Pattern matching in an indefinite integral over a sum involving Gamma functions and exponentials. Suppose that we have the following fact: For $h > 0$,
$$\int_0^{\infty} z^{-\frac{3}{2}} \sum_{n=0}^{\infty} (-1)^n \frac{2^h}{\Gamma(h)} \frac{\Gamma(n+h)}{\Gamma(n+1)} \frac{(2n+h)}{\sqrt{2\pi}} \exp\left\{ 
-\frac{(2n+h)^2}{2z} \right\} \, dz = 1 \;,$$ where $\Gamma(t) = \int_0^{\infty} x^{t-1} e^{-x} \, dx$ is the usual Gamma function.
Does this imply that, for $\delta > 0$, 
$$\int_0^{\infty} z^{-\frac{3}{2}} \sum_{k=0}^{\infty} (-1)^k \frac{\Gamma(k+2\delta)}{\Gamma(k+1)} \frac{(k+\delta)}{\Gamma(\delta)\Gamma(\delta)} \exp\left\{ 
-\frac{(k+\delta)^2}{2z} \right\} \, dz < \infty \;\;?$$
The two expressions look very similar. I suspect that it is true, but I'm having a hard time pattern matching the $n$'s with the $k$'s and the $h$'s with the $\delta$'s. I would really appreciate any thoughts on this.
 A: Starting from your expression
$$\int_0^{\infty} z^{-\frac{3}{2}} \sum_{n=0}^{\infty} (-1)^n \frac{2^h}{\Gamma(h)} \frac{\Gamma(n+h)}{\Gamma(n+1)} \frac{(2n+h)}{\sqrt{2\pi}}
\exp\left\{ -\frac{(2n+h)^2}{2z} \right\} \, dz = 1 \;,\tag{1}$$
you can rewrite the exponential part as
$$\exp\left\{ -\frac{(2n+h)^2}{2z} \right\}=\exp\left\{-\frac{(n+\tfrac h2)^2}{\tfrac z2}\right\},$$
and then substitute $y:=\tfrac z4$ to turn $(1)$ into
$$\int_0^{\infty}(4y)^{-\frac32}\sum_{n=0}^{\infty}
(-1)^n\frac{2^h}{\Gamma(h)}\frac{\Gamma(n+h)}{\Gamma(n+1)}\frac{(2n+h)}{\sqrt{2\pi}}
\exp\left\{-\frac{(n+\tfrac h2)^2}{2y}\right\} \, d(4y) = 1.$$
This can be simplified a bit, and setting $k:=n$ and $\delta:=\tfrac h2$ yields
$$\int_0^{\infty}y^{-\frac32}\sum_{k=0}^{\infty}
(-1)^k\frac{4^\delta}{\Gamma(2\delta)}\frac{\Gamma(k+2\delta)}{\Gamma(k+1)}\frac{(k+\delta)}{\sqrt{2\pi}}
\exp\left\{-\frac{(k+\delta)^2}{2y}\right\} \, dy = 1,$$
which is almost what we want. Multiplying through by the constant factor
$$\frac{\Gamma(2\delta)}{\Gamma(\delta)\Gamma(\delta)}\frac{\sqrt{2\pi}}{4^{\delta}},$$
yields the other expression. Hence the other expression converges to the constant above.
