Integral of Modified Bessel Function 
Evaluate:
  $$\int\limits_{x=0}^\infty e^{-\alpha x^2}I_0(x) \ln(I_0(x))x \, dx$$
where $I_0(x)$ is the modified Bessel function of the first kind and zeroth order, and $\alpha>0$.

I can find upper bounds and lower bounds using the expansion of $I_0(x)$; however, they are not enough. I need an exact value, no bounds.
 A: This is more a comment than an answer, but too long to be edited in the comments section.
$$S(\alpha)=\int\limits_{x=0}^\infty e^{-\alpha x^2}I_0(x) \ln(I_0(x))x \, dx$$
$$S(\alpha)=\frac12\int\limits_{x=0}^\infty e^{-\alpha x}I_0(\sqrt{x}) \ln(I_0(\sqrt{x})) \, dx$$
In fact $S(\alpha)$ is the Laplace transform of the function $I_0(\sqrt{x}) \ln(I_0(\sqrt{x}))$.
$$S(\alpha)=\mathcal{L}_x\Big(I_0(\sqrt{x}) \ln(I_0(\sqrt{x}))\Big)(\alpha) $$
In the extended tables of Laplace transform (for example from H.Bateman) one find the Laplace transforms of a lot of functions involving combinations of Bessel functions, such as $I_\nu(c_1\sqrt{x}) I_\nu(c_2\sqrt{x}))$ for example. But not the function with logarithm $I_0(\sqrt{x}) \ln(I_0(\sqrt{x}))$.
Also math software such as WolframAlpha find no result in terms of standard mathematical function.
This draw to think that nowadays $S(\alpha)$ cannot be expressed in terms of standard mathematical functions. Of course, this is not a definitive proof. Moreover, new special functions might be defined in the future and among them one allowing to express $S(\alpha)$ in closed form.
