Definite integral involving modified Bessel function

Question

I'm trying to solve the following integrals: $$ I:=\int_{0}^{\infty}xe^{-x}I_{0}\left(-\frac{x}{2}\right)dx\quad II:=\int_{0}^{\infty}xe^{-x}I_{1}\left(-\frac{x}{2}\right)dx $$ where $I_{n}$ denotes the modified Bessel function of first kind.

My question : Is my answer finished and true?

My answer : $$ I=\sum_{k=0}^{\infty}\frac{(2k+1)2k\cdots(k+1)}{k!}\left(\frac{1}{4}\right)^{2k} $$ $$ II=-\sum_{k=0}^{\infty}\frac{(2k+2)(2k+1)\cdots (k+2)}{k!}\left(\frac{1}{4}\right)^{2k+1} $$

I wonder if I can get more simple form of solution but I don't know. If there are some formula, I'm glad if you tell me.

Thank you.

Answer 1

Now a full solution using Elementary methods (Thanks to Mickep for the advice) $$I_n(z) = \frac{1}{\pi} \int_0^{\pi} e^{z\cos\theta}\cos(n\theta)d\theta$$ $$I_0(-\frac{x}{2}) = \frac{1}{\pi} \int_0^{\pi} e^{-\frac{x} {2}\cos\theta}d\theta $$ $$***$$ $$\frac{1}{\pi}\int_{0}^{\infty}xe^{-x}I_{0}\left(-\frac{x}{2}\right)dx$$ $$= \frac{1}{\pi}\int_{0}^{\infty}xe^{-x}\int_0^{\pi} e^{-\frac{x}{2}\cos\theta}d\theta dx$$ $$= \frac{1}{\pi}\int_{0}^{\infty}\int_0^{\pi} xe^{-x(\frac{\cos\theta}{2}+1)}d\theta dx$$ $$= \frac{1}{\pi}\int_0^{\pi} \frac{4}{(\cos\theta+2)^2} d\theta$$ $$= \frac{1}{\pi}\frac{8\pi}{3\sqrt{3}} = \color{red}{\frac{8}{3\sqrt{3}}}$$ $$***$$ We follow a similar process for the second integral $$\frac{1}{\pi}\int_{0}^{\infty}xe^{-x}I_{1}\left(-\frac{x}{2}\right)dx$$ $$= \frac{1}{\pi}\int_{0}^{\infty}xe^{-x}\int_0^{\pi} e^{-\frac{x}{2}\cos\theta}\cos\theta \,d\theta \,dx$$ $$= \frac{1}{\pi}\int_{0}^{\infty}\int_0^{\pi} xe^{-x(\frac{\cos\theta}{2}+1)}\cos\theta \,d\theta \,dx$$ $$= \frac{1}{\pi}\int_0^{\pi} \frac{4\cos\theta}{(\cos\theta+2)^2} d\theta$$ $$= \frac{1}{\pi}\int_0^{\pi} \frac{4}{(\cos\theta+2)^2} d\theta$$ $$= \frac{1}{\pi}\frac{-4\pi}{3\sqrt{3}} = \color{red}{\frac{-4}{3\sqrt{3}}}$$

Answer 2

There are way nicer closed forms, since: $$ \mathcal{L}\left(I_0(x)\right) = \frac{1}{\sqrt{s^2-1}}\cdot\mathbb{1}_{s>1}(s),\qquad \mathcal{L}\left(I_0(x)\right) = -1+\frac{s}{\sqrt{s^2-1}}\cdot\mathbb{1}_{s>1}(s)$$ and: $$ \mathcal{L}^{-1}\left(x e^{-2x}\right) = \delta'(s-2),$$ hence the first integral equals:

$$ I = 4\int_{0}^{+\infty}x e^{-2x} I_0(x)\,dx =4\cdot\left.\frac{d}{ds}\frac{1}{\sqrt{s^2-1}}\right|_{s=2}=\color{red}{\frac{8}{3\sqrt{3}}}$$

and the second integral equals:

$$ II = -4\int_{0}^{+\infty}x e^{-2x} I_1(x)\,dx = -4\cdot\left.\frac{d}{ds}\frac{s}{\sqrt{s^2-1}}\right|_{s=2} = \color{red}{-\frac{4}{3\sqrt{3}}}.$$