Tough Nested Integral

Solution 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}