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Is there any simplified closed form for the following integral?

$\int_0^{\tau}\int_0^{\tau-x_1}\int_0^{\tau-x_2}\dots\int_0^{\tau-x_{n-1}}e^{-\lambda x_1}e^{-\lambda x_2}\dots e^{-\lambda x_n}\ d{x_n}\ \dots d{x_2}\ d{x_1}$

in which

$x_i\ge 0; 1\le i\le n$

We can also define it recursively as follows if it helps:

$F(i,\tau)=\int_{0}^{\tau-x_{i-1}}e^{-\lambda x_{i}}F(i-1,\tau)$

$F(3,\tau)=\int_{0}^{\tau}\int_{0}^{\tau-x_1}\int_{0}^{\tau-x_2}e^{-\lambda x_1}e^{-\lambda x_2}e^{-\lambda x_3}\ dx_3\ dx_2\ dx_1$

$x_i\ge 0; 1\le i\le n$

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    $\begingroup$ I don't think it makes sense to have your dummy variable the same as the variable in your limits of integration. I'm happy to stand corrected though. $\endgroup$ – mattos Jun 7 '16 at 4:23
  • $\begingroup$ I just fixed it. Does it make sense now? @Mattos $\endgroup$ – Alireza Montazeri Gh Jun 7 '16 at 4:29
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    $\begingroup$ "Does it make sense now?" Not really: the integration bounds for $x_2$, say, are really $0$ and $x_1-\tau<0$, or do you mean $\tau-x_1$? Likewise for the other $x_k$-bounds. $\endgroup$ – Did Jun 7 '16 at 7:01
  • $\begingroup$ OMG, I meant $\tau -x_i$ @Did $\endgroup$ – Alireza Montazeri Gh Jun 7 '16 at 7:07
  • $\begingroup$ And now, are you sure that $\tau-x_2$ should not be $\tau-x_2-x_1$ instead? Likewise for the other upper bounds, until $\tau-x_{n-1}-\cdots-x_2-x_1$ instead of $\tau-x_{n-1}$? $\endgroup$ – Did Jun 7 '16 at 7:09
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Since this is a continuous function on a smooth boundary in $R^n$, you can simply use Fubini's theorum to express the n-dimensional multiple integral into a product of n-single integrals on $R$.

$\int_0^{\tau}\int_0^{x_1-\tau}\int_0^{x_2-\tau}\dots\int_0^{x_{n-1}-\tau}e^{-\lambda x_1}e^{-\lambda x_2}\dots e^{-\lambda x_n}\ d{x_n}\ \dots d{x_2}\ d{x_1}$

= $\int_0^{\tau}e^{-\lambda x_1}\ d{x_1}$ *$\int_0^{x_1-\tau}e^{-\lambda x_2}d{x_2}$ *$ \dots$ $\int_0^{x_{n-1}-\tau}e^{-\lambda x_n}\ d{x_n}$

We simply then integrate each expression beginning with the one on the far left.

Ick.

Very tedious, but not difficult.

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  • $\begingroup$ thanks for your answer. Could you please say what * means in your solution? @Mathemagician1234 $\endgroup$ – Alireza Montazeri Gh Jun 7 '16 at 6:56
  • $\begingroup$ Oh,that's just a multiplication sign,like ordinary multiplication of real numbers. $\endgroup$ – Mathemagician1234 Jun 7 '16 at 22:18

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