Integration on Two Independent Poisson Process

$$\int_{\theta =0}^{T}{\left\{ \sum\limits_{u=0}^{{{S}_{1}}-1}{\sum\limits_{m=0}^{{{S}_{1}}-u-1}{p\left( m,{{\lambda }_{0,1}}\left( T-\theta \right) \right)}p\left( u,{{\lambda }_{1}}\theta \right)} \right\}{{\lambda }_{0}}p({{S}_{0}}-1,{{\lambda }_{0}}\theta )d\theta }$$

I am trying to work out this integral, I will appreciate if someone help. Thanks!

There are two independent Poisson process with rate of $\lambda_1$ and $\lambda_0$.

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Rearrange the order of sum and integral, you should get

$$\frac{\lambda_0^{S_0} e^{-\lambda_{01} T}}{(S_0-1)!} \sum_{u=0}^{S_0-1} \frac{\lambda_1^u}{u!} \sum_{m=0}^{S_1-u-1}\frac{\lambda_{01}^m}{m!} \int_0^T d\theta \: (T-\theta)^m \theta^{S_0+u-1} e^{-(\lambda_0 + \lambda_1 - \lambda_{01}) \theta}$$

The integral may be expressed in terms of a confluent hypergeometric function $M$:

$$\frac{(\lambda_0 T)^{S_0} e^{-\lambda_{01} T}}{(S_0-1)!} \sum_{u=0}^{S_0-1} \frac{(\lambda_1 T)^u}{u!} \sum_{m=0}^{S_1-u-1}\frac{(\lambda_{01} T)^m}{m!} \frac{ m! (S_0+u-1)!}{(m+S_0+u-1)!} M[m,m+S_0+u-1,-(\lambda_0 + \lambda_1 - \lambda_{01})T]$$

From here, I do not see any path to simplification.

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Thanks! does $\lambda_0+\lambda_1=\lambda_{01}$ helps to simplify? thanks again –  Eln Feb 15 '13 at 20:52
I imagine it might. I'll see. –  Ron Gordon Feb 15 '13 at 21:01
In fact, $M=1$ and you can get rid of one of the sums, as the inner one evaluates to a pair of incomplete gamma functions. From there, though, the going likely gets really tough. –  Ron Gordon Feb 15 '13 at 21:13