I need to calculate this integral

$$\sum\limits_{k=1}^\infty\int\limits_I \frac{\lambda^k}{(k-1)!}t_k^{k-1}e^{-\lambda t_k} dt_k$$

where $I=(a,b)$.

Someone told me that summation equals $\lambda$, but I don't see why. I wrote the summation as

$$\sum\limits_{n=0}^\infty \frac{(\lambda t_{n+1})^n}{n!}\lambda e^{-\lambda t_{n+1}} $$

and I was trying to see if the fact that $\sum\limits_{n=0}^\infty \frac{(\lambda t)^n}{n!}=e^{\lambda t} $ could help but I'm not seeing how.

Any ideas?

  • 2
    $\begingroup$ Does $t_n$ depend on $n$? $\endgroup$ – Ahmed S. Attaalla Jul 1 '16 at 20:17
  • $\begingroup$ @AhmedS.Attaalla Yes, $T_n$ are the arrival times of a Poisson process so they are different from each other and random variables each. The $t_n$ in the summation above appeared because I was trying to get the expectation of an expression involving those $T_n$, so I had to integrate its probability density function. I'm editing my question to clarify this. $\endgroup$ – Tendero Jul 1 '16 at 20:20
  • $\begingroup$ It looks me odd that $I=(t_1,t_2)$, $t_1$ and $t_2$ being the two first integration variables $t_k$, but may be I am wrong. $\endgroup$ – Jean Marie Jul 1 '16 at 20:33
  • $\begingroup$ @JeanMarie Yes that was an unfortunate naming, I just edited it. Thanks $\endgroup$ – Tendero Jul 1 '16 at 20:34

The confusion arises because of the bad choice of the name of the integration variable(s), as if it actually depended on $k$. Write the first few terms explicitly: $$ \sum\limits_{k=1}^\infty\int\limits_I \frac{\lambda^k}{(k-1)!}t_k^{k-1}e^{-\lambda t_k} dt_k=\int_I \frac{\lambda}{0!}t_1^{0}e^{-\lambda t_1}dt_1+ \int_I \frac{\lambda^2}{1!}t_2 e^{-\lambda t_2}dt_2+\cdots $$ You can easily see that calling the first integration variable $t_1$ and the second $t_2$ is unnecessary. Rename all the $t_k$'s as $t$. Your object is just a sum of one-dimensional integrals. Then $$ \sum_{k=1}^\infty\frac{\lambda^k}{(k-1)!}t^{k-1}=\lambda e^{t\lambda}\ . $$ Therefore the final result is $$ \lambda\int_a^b dt\ e^{-\lambda t}e^{t\lambda}=\lambda(b-a). $$ Note that whoever told you that the solution is just $\lambda$ ignores the fact that it is very unlikely that $a$ and $b$ (the integration range) drop out of the game!


There's no need for having different variables for integration. So we aim to solve $$ \sum_{k=1}^\infty\int_I\frac{\lambda^k}{(k-1)!}t^{k-1}e^{-\lambda t}\ dt =\int_I\sum_{k=0}^\infty\frac{\lambda^{k+1}}{k!}t^ke^{-\lambda t}\ dt$$ Pulling out the parts which are unnecessary for summation, we have \begin{align*} \sum_{k=1}^\infty\int_I\frac{\lambda^k}{(k-1)!}t^{k-1}e^{-\lambda t}\ dt &=\int_I\lambda e^{-\lambda t}\sum_{k=0}^\infty\frac{(\lambda t)^k}{k!}\ dt \\ &=\int_I\lambda e^{-\lambda t}e^{\lambda t}\ dt \\ &=\lambda(b-a) \end{align*}


You cannot a priori write the summation as you have done, because each time it's different variables. But you can do it a posteriori as I will show, using an appropriate change of variable which will "neutralize" the differences between the different $t_k$s. Here is how:

Let $u:=\lambda t_k$ and $J=(\lambda a, \lambda b)$:

$$\int\limits_I \frac{\lambda^k}{(k-1)!}t_k^{k-1}e^{-\lambda t_k} dt_k=\frac{1}{(k-1)!}\int\limits_J u^{k-1}e^{-u} du$$

Thus, setting $A:=\int\limits_J u^{k-1}e^{-u} du$

$$\sum\limits_{k=1}^\infty\int\limits_J \frac{\lambda^k}{(k-1)!}t_k^{k-1}e^{-\lambda t_k} dt_k=\int\limits_J \sum\limits_{k=1}^\infty\frac{u^{k-1}}{(k-1)!}e^{-u} du=\int\limits_J e^{u}e^{-u} du=\int\limits_J 1 du=\lambda b- \lambda a=\lambda(b-a)$$

  • 1
    $\begingroup$ Isn't your $A$ depending on $k$? Why can you pull it out of the summation over $k$? $\endgroup$ – Pierpaolo Vivo Jul 1 '16 at 20:59
  • $\begingroup$ You are write : I have done a big mistake... I correct it... $\endgroup$ – Jean Marie Jul 1 '16 at 21:11

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