How to get the following integral $$\int_0^\infty E_{\lambda\sim \mu}[\lambda e^{2 r\lambda}]e^{-4r}dr$$ where $\mu$ is the semi-circle law $d\mu(x)=\frac{1}{2\pi}\sqrt{4-x^2}1_{|x|\le 2}dx$.
Note that if define the moment generating function $M(t)=E[e^{t\lambda}]$ and then $E_{\lambda\sim \mu}[\lambda e^{2 t\lambda}]=M'(2t)$. Also, we can express it by the modified Bessel function $$ M'(2t)=\frac{I_1(2t)}{t} $$ So that integral becomes the Laplace transform of the modified Bessel function $$ \int_0^\infty \frac{I_1(2r)}{r}e^{-4r}dr $$
But how to get the result?
The integral can be simplified as \begin{equation} \int_0^\infty \frac{I_1(ar)}{r}e^{-pr}\,dr=\int_0^\infty\frac{I_1(z)}{z}e^{-zp/a}\,dz \end{equation} Denoting $s=p/a$ we must calculate \begin{equation} J(s)=\int_0^\infty\frac{I_1(z)}{z}e^{-zs}\,dz \end{equation} which converges when $s>1$. We must take $s=2$ for the proposed case. One use the series expansion for the modified Bessel function: \begin{equation} I_{\nu}\left(z\right)=(\tfrac{1}{2}z)^{\nu}\sum_{k=0}^{\infty}\frac{(\tfrac{1} {4}z^{2})^{k}}{k!\Gamma\left(\nu+k+1\right)} \end{equation} with $\nu=1$, \begin{align} J(s)&=\int_0^\infty\frac 12\sum_{k=0}^{\infty}\frac{(\tfrac{1} {4}z^{2})^{k}}{k!(k+1)!} e^{-zs}\,dz\\ &=\frac 12\sum_{k=0}^{\infty}\frac{2^{-2k}}{k!(k+1)!} \int_0^\infty z^{2k} e^{-zs}\,dz\\ &=\frac 12\sum_{k=0}^{\infty}\frac{2^{-2k}(2k)!}{k!(k+1)!} s^{-2k-1} \end{align} This expression looks similar to the generating function for the central binomial coefficients: \begin{equation} \sum_{k=0}^\infty\binom{2k}kx^k=\sum_{k=0}^\infty\frac{(2k)!}{(k!)^2} x^k=\frac{1}{\sqrt{1-4x}} \end{equation} To proceed, we perform an integration between $0$ and $0<X<1/4$ which gives \begin{align} \int_0^X\sum_{k=0}^\infty\frac{(2k)!}{(k!)^2} x^{k}\,dx&=\sum_{k=0}^\infty\frac{(2k)!}{k!(k+1)!} X^{k+1}\\ &=\int_0^X\frac{dx}{\sqrt{1-4x}}\\ &=\frac12(1-\sqrt{1-4X}) \end{align} With \begin{equation} J(s)=2s\sum_{k=0}^{\infty}\frac{(2k)!}{k!(k+1)!} \left( \frac{1}{4s^2} \right)^{k+1} \end{equation} taking $X=1/(4s^2)$, it comes \begin{equation} J(s)=s\left( 1-\sqrt{1-\frac{1}{s^2}} \right) \end{equation} or \begin{align} J(s)&=\frac{p}{a}\left( 1-\sqrt{1-\frac{a^2}{p^2}} \right)\\ &=\frac1a\left( p-\sqrt{p^2-a^2} \right)\\ &=\frac{\sqrt{p+a}-\sqrt{p-a}}{\sqrt{p+a}+\sqrt{p-a}} \end{align} The Laplace transform is given in this form in Ederlyi TI (4.16.3).
Here, $s=2$, $J(2)=2-\sqrt3$.