Proving integration identity Can you please help me prove this identity below? I have tried integration by parts and I am able to see where the numerator of the RHS comes from; however, I am unable to get the denominator. Additionally, I am unsure of how to use induction to show this, because I know that I would have to use integration by parts $k-1$ times to prove this... thanks for the help!
$$\int_{\alpha w}^\infty \frac{z^{k-1}e^{-z}}{(k-1)!}\mathrm dz=\sum_{x=0}^{k-1} \frac{(\alpha w)^x e^{-\alpha w}}{x!}$$
 A: $$\int_{\alpha w}^\infty \frac{z^{k-1}e^{-z}}{(k-1)!}\mathrm dz=\sum_{x=0}^{k-1} \frac{(\alpha w)^x e^{-\alpha w}}{x!}$$
The right side is the probability that the number of occurrences in a Poisson process before time $w$ is less than $k$ (with an average of $\alpha$ occurrences per unit of time).
The left side is the probability that the waiting time until the $k$th occurrence is more than $w$.
One side looks at the discrete probability distribution of the number of occurrences in a given time interval; the other looks at the continuous probability distribution of the the time until the $k$th occurrence.
But the two events are the same event.  So they must have the same probability.  Therefore they are equal, quod erat demonstrandum.
Later edit: I find a complaint of cirucularity in the comments.  That is mistaken.  Find the probability distribution of the number of successes in $n$ independent Bernoulli trials with probability $p$ of success on each trial; take the limit as $n\to\infty$ while $np$ remains fixed at $\lambda$.  The limiting probability of $x$ successes is $e^{-\lambda} \lambda^x/x!$.  Say this is in unit time; the numbers of successes in non-overlapping time intervals are independent; then the number of successes before time $t$ has expected value $t\lambda$ and the probability of exactly $x$ successes is $e^{-t\lambda}(t\lambda)^x/x!$.  Now the probability of fewer than $x$ successes before time $t$ is the sum of $x$ terms given above (but I'm using "$x$" where "$n$" was used).  What then is the probability distribution of the waiting time until the $n$th success?  The value of the cumulative distribution function at time $t$ is just $1$ minus the sum given above.  Differentiate that to get the density.  The integral of the density is the integral above.
So there's no circularity.
A: This one cries out for a generating function.  For $|t|<1$, 
$$\sum_{k=0}^\infty t^k \int_{\alpha w}^\infty \frac{z^{k}}{k!} e^{-z} \ dz = \int_{\alpha w}^\infty \sum_{k=0}^\infty \frac{(tz)^k}{k!} e^{-z} \ dz $$
$$ = \int_{\alpha w}^\infty e^{t z - z}\ dz  = \frac{e^{(t-1) \alpha w}}{1-t} $$
$$ = e^{-\alpha w} \left(\sum_{k=0}^\infty \frac{(\alpha w t)^k}{k!}\right)\left(\sum_{k=0}^\infty t^k\right)$$
$$ = \sum_{k=0}^\infty t^k \sum_{j=0}^k \frac{(\alpha w)^j}{j!}$$
A: Here's one way of proceeding: first, verify that both sides of the purported identity agree for $k=1$. Now, if you carry out integration by parts on the left hand side, you should obtain the recursion relation
$$\beta_{k-1}=\beta_{k-2}+\frac{(\alpha w)^{k-1}e^{-\alpha w}}{(k-1)!}$$
where
$$\beta_{k-1}=\int_{\alpha w}^\infty \frac{z^{k-1}e^{-z}}{(k-1)!}\mathrm dz=\frac{\Gamma(k,\alpha w)}{\Gamma(k)}=Q(k,\alpha w)$$
and $\Gamma(k,u)$ and $Q(k,u)$ are various notations for the (upper) incomplete gamma function.
The trick now is to replace $\beta_{k-1}$ in the recursion relation with the sum on the right hand side, and verify that the recursion still holds. This along with the verified initial condition $k=1$ proves your identity.

As a variation of Didier's strategy in the comments: after letting $\alpha w=0$, to verify the equation
$$\int_0^\infty \frac{z^{k-1}e^{-z}}{(k-1)!}\mathrm dz=e^{0}\left(1+\sum_{x=1}^{k-1} \frac{0^x}{x!}\right)$$
one can demonstrate that the expression on the right does simplify to $1$; showing that
$$(k-1)!=\int_0^\infty z^{k-1}e^{-z}\mathrm dz=\Gamma(k)$$
can be done by verifying that both sides of the equation agree when $k=1$ and establishing the recursion $(k-1)!=(k-1)(k-2)!$ through integration by parts.
Differentiating both sides of the original identity with respect to $u=\alpha w$ yields the relation
$$-\frac{u^{k-1}e^{-u}}{(k-1)!}=e^{-u}\left(\sum_{x=1}^{k-1} \frac{xu^{x-1}}{x!}-\sum_{x=0}^{k-1} \frac{u^x}{x!}\right)$$
I'll leave the simplification to the reader.
A: Another answer, my second one to this question:
Differentiate both sides with respect to $\omega$.  On the left you need the chain rule in a trivial way and the fundamental theorem of calculus, and you get only one term.  On the right you get $2k-1$ terms (after apply the product rule to all but one of the $k$ terms) and $2k-2$ of them cancel out.  The highest-degree term is equal to the expression you get on the right side.  That proves they differ by a constant; finding the constant is the remaining task.
