Different ways to evaluate $\int_0^\infty xI_0(2x)e^{-x^2}\,dx$

A simpler way is to exploit Fubini's theorem and polar coordinates:

$$ \int_{0}^{+\infty}x\,I_0(2x) e^{-x^2}\,dx = \frac{1}{\pi}\int_{0}^{\pi}\int_{0}^{+\infty}\rho e^{-2\rho\cos\theta} e^{-\rho^2}\,d\rho\,d\theta $$ where the last integral equals: $$ \frac{1}{\pi}\int_{0}^{+\infty}\int_{-\infty}^{+\infty} e^{-2x} e^{-x^2-y^2}\,dx\,dy \stackrel{x\mapsto x-1}{=}\frac{e}{\pi}\int_{0}^{+\infty}\int_{-\infty}^{+\infty}e^{-x^2-y^2}\,dx\,dy$$ that simplifies to: $$ \frac{e}{2\pi}\iint_{\mathbb{R}^2} e^{-x^2-y^2}\,dx\,dy = \color{red}{\frac{e}{2}}.$$

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This also gives the straightforward generalization $$ \int_{0}^{+\infty} x\,I_0(\rho x)e^{-x^2}\,dx = \color{red}{\frac{1}{2}\,e^{\rho^2/4}}.$$

Note that $$I=\int_{0}^{\infty}xI_{0}\left(2x\right)e^{-x^{2}}dx\stackrel{e^{-x^{2}}=u}{=}\frac{1}{2}\int_{0}^{1}I_{0}\left(2\sqrt{-\log\left(u\right)}\right)du$$ $$=\frac{1}{2}\sum_{k\geq0}\frac{\left(-1\right)^{k}}{\left(k!\right)^{2}}\int_{0}^{1}\log^{k}\left(u\right)du=\frac{1}{2}\sum_{k\geq0}\frac{1}{k!}=\color{red}{\frac{e}{2}}.$$

$$I = \int_0^\infty xI_0(2x)e^{-x^2}\,dx$$

$$I_0(x)=\frac 1\pi\int_0^\pi e^{x\cos\theta}d\theta \stackrel{\theta\rightarrow\theta/2}{=}\frac 1{2\pi}\int_0^{2\pi} e^{x\cos(\theta/2)}d\theta \stackrel{z=\exp(i\theta/2)}{=}\frac 1{2\pi i}\int_{|z|=1} \frac{xe^{[z+1/z]/2}}{z}dz$$

To evaluate this, we proceed by expanding the function as a series and finding the residue, using the residue theorem. We have that $$I_0(x)=\text{Res}(\frac{{e^{x[z+1/z]/2}}}z,z=0) = \text{constant term of } 1+x\frac{z+z^{-1}}2+x^2\frac{(z+z^{-1})^2}{2^{2} 2!}+x^3\frac{(z+z^{-1})^3}{2^3 3!} +\cdots$$

Noting that we only get a constant term contribution from the even terms, and the constant of $(z+z^{-1})^{2n}$ is $\binom{2n}n,$ we have that $$I_0(x)=\text{Res}(\frac{{e^{z+1/z}}}z,z=0) = \sum_{n\geq0} \frac{x^{2n}}{2^{2n}(2n)!}\binom{2n}{n}=\sum_{n\geq0} \frac{x^{2n}}{2^{2n}(n!)^2}$$ Then $$I_0(2x)=\sum_{n\geq0} \frac{x^{2n}}{(n!)^2}$$ So $$I = \int_0^\infty \sum_{n\geq0} \frac{x^{2n}}{(n!)^2} xe^{-x^2}\,dx\stackrel{\text{terms are all positive}}= \sum_{n\geq0}\frac{1}{(n!)^2} \int_0^\infty{x^{2n+1}}e^{-x^2}\,dx\stackrel{u=x^2}=\frac12 \sum_{n\geq0}\frac{1}{(n!)^2} \int_0^\infty{x^{n}}e^{-u}\,du = \frac 12\sum_{n\geq0}\frac{1}{(n!)^2} \Gamma(n+1) = \frac 12\sum_{n\geq0}\frac{1}{(n!)} =\boxed{\frac e2}$$