While doing some calculations for a random matrix model I arrived at the following formula
$$ \int_0^x\int_x^∞ e^{−t−s} I_0(2\sqrt{ts}) \mathrm{d}t\mathrm{d}s
= xe^{−2x} (I_0(2x) + I_1(2x)) $$
where $I_ν$ denotes the modified Bessel function of first kind, i.e.
$$I_ν(x) = \sum_{n=0}^{\infty} \frac{1}{n!Γ(n+ν+1)}\left(\frac{z}{2}\right)^{2n+ν}.$$
I am now looking for some (short) analytic proof for it? I tried a bit around but could not find any.
Specific integral of modified Bessel function of first kind
bessel functionsdefinite integrals
Related Solutions
From $$ I_n(x) ~=~\frac{1}{\pi}\int_0^{\pi} \! \mathrm{d}\theta ~\exp\left(x\cos\theta\right)\cos n\theta, \qquad n\in~\mathbb{N}_0,\tag{A} $$ we calculate $$\begin{align} \sqrt{x}\pi e^{-x}I_n(x) &~~=~\sqrt{x} \int_0^{\pi} \! \mathrm{d}\theta ~\exp\left(- x(1-\cos\theta)\right)\cos n\theta \cr &\stackrel{t=\sqrt{x}\theta}{=}~ \int_0^{\pi\sqrt{x}} \! \mathrm{d}t ~\exp\left(- x(1-\cos\frac{t}{\sqrt{x}})\right)\cos \frac{nt}{\sqrt{x}} \cr &~~=~ \int_0^{\infty} \! \mathrm{d}t ~\exp\left(- \frac{t^2}{2} + \frac{t^4}{24 x} + O(x^{-2})\right) \left(1- \frac{(nt)^2}{2x} + O(x^{-2})\right) \cr &~~=~ \int_0^{\infty} \! \mathrm{d}t ~\exp\left(- \frac{t^2}{2}\right) \left(1- \frac{(nt)^2}{2x}+ \frac{t^4}{24 x} + O(x^{-2})\right) \cr &\stackrel{u=t^2/2}{=}~ \int_0^{\infty} \! \frac{\mathrm{d}u}{\sqrt{2u}} ~\exp\left(- u\right) \left(1- \frac{n^2u}{x}+ \frac{u^2}{6x} + O(x^{-2})\right) \cr &~~=~\frac{1}{\sqrt{2}}\left( \Gamma(\frac{1}{2}) - \frac{n^2}{x}\Gamma(\frac{3}{2}) +\frac{1}{6x}\Gamma(\frac{5}{2}) + O(x^{-2})\right)\cr &~~=~\sqrt{\frac{\pi}{2}}\left( 1+ \frac{1-4n^2}{8x} + O(x^{-2})\right), \end{align}\tag{B}$$ which agrees with OP's sought-for formulas (1).
This is not a complete answer to your question that, in the way it is stated, appears very hard. But, as said in the comments, it is easily amenable to asymptotic treatment and the approximation is not that bad. Firstly, it is known that, for $x\rightarrow\infty$, $$ I_0(x)\sim\frac{e^x}{\sqrt{2\pi x}}. $$ So, I approximate your integral as $$ Z(\kappa)=\int_0^\frac{\pi}{2}tI_0(2\kappa\cos t)dt\sim\int_0^\frac{\pi}{2}t\frac{e^{2\kappa\cos t}}{\sqrt{4\pi\kappa\cos t}}dt. $$ The last integral can be managed with the Laplace method by noting that it takes the great part of contributions at $t=0$. So, I do a Taylor series for the cosine obtaining $$ Z(\kappa)\sim \frac{e^{2\kappa}}{\sqrt{4\pi\kappa}}\int_0^\frac{\pi}{2}te^{-\kappa t^2}\left(1-\frac{t^2}{16\pi\kappa}\right) $$ and we see that the next-to-leading correction can be neglected. We are left with a very easy integral and the final result will be $$ Z(\kappa)\sim\frac{e^{2\kappa}}{\sqrt{4\pi\kappa}}\frac{1}{2\kappa}\left(1-e^{-\kappa\frac{\pi^2}{4}}\right). $$ Of course, this is not defined for $\kappa=0$ but we know that in that case the integral has the exact value $\frac{\pi^2}{8}$.
So, how good is this approximation? It is fairly good indeed. Let me show some values
$Z(1)\sim 0.9538227748$ the exact value is $1.658067328$.
$Z(4)\sim 52.55432675$ the exact value is $61.08994014$.
$\vdots$
$Z(20)\sim 3.711926385\cdot 10^{14}$ the exact value is $3.804956771\cdot 10^{14}$.
$\vdots$
$Z(10 0)\sim 1.019204783\cdot 10^{83}$ the exact value is $1.024131055\cdot 10^{83}$.
To have a clear idea, in the range $\kappa=0.01\ldots 20$, I plotted the following log-log graph.
I should say that the agreement is excellent. The red curve is the exact one. I hope this will be of some help to you.
Best Answer
A few thoughts that may propel you onto a full proof.
Consider taking the following integral over the rectangle $0 \leq u \leq x, 0 \leq t \leq y$ $$I(x, y) = \int_0^x \int_0^ye^{-u-t}I_0(2 \sqrt{ut})\,\mathrm{du}\, \mathrm{dt}$$
I aim to show that $$I(x, y) = x-\frac{1}{2}e^{-x-y}[(x+y)I_0(\xi)+\xi I_1(\xi)]+(y-x)F(x, y) \tag{1}$$ Where $$F(x, y) = \frac{1}{2}\kappa\int_\xi^\infty e^{- \sigma t}f(t)\, \mathrm{dt}$$ and \begin{align} f(t) &= e^{-t}I_0(t)\\ \xi &=2\sqrt{xy} \\ \sigma &= \frac{(\sqrt{y}-\sqrt{x})^2}{\xi}\\ \kappa &= \frac{y-x}{\xi} \end{align}
So, write \begin{align} I(x, y)&=x+(y-x)K(x, y)-e^{-x-y}[\frac{1}{2}\xi I_1(\xi)+xI_0(\xi)] \tag{2}\\ K(x, y) &=\int_0^xe^{-(t+y)}I_0(2 \sqrt{ty})\,\mathrm{dt} \tag{3} \end{align} This relation comes straight from Lassey On the computation of certain integrals containing the modified Bessel function $I_0(\xi)$. The second relation here can be verified by applying $\partial^2 / \partial x \partial y$ to the first relation and using the fact that \begin{align} \frac{\partial K}{\partial y} &= -e^{-x-y}\sqrt{x/y}I_1(\xi)\\ \frac{\partial}{\partial x}\sqrt{x/y}I_1(\xi)&=I_0(\xi) \end{align} Now, proceeding with $(3)$ and replacing the Bessel function with $$I_0(2\sqrt{ty}) = \frac{1}{2 \pi i}\int_L e^{t/s + ys}\,\frac{\mathrm{ds}}{s}\tag{4}$$ Where $L$ is the unit circle around $s=0$. Inserting $(4)$ into $(2)$ yields $$K(x, y) = \frac{e^{-x-y}}{2 \pi i}\int_L \frac{e^{ys}(e^{x/s}-e^x)}{1-s}\, \mathrm{ds}$$ Integrating this along $|s| = \rho = \sqrt{x/y}$ (assuming that $\rho < 1$ then $$K(x,y)=- \frac{e^{-x-y}}{2 \pi}\int_0^\pi e^{\xi \cos \theta}\frac{2 \rho (\rho-\cos \theta)}{\rho^2 - 2\rho \cos \theta +1}\, \mathrm{d \theta} $$ Writing $2 \rho (\rho - \cos \theta) = \rho^2 - 2 \rho \cos \theta + 1 + \rho^2-1$ gives $$K(x, y) = F(x, y)-\frac{1}{2}e^{-x-y}I_0(\xi)\tag{5}$$ In the above we have integrated $(4)$ along the circle $|s|=\sqrt{t/y}$ to obtain $$I_0(\xi) = \frac{1}{\pi}\int_0^\pi e^{\xi \cos \theta}\, \mathrm{d \theta}$$ Returning to $(5)$ we define $$F(x, y) = \frac{(1-\rho^2)e^{-x-y}}{2 \pi} \int_0^\pi \frac{e^{\xi \cos \theta}}{\rho^2-2\rho\cos \theta +1}\, \mathrm{d \theta}$$ Combining $(5)$ and $(2)$ you should get to $(1)$. I hope that this gives some help, please comment and critique as you see fit!