How to get the following integral which is the Laplace transform of the modified Bessel function? 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?
 A: 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, $s=2$, $J(2)=2-\sqrt3$.
