# Specific integral of modified Bessel function of first kind

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.

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!

• Thank's a lot. Actually, already (2) solves my question by using $x=y$. Notice that in (2) there is a small typo: the first term should be $\frac{1}{2}\xi I_1(\xi)$. – Marcel Jul 18 '18 at 15:23
• Do you have any hints on how you guessed/knew formula (2)? Even with your hint of taking the derivatives it is straight forward but far from obvious to me. Also I do not completely get the advantage of (1) compared to (2). They both seem to me of same difficulty. Do you have any comments on this as well? – Marcel Jul 18 '18 at 15:30
• Hello Marcel. Thanks for the feedback. I will address the typos and your other question later today. Sorry for delay. Best, Kevin – higgs Jul 19 '18 at 6:57
• Hi Kevin, could you please explain me quickly how you came up with equation (2)? Until now, this seems to be kind of magic guessing. – Marcel Jul 24 '18 at 7:46
• @Marcel Hello! Apologies for the lateness, I had to do my job over the last few days! The result in (2) comes from a result in the paper of Lassey - I should have been more careful to express this at the outset. I have looked online for a copy of the paper but cannot source a pdf, only a reference to the AMS journal entry. I will re-write the section above to inlcude this as a reference. Please check it out, the apper itself is teriffic and considers lots of other cool relation involving $I_0(\xi)$. – higgs Jul 24 '18 at 8:13