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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.

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    $\begingroup$ A not-so-terrible upper bound can be derived from $\log I_0(x)\leq x$. We have $$\int_{0}^{+\infty} x\cdot I_0(x) e^{-ax^2}\,dx = \frac{1}{2a}e^{\frac{1}{4a}}.$$ $\endgroup$ – Jack D'Aurizio Jul 25 '18 at 20:39
  • $\begingroup$ @JackD'Aurizio Thanks. The diamond means you are one of the moderators? $\endgroup$ – Susan_Math123 Jul 25 '18 at 21:05
  • $\begingroup$ You're welcome, and... yup. $\endgroup$ – Jack D'Aurizio Jul 25 '18 at 21:16
  • $\begingroup$ @JackD'Aurizio To get another bound I can use $\ln(x)\leq x-1$. Then I will need to evaluate $\int_{0}^{\infty}x I_0(x)^2 e^{-a x^2} dx$. Do you have an answer for this as well? $\endgroup$ – Susan_Math123 Jul 26 '18 at 16:30
  • $\begingroup$ Such bound is much worse than the previous one, but in any case $$\int_{0}^{+\infty}x I_0(x)^2 e^{-ax^2}\,dx = \frac{1}{2a} e^{1/(2a)} I_0(1/(2a)).$$ $\endgroup$ – Jack D'Aurizio Jul 26 '18 at 16:42
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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.

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