Consider the following density function (unnormalized) $$p(y;b)=\int_{0}^{\infty}\dfrac{\exp\left(-\frac{y^2}{2\left(b+\frac 5u\right)}\right)}{\sqrt{b+\frac 5u}} u^{1/2}\exp\left(-\dfrac{2u}{3}\right)du.$$
Is it true that $p(y;b)$ is a log-concave function in $b$ when $y$ is considered fixed?
My attempt: First of all, It is easy to plot $\log p(y;b)$ as a function of $b$ for a fixed value of $y$, and I've gotten a good hint that the curve is indeed concave. Note that one can think of $p(y;b)$ as a scale mixture of normal-gamma densities. I am aware that normal and gamma both are log-concave densities. However, I am not immediately able to see why that makes $p(y;b)$ log-concave as well. It is useful to know that the integral cannot be put into any standard closed form. One idea I had was to look at the integral as an expectation $p(y;b)=\mathbb{E}_u[\text{integrand}]$, where $u\sim \text{Gamma}(3/2,3/2)$, but this route requires the joint (log)concavity of the integrand in $(b,u)$ which seems like a formidable task to check through the means of the Hessian. Is there any other approach one can take to prove the proposed log-concavity?
PS: there was a trivial error in my previous argument (where I said those Hessians are negative definite, in fact, they aren't). The question remains open.
I realized that the function $p(y;b)$ cannot be a log-concave for all fixed $y$. Let me show this for $y=0$. In this case, $$p(0;b)=\int_{0}^{\infty}\dfrac{1}{\sqrt{b+\frac 5u}} u^{1/2}\exp\left(-\dfrac{2u}{3}\right)du.$$
is both convex and log-convex over $b>0$.
Considering that the function $$\frac{1}{\sqrt{b+\frac 5u}}$$ is well-defined and convex over $b>0$ for any fixed value of $u>0$, $p(0;b)$ is also convex in $b$.
Moreover, after applying the transformation $x=\frac{5}{u}$, the integration used in the definition of $p(0;b)$ can be calculated explicitly, see this Wolfram link: explicit formula of $p(0;b)$. The log of this function, which is plotted here: plot of $\log p(0;b)$, is clearly a convex function over $b>0$.
Responding to a comment below on the unimodality of the following function:
$$f(b)=\mathbb E_{N} \left [ -\log \mathbb E_{G} \left[\dfrac{\exp\left(-\frac{N^2}{2\left(b+\frac 5G\right)}\right)}{\sqrt{b+\frac 5G}} \right ]\right ] $$
with $N \sim \text{Nromal}(0,1)$ and $G \sim \text{Gamma}(3/2,3/2)$, My numerical checks show that the function:
$$ g(b,y,u)= \dfrac{\exp\left(-\frac{y^2}{2\left(b+\frac 5u\right)}\right)}{\sqrt{b+\frac 5u}} $$
for fixed values of $y \in \mathbb R $ and $u>0$ is quasi-concave, which can be increasing, decreasing, or unimodal with a single maximum for different values of $y$ and $u$. However, unfortunately, the summation of quasi-concave functions is not necessarily a quasi-concave function, and thus we cannot use this property. The above observations indicate that analytically proving that $f(b)$ is unimodal seems very difficult. Hence, the best way to realize whether $f(b)$ is unimodal or not is to numerically check it.