# integral involving modified bessel function

I would like to put in closed form the integral:

$\int{e^{-k x} I_0(x) dx }$

where $I_\alpha(x)$ is the modified Bessel function of the first kind.

The closest I have found in tables is for k=1

$\int{e^{-x} I_0(x) dx } = x e^{-x} (I_0(x)+I_1(x))$

It would be interesting to see if it is solvable, at least for a numerable set of k values.

The integral arises in a rather fundamental problem: the probability of being inside a ball of defined radious for a bivariate normal. So if it is not solvable, perhaps it worth to define a ad hoc function for it.

For any $\alpha\in\mathbb{N}$ we have: $$I_\alpha(x) = \frac{1}{\pi}\int_{0}^{\pi}e^{x\cos\theta}\cos(\alpha\theta)\,d\theta,$$ $$I_\alpha'(x) = \frac{1}{\pi}\int_{0}^{\pi}e^{x\cos\theta}\cos(\alpha\theta)\cos\theta\,d\theta=\left\{\begin{array}{rcl}\frac{1}{2}\left(I_{\alpha-1}(x)+I_{\alpha+1}(x)\right)&\text{if}&\alpha\geq 1,\\I_1(x)&\text{if}&\alpha=0.\end{array}\right.$$
$$I_{\alpha+1}(x)=\frac{x}{2\alpha+2}\left(I_{\alpha}(x)-I_{\alpha+2}(x)\right).$$ With these identities, it is straightforward to check that the primitive of $I_0(x)e^{-x}$ is $xe^{-x}(I_0(x)+I_1(x))$. Differentiating both sides, we have that the primitive of $(I_1(x)-I_0(x))e^{-x}$ is $I_0 e^{-x}$, hence the primitive of $I_1(x)e^{-x}$ is $I_0(x) e^{-x}+xe^{-x}(I_0(x)+I_1(x))$. Along the same lines, we can compute the primitive of $I_{\alpha}(x)e^{-x}$ with a recursive argument.
Since for any $k>1$ we have: $$\int f(x)e^{-kx}\,dx = -e^{-kx}\sum_{j=0}^{+\infty}\frac{f^{(j)}(x)}{k^{j+1}}$$ it follows that: $$\mathcal{J}_k(x)=\int I_0(x)e^{-kx}=-\frac{e^{-kx}}{\pi}\int_{0}^{\pi}e^{-x\cos\theta}\sum_{j=0}^{+\infty}\frac{\cos^j\theta}{k^{j+1}}\,d\theta=-\frac{e^{-kx}}{\pi}\int_{0}^{\pi}\frac{e^{-x\cos\theta}}{k-\cos\theta}\,d\theta,$$ or, by exploiting the Fourier cosine series of $e^{-x\cos\theta}$: $$-\pi e^{kx}\mathcal{J}_k(x)=\int_{0}^{\pi}\frac{I_0(x)}{k-\cos\theta}d\theta+2\sum_{m=1}^{+\infty}(-1)^m\int_{0}^{\pi}\frac{I_m(x)\cos(m\theta)}{k-\cos\theta}\,d\theta,$$ $$-e^{kx}\mathcal{J}_k(x)=\frac{I_0(x)}{\sqrt{k^2-1}}+2\sum_{j=1}^{+\infty}(-1)^j I_j(x)\int_{-\infty}^{+\infty}\frac{\cos(2j\arctan u)}{k(1+u^2)-(1-u^2)}\,du$$ where the last integrals can be computed through the residue theorem and they always equal $$\left(A+\frac{B}{\sqrt{k^2-1}}\right)$$ for some $A,B\in\mathbb{Z}$.
Start with $$\int e^{-k x}\;dx = -\frac{e^{-k x}}{k}$$ by inspection $$\frac{d^n}{dk^n}\left[-\frac{e^{-k x}}{k}\right]=\sum_{i=1}^{n+1}\frac{(-1)^{n+1}n!}{(n-i+1)!}\frac{x^{n+1-i}e^{-kx}}{k^i}$$ we also have $$\frac{d^n}{dk^n}\left[e^{-k x}\right]=(-x)^n e^{-kx}$$ generate the operator $$\hat{O}=\sum_{n=0}^\infty \frac{1}{4^nn!^2}\frac{d^{2n}}{dk^{2n}}$$ apply to both sides of the first equation $$\hat{O}\left[\int e^{-k x}\;dx\right] = \hat{O}\left[-\frac{e^{-k x}}{k}\right]$$ $$\int \sum_{n=0}^\infty \frac{1}{4^nn!^2}\frac{d^{2n}}{dk^{2n}}\left[e^{-k x}\right]\;dx = \sum_{n=0}^\infty \frac{1}{4^nn!^2}\frac{d^{2n}}{dk^{2n}}\left[-\frac{e^{-k x}}{k}\right]$$ $$\int \sum_{n=0}^\infty \frac{1}{4^nn!^2}(-x)^{2n}e^{-kx}\;dx = \sum_{n=0}^\infty \frac{1}{4^nn!^2}\sum_{i=1}^{2n+1}\frac{(-1)^{2n+1}(2n)!}{(2n-i+1)!}\frac{x^{2n+1-i}e^{-kx}}{k^i}$$ we have $$I_0(x)=\sum_{n=0}^\infty \frac{x^{2n}}{4^nn!^2}$$ $$\int e^{-kx}I_0(x)\;dx = -e^{-kx}\sum_{n=0}^\infty \frac{(2n)!x^{2n}}{4^nn!^2}\sum_{i=1}^{2n+1}\frac{x^{1-i}}{k^i(2n-i+1)!}$$ $$\int e^{-kx}I_0(x)\;dx = -e^{-kx}\sum_{n=0}^\infty \frac{(2n)!x^{2n}}{4^nn!^2}\left(-\frac{x}{(1+2n)!}+e^{kx}\frac{\Gamma(2+2n,k x)}{k(kx)^{2n}(1+2n)!} \right)$$ $$\int e^{-kx}I_0(x)\;dx = xe^{-kx}\,_1F_2\left(\frac{1}{2};1,\frac{3}{2},\frac{x^2}{4}\right) - \sum_{n=0}^\infty \frac{\Gamma(2+2n,k x)}{4^nn!^2 (1+2n)k^{2n+1}}$$ The incomplete gamma function does not help reduce the last sum to a hypergeometric function.