Computing $\log\Bigg(\sum_{n=0}^{\infty}(-1)^n\frac{\Gamma(n + b)}{\Gamma(n + 1)}\frac{(2n + b)}{\sqrt{2\pi x^3}}e^{-\frac{(2n + b)^2}{8x}}\Bigg)$

Question

I have been trying to think of ways to evaluate the following expression and hope that I will find some guidance here.

$$\log\Bigg(\frac{2^{b-1}}{\Gamma(b)}\sum_{n=0}^{\infty}(-1)^n\frac{\Gamma(n + b)}{\Gamma(n + 1)}\frac{(2n + b)}{\sqrt{2\pi x^3}}e^{-\frac{(2n + b)^2}{8x}}\Bigg)$$

where $b, x \in \mathbb{R}^{+}$.

The most obvious way I tried calculating this is by taking logs of the terms in the summation using a partial sum of around 100 terms. Then apply the logsumexp trick to avoid overflow when taking the exponential of the logarithm terms. This worked fine until I realized this method is very unstable when $b>>50$. At such values, the expression starts to overflow or return the wrong values. As shown below:

I am not strong mathematically so I am sure I am missing some trick or key theorems that could help evaluate this expression reliably at larger values of $b$. I tried using the pochhammer symbol to evaluate the ratio of gammas as $(n + 1)_{(b - 1)}$ but that did not help. I am also aware that $$ \frac{2^{b-1}}{\Gamma(b)\sqrt{\pi}} = \dfrac{1}{\Gamma(\frac{b}{2})\Gamma(\frac{b + 1}{2})}$$ and $$ \frac{\Gamma(n + b)}{\Gamma(b)\Gamma(n + 1)} = \frac{1}{nB(n, b)} $$ where $B(x, y)$ is the Beta function; but I don't know how that could be helpful.

EDIT: It is worth noting that the expression can also be re-written as

$$\log\Bigg(\frac{2^{b}}{\Gamma(b)}\sum_{n=0}^{\infty}(-1)^n\frac{\Gamma(n + b)}{\Gamma(n + 1)}f\Big(x| 0.5, (2n + b)^2 / 8)\Big)\Bigg)$$ where $f$ is the density function of an Inverse-Gamma distribution with shape parameter $0.5$ and scale $\frac{(2n + h)^2}{8}$.

EDIT2: It is also worth noting that the absolute value of the coefficients of the series increase and then decrease rapidly after a point. This point seems to depends on $b$ and $x$.

EDIT3: Page 10 of this text contains similar expressions but I don't know how to take advantage of them in this particular case: https://empslocal.ex.ac.uk/people/staff/mrwatkin//zeta/biane-pitman-yor.pdf

I have ran out of ideas so that is why I have come here hoping that someone can help guide me to the right way of computing this expression. Thanks in advance.

Answer 1

Denoting your function by $\log f(x)$, it follows from the Biane-Pitman-Yor paper that $$ \int_0^\infty \mathrm{d}x\,e^{-\lambda x} f(x) = \left(\cosh\sqrt{\tfrac{\lambda}{2}}\right)^{-b}.$$ So for integer $b\geq 1$, $f(x)$ is the density of a sum of $b$ i.i.d. random variables with Laplace transform $\left(\cosh\sqrt{\tfrac{\lambda}{2}}\right)^{-1}$. These variables have mean $1/4$ and finite variance $1/24$, so a (local) central limit theorem then gives $$ f(x) = \sqrt{\frac{12}{\pi b}} e^{-\frac{3}{4b}(b-4x)^2} + o(b^{-1/2})$$ with error uniform in $x$.