# Definite integral involving modified Bessel function

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.

• Wolfram Alpha gives the first integral as $\frac{8}{3 \sqrt{3}}$ – Brevan Ellefsen Sep 27 '15 at 4:15
• Likewise, Wolfram Alpha gives the second integral as $\frac{4}{3 \sqrt{3}}$, exactly half of the first integral – Brevan Ellefsen Sep 27 '15 at 4:17
• I couldn't find the formula in Wolfram Alpha. Could you tell me the url? – user Sep 27 '15 at 4:20
• wolframalpha.com/input/… – Brevan Ellefsen Sep 27 '15 at 4:21
• We do know that $I_0(z) = \sum_{k=0}^{\infty} \frac{(\frac{1}{4}z^2)^k}{(k!)^2}$... Maybe I can solve the first in terms of this. I'm trying to figure out what process WA uses – Brevan Ellefsen Sep 27 '15 at 4:22

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:

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

• I finally finished my proof with Mickep's help, but I missed the beauty of this transform. As is most of your work, beautifully done. – Brevan Ellefsen Sep 27 '15 at 23:50
• @Jack D'Aurizio I'm sorry for my late reply. You told me your technique using Laplace transform but I forgot... Thank you again. When you get a chance, I'm glad if you think about this question. – user Oct 1 '15 at 13:21

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 $$\frac{1}{\pi}\int_{0}^{\infty}xe^{-x}I_{1}\left(-\frac{x}{2}\right)dx$$ $$= \frac{1}{\pi}\int_{0}^{\infty}xe^{-x}\int_0^{\pi} e^{-\frac{x}{2}\cos\theta}\cos\theta \,d\theta \,dx$$ $$= \frac{1}{\pi}\int_{0}^{\infty}\int_0^{\pi} xe^{-x(\frac{\cos\theta}{2}+1)}\cos\theta \,d\theta \,dx$$ $$= \frac{1}{\pi}\int_0^{\pi} \frac{4\cos\theta}{(\cos\theta+2)^2} d\theta$$ $$= \frac{1}{\pi}\int_0^{\pi} \frac{4}{(\cos\theta+2)^2} d\theta$$ $$= \frac{1}{\pi}\frac{-4\pi}{3\sqrt{3}} = \color{red}{\frac{-4}{3\sqrt{3}}}$$

• Just change order of integration, and use that $\int_0^{+\infty} x e^{-ax}\,dx=1/a^2$. Then, you have a "simple" integral of a rational function in $\cos \theta$ left. – mickep Sep 27 '15 at 7:58
• @mickep Thanks for the hint! It seems obvious now, but I guess I was just too tired last night to notice that I could simply flip the integrals. – Brevan Ellefsen Sep 27 '15 at 23:52
• @BrevanEllefsen I'm sorry for my late reply. I can understand obediently your calculation. Thank you very much :) – user Oct 1 '15 at 13:14