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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.

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Your expression for I'_r(t) immediately gives you an expression for I_r(t); just expand and integrate term-by-term. Where is the issue? –  Qiaochu Yuan Apr 26 '10 at 18:52
    
I am seeking a solution where I do not have to evaluate $I_{r-1}(t)$ to find $I_r(t)$. Notice that I'_r(t) is linear in r, but for any r I know what the repeated integral evaluates to. I am seeking the same "happy" solution for I_r(t) which is quadratic in r. –  Ed Gorcenski Apr 26 '10 at 18:57
    
Oh, I see; I'_r(t) is the answer to a different problem, not the derivative with respect to t. In that case, you should try to guess the coefficients of q and then prove your guess by induction. –  Qiaochu Yuan Apr 27 '10 at 1:30
    
Yes, I apologize, it was a poor choice of nomenclature! I have not had much luck in determining a general rule for the coefficients, but they are annoyingly close to matching certain sequences. Thank you for your comments. –  Ed Gorcenski Apr 27 '10 at 13:12
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1 Answer

up vote 2 down vote accepted

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.

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Brilliant, I should have known to use the Laplace transform, foolish me. Thank you! –  Ed Gorcenski Apr 27 '10 at 18:22
    
On reflection, the appearance of all those squares is, to my mind, surprisingly pretty. I hope they have a nice interpretation in whatever your next step is. –  Dan Piponi Apr 27 '10 at 18:53
    
Indeed, I had hoped for something of the sort. I am trying to derive a probability distribution for a discrete random process wherein the transition rate function is a nonlinear polynomial in terms of the number of events per cell. For instance, if the transition rate function was f(r,t) = c+br, (without writing out all the equations behind it), we could easily derive the negative binomial distribution, which leads to a clustered grouping of points (the existence of k events in a cell has a linearly positive influence on the induction of another event in that cell). –  Ed Gorcenski Apr 27 '10 at 19:01
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