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Does anyone know of any clever tricks that solve

$$\large G_n(t) = \int_0^t dt_1 \int_0^{t_1} dt_2 \cdots \int_0^{t_{n-2}}dt_{n-1}\int_0^{t_{n-1}}dt_{n} e^{i\lambda(t_1-t_2+t_3-\cdots + t_{n-1}-t_n)}$$

I've come up with a few recursion relations but I'm finding it hard to pin down the exact answer.

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You can separate the exponential. Why don't you do that? –  Pedro Tamaroff Mar 7 '12 at 22:35
    
That doesn't help much. If you separate and then evaluate the first integral you wind up with $-\frac{1}{i\lambda}\left(e^{-i\lambda t_{n-1}} - 1\right)$. The exponent term will cancel the exponent in the next integral, which will produce ultimately polynomial terms in $t$. I didn't make too much progress with that approach. –  fatbox Mar 7 '12 at 22:45
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I get the recursion $$G_n(t) = \frac{t_{n-2} G_{n-2} - G_{n-1}}{i\lambda (-1)^{n+1}}$$ which doesn't seem to yield a nice answer unless there is some relation between the $t_i$'s. –  user17762 Mar 7 '12 at 22:47
    
If the bounds in the inner integrals didn't depend on the varaibles with respect to which one integrates in the outer integrals, then "separating" the exponential would pay off. But it's less clear that it will in the present situation. –  Michael Hardy Mar 7 '12 at 23:46
    
@fatbox May I ask where does such an integral arise? –  Sasha Mar 8 '12 at 14:49
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1 Answer 1

By rescaling $t_{k} = t u_{k}$ we get $$ \begin{eqnarray} G_n(\lambda, t) &=& t^n n! \int_0^1 \mathrm{d} u_1 \frac{1}{n!} \int_0^{u_1} \mathrm{d} u_2 \cdots \int_0^{u_{n-1}} \mathrm{d} u_n \exp\left(i \lambda t \sum_{k=1}^n (-1)^{k-1} u_k \right) \\ &=& t^n n! \mathbb{E}\left( \exp\left(i \lambda t \sum_{k=1}^n (-1)^{k-1} U_{k:n} \right) \right) \end{eqnarray} $$ where $U_{k:n}$ is the $k$ out of $n$ order statistics on a sample from uniform distribution. It is well known that the vector $(U_{1:n},U_{2:n},\ldots,U_{n:n})$ is equal in distribution to $$ \left( \frac{X_1}{\sum_{k=1}^{n+1} X_k}, \frac{X_1+X_2}{\sum_{k=1}^{n+1} X_k}, \ldots, \frac{X_1+X_2 + \cdots + X_n}{\sum_{k=1}^{n+1} X_k} \right) $$ where $X_i$ are i.i.d. exponential random variables with unit mean. Thus $G_n$ is related to the characteristic function of the beta distribution: $$ \begin{eqnarray} \sum_{k=1}^n (-1)^{k-1} U_{k:n} &=& \frac{1}{\sum_{m=1}^{n+1} X_m} \sum_{k=1}^n (-1)^{k-1} \sum_{p=1}^{k} X_p = \frac{1}{\sum_{m=1}^{n+1} X_m} \sum_{k=1}^n (-1)^{k-1} \sum_{p=1}^{k} X_p \\ &=& \frac{1}{\sum_{m=1}^{n+1} X_m} (-1)^{n-1} \sum_{0\leqslant 2k < n} X_{n-2k} \\ &=& (-1)^{n-1} \frac{\sum_{0\leqslant 2k < n} X_{n-2k} }{ \sum_{0\leqslant 2k < n} X_{k} + \sum_{0\leqslant 2k < n+1} X_{n+1-2k} } = (-1)^{n-1} \frac{g_1}{g_1+g_2} \end{eqnarray} $$ where $g_1$ follows $\Gamma\left( \left\lfloor \frac{n+1}{2} \right\rfloor\right)$ distribution and $g_2$ follows $\Gamma\left( \left\lceil \frac{n+1}{2} \right\rceil\right)$. The ratio $V = g_1/(g_1+g_2)$ is a $\operatorname{Beta}\left(\left\lfloor \frac{n+1}{2} \right\rfloor, \left\lceil \frac{n+1}{2} \right\rceil \right)$ random variable. Thus

$$ G_n(t) = n! t^n \phi_V( (-1)^{n-1} \lambda t ) = n! t^{n} \cdot {}_1 F_1\left( \left\lfloor \frac{n+1}{2} \right\rfloor ; n+1; (-1)^{n-1} i \lambda t \right) $$

Here is verification in Mathematica v8:

In[1]:= Table[
  Hypergeometric1F1[Floor[(n + 1)/2], 1 + n, (-1)^(n - 1) I la t] - 
   Expectation[ Exp[I la t Sum[(-1)^(k - 1) x[k], {k, 1, n}]], 
    Array[x, n] \[Distributed] 
     OrderDistribution[{UniformDistribution[], n}, Range[n]], 
    Assumptions -> la > 0 && t > 0], {n, 2, 6}] // Expand

Out[1]= {0, 0, 0, 0, 0}
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Very nice solution! –  Antonio Vargas Mar 8 '12 at 4:47
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