I have the following relation: $$\int_0^\infty \sum_{n=0}^{\infty} \frac{\lambda^{n}e^{-\lambda}}{n!}g(n)e^{-g(n)x}x\; \mathrm{d}x= \sum_{n=0}^{\infty}\tfrac{\lambda^{n}e^{-\lambda}}{n!}g(n)$$
Is it possible to isolate the unknown function $g(n)$, assuming that $g$ is a discrete function?
I tried taking the log of both sides and I couldn't really get anywhere - just a bunch of diverging integrals. I'm guessing there's a good chance that it's not possible to simplify this equation any further, but I'm interested in trying to approximate what $g(n)$ could be. I have no reason to believe $g(n)$ is anything in particular, but I'm not sure how to use that fact.
(One could also re-write the expression: $$\int_0^\infty \sum_{n=0}^{\infty} f(n)g(n)e^{-g(n)x}x\; \mathrm{d}x= \sum_{n=0}^{\infty}f(n)g(n)$$ where f is Poisson distributed with parameter $\lambda$
or even further: $$\int_0^\infty \sum_{n=0}^{\infty} f(n)B(x)x\; \mathrm{d}x= \sum_{n=0}^{\infty}f(n)g(n)$$
where $B(x)$ is Exponentially distributed with parameter $g(n)$, where $g(n)$ is some unknown (presumably simple/nice) function of $n$)
Thanks!
If as eyeballfrog says, we can switch the integral and the summation:
$$ \int_0^\infty \sum_{n=0}^{\infty} \frac{\lambda^{n}e^{-\lambda}}{n!}g(n)e^{-g(n)x}x\; \mathrm{d}x = \sum_{n=0}^{\infty} \frac{\lambda^{n}e^{-\lambda}}{n!}g(n) \int_0^\infty e^{-g(n)x}x\; \mathrm{d}x $$
Now, by comparing both sides of the equation: $$ \sum_{n=0}^{\infty} \frac{\lambda^{n}e^{-\lambda}}{n!}g(n) \int_0^\infty e^{-g(n)x}x\; \mathrm{d}x = \sum_{n=0}^{\infty}\frac{\lambda^{n}e^{-\lambda}}{n!}g(n) $$
And assuming both sides need to be identical no matter the value of $\lambda$, then we have for $g(n)\neq 0$:
$$ \int_0^\infty e^{-g(n)x}x\; \mathrm{d}x = 1 $$
Solving the integral:
$$ \int_0^\infty e^{-g(n)x}x\; \mathrm{d}x = \left(\frac{1}{g(n)} \right)^2 $$
If the real part of $g(n)$ is greater than 0. Thus the only solution would be $g(n)=1 \, (\forall n)$