Has this summation a specific result? 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?
 A: 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!
A: 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*}
A: 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 $A:=\int\limits_J u^{k-1}e^{-u} du$
$$\sum\limits_{k=1}^\infty\int\limits_J \frac{\lambda^k}{(k-1)!}t_k^{k-1}e^{-\lambda t_k} dt_k=\int\limits_J \sum\limits_{k=1}^\infty\frac{u^{k-1}}{(k-1)!}e^{-u} du=\int\limits_J e^{u}e^{-u} du=\int\limits_J 1 du=\lambda b- \lambda a=\lambda(b-a)$$
