4
$\begingroup$

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.

$\endgroup$
12
  • $\begingroup$ Wolfram Alpha gives the first integral as $ \frac{8}{3 \sqrt{3}}$ $\endgroup$ Sep 27 '15 at 4:15
  • $\begingroup$ Likewise, Wolfram Alpha gives the second integral as $\frac{4}{3 \sqrt{3}}$, exactly half of the first integral $\endgroup$ Sep 27 '15 at 4:17
  • $\begingroup$ I couldn't find the formula in Wolfram Alpha. Could you tell me the url? $\endgroup$
    – user
    Sep 27 '15 at 4:20
  • $\begingroup$ wolframalpha.com/input/… $\endgroup$ Sep 27 '15 at 4:21
  • $\begingroup$ 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 $\endgroup$ Sep 27 '15 at 4:22
3
$\begingroup$

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:

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

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ Sep 27 '15 at 23:50
  • $\begingroup$ @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. $\endgroup$
    – user
    Oct 1 '15 at 13:21
1
$\begingroup$

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

$\endgroup$
3
  • 2
    $\begingroup$ 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. $\endgroup$
    – mickep
    Sep 27 '15 at 7:58
  • $\begingroup$ @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. $\endgroup$ Sep 27 '15 at 23:52
  • $\begingroup$ @BrevanEllefsen I'm sorry for my late reply. I can understand obediently your calculation. Thank you very much :) $\endgroup$
    – user
    Oct 1 '15 at 13:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.