Question
How can this integral, plotted in Fig. 1, be expressed more simply, for evaluation, perhaps in terms of common special functions?
$$H = \int_0^\infty \log(a)\,e^{-a^2}I_0(ba)\,da,\tag{1}$$
where $I_0$ is a modified Bessel function of the first kind and $b > 0.$
Figure 1. Plot of the integral $H$ in Eq. (1) as function of variable $b$.
My attempts
I have not been able to find such integrals in tables of integrals and Wolfram Alpha times out. Mathematica gives:
$$H = \frac{1}{4}\, \sqrt{\pi}\Bigg(L_{-1/2}\left(\frac{b^2}{4}\right)\Psi\left(\frac{1}{2}\right) - L^{(1,\,0)}_{-1/2}\left(\frac{b^2}{4}\right)\Bigg),\tag{2}$$
where $L_n(x)$ is Laguerre's L function, $\gamma$ is the Euler–Mascheroni constant, $\Psi$ is the digamma function, and the superscript $(1,0)$ denotes taking the derivative of $L_n(x)$ with respect to the variable $n$. $L$ is defined in terms of the confluent hypergeometric function of the first kind, ${_1F_1}:$
$$L_n(x) = {_1F_1}(-n;1;x) = \sum_{k=0}^\infty \frac{(-n)_k\,x^k}{(1)_k\,k!},\tag{3}$$
where $(x)_k$ is a Pochhammer symbol with special cases:
$$\begin{eqnarray}(1)_k &=& k!,\\ (1/2)_k &=& (2k - 1)!!/2^k,\end{eqnarray}\tag{4}$$
where $!!$ denotes a double factorial rather than a factorial of a factorial. The first special function in Eq. (2) has a series ($x = b^2/4$ to be plugged in):
$$L_{-1/2}(x) = {_1}F_1\left(\frac{1}{2};1;x\right) = \sum_{k=0}^\infty\frac{(1/2)_k\,x^k}{(1)_k\,k!}.\tag{5}$$
$${_1F_1}\left(\frac{1}{2} + \nu;\,2\nu + 1;\,x\right) = x^{-\nu-1/2}\,e^{x/2}\,4^\nu\sqrt{x}\,I_\nu\left(\frac{x}{2}\right)\,\Gamma(\nu+1).\tag{6}$$
At $\nu = 0,$ we get:
$$L_{-1/2}(x) = e^{x/2}\,I_0\left(\frac{x}{2}\right).\tag{7}$$
The second special function in Eq. (2), the derivative of $L_n(x)$ with respect to $n$, with $n=-1/2,$ does not seem to be very common. A good source of identities is (Ancarani & Gasaneo, 2008), for example their Eq. (5a) helps to get:
$$\begin{array}{l}\displaystyle L_{-1/2}^{(1, 0)}(x) = -\,{_1F_1}^{(1, 0, 0)}\left(\frac{1}{2};1;x\right)\\ \displaystyle = -\sum_{k=0}^\infty\frac{(1/2)_k}{(1)_k}\Psi\left(\frac{1}{2}+k\right)\frac{x^k}{k!} + \Psi\left(\frac{1}{2}\right)\,{_1F_1}\left(\frac{1}{2};1;x\right)\\ \displaystyle = -\sum_{k=0}^\infty\frac{(1/2)_k\,x^k}{(1)_k\,k!}\Psi\left(\frac{1}{2}+k\right)+\Psi\left(\frac{1}{2}\right)\,e^{x/2}\,I_0\left(\frac{x}{2}\right)\\ \displaystyle = -\sum_{k=0}^\infty\frac{(1/2)_k\,x^k}{(1)_k\,k!}\left(\Psi\left(\frac{1}{2}\right) + 2\sum_{n=1}^k \frac{1}{2n - 1}\right)+\Psi\left(\frac{1}{2}\right)\,e^{x/2}\,I_0\left(\frac{x}{2}\right).\end{array}\tag{8}$$
Recognizing the series of $e^{z/2}\,I_0\left(\frac{x}{2}\right),$ see Eqs. (5) and (7), some terms cancel and we get:
$$L_{-1/2}^{(1, 0)}(x) = -\sum_{k=0}^\infty\left(\frac{(1/2)_k\,x^k}{(1)_k\,k!}\times 2\sum_{n=1}^k \frac{1}{2n - 1}\right)\\ = 0 - x - \frac{1}{2}x^2 - \frac{161}{144}x^3 - \frac{23595}{32}x^4 - \frac{1524360005233065}{1024}x^5 - \frac{26034220296741617347072389763198341076875}{2048}x^6 - \ldots.\tag{9}$$
That doesn't seem to converge. Combining the series of Eqs. (5) and (9) into Eq. (2) gives:
$$\begin{eqnarray}H &=& \frac{\sqrt{\pi}}{4}\,\left(\Psi\left(\frac{1}{2}\right)\sum_{k=0}^\infty\frac{(1/2)_k\,\left(\frac{b^2}{4}\right)^k}{(1)_k\,k!} + \sum_{k=0}^\infty\Bigg(\frac{(1/2)_k\,\left(\frac{b^2}{4}\right)^k}{(1)_k\,k!}\times 2\sum_{n=1}^k \frac{1}{2n - 1}\Bigg)\right)\\ &=&\frac{\sqrt{\pi}}{2}\,\sum_{k=0}^\infty\left(\frac{(1/2)_k\,\left(\frac{b^2}{4}\right)^k}{(1)_k\,k!}\left(\sum_{n=1}^k \frac{1}{2n - 1} + \frac{\Psi(1/2)}{2}\right)\right),\end{eqnarray}\tag{10}$$
where $\Psi\left(\frac{1}{2}\right) = \psi^{(0)}\left(\frac{1}{2}\right) = -\gamma - \log 4 \approx -1.96351.$ The combined series seems to converge well, and seems identical to @user150203's series, with some of the early terms given in my edit to his answer.
Problem background
I encountered this integral in the context of calculating a minimum mean square log-magnitude error estimator of the magnitude of a complex number from its noisy observation polluted by additive circularly symmetric complex Gaussian noise of known variance, assuming an improper uniform prior of the magnitude. From this context the full expression I want to simplify is:
$$G = \exp\left(\frac{2\int_0^\infty \log(a) e^{-a^2}I_0(2ma)da}{\sqrt{\pi} e^{m^2/2} I_0(m^2/2)}\right).\tag{11}$$
$G$ approaches $m$ asymptotically as $m\to\infty$ (Fig. 2).
Figure 2. the desired estimator $G(m).$
References
Ancarani L. U. and Gasaneo G., Derivatives of any order of the confluent hypergeometric function ${_1F_1}(a,b,z)$ with respect to the parameter $a$ or $b,$ Journal of Mathematical Physics 49, 063508 (2008); https://doi.org/10.1063/1.2939395