, the denominator is made up of factors of the form $s-(r^2-i^2)a$. So it's easy to write the Laplace transform of $I_r$ using product notation. You now simply need to invert the Laplace transform. To do this, we need to write our rational function in $s$ as partial fractions. The factors in the denominator are all distinct so this is the easy case. After the inverse transform you should end up with something like
$e^{r^2t}\sum_{i=0}^r\exp(-i^2t)/\prod_{j\ne i}(i^2-j^2)$
I expect that I've made a slip somewhere as I only have a few moments spare. That's why I said 'something like'. But the method is sound and if you put in more work than me you'll get a definite result.
Update: Did some testing in Mathematica. I think the expression above is correct.
