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Let $j_{n}(x)$ be the spherical Bessel function of the first kind of nonnegative integer order $n$.

How can I prove the equation below using the spherical Bessel function recurrence relation?

$$ \int_0^\infty x^{-n} j_{n+1}(x) \, dx = \frac1{(2n+1)(2n-1)(2n-3)\cdots3\cdot1} = \frac1{(2n+1)!!} \tag{1}$$

We can express $j_{n}(x)$ in terms of the Bessel function of the first kind: $$ j_n(x)=\sqrt\frac{\pi}{2x} J_{n+\frac12}(x) $$

And Rayleigh's formula gives us a way to express $j_{n}(x)$ in terms of elementary functions:

$$j_n(x) = (-x)^n\left(\frac{1}{x}\frac{d}{dx}\right)^n\frac{\sin(x)}{x}$$

(See HERE for an explanation of the formula.)

But I don't know if these facts help us prove $(1)$.

EDIT:

The OP is not an active user, but I think he or she was referring to the recurrence relation $$\frac{d}{dx} \left(x^{-n}j_{n}(x) \right)=-x^{-n}j_{n+1}(x),\tag{1} $$ which reduces the evaluation to calculating a limit.

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user152691
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1 Answers1

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EDIT:

I think the OP was referring to the recurrence relation $$\frac{d}{dx} \left(x^{-n}j_{n}(x) \right)=-x^{-n}j_{n+1}(x). \tag{1} $$

From $(1)$ it follows immediately that $$\begin{align}\int_{0}^{\infty}x^{-n} j_{n+1}(x) \, dx &= -x^{-n} j_{n}(x) \Bigg|^{\infty}_{0} \\ &=-x^{n} \sqrt{\frac{\pi}{2x}} \, J_{n+\frac{1}{2}}(x) \Bigg|^{\infty}_{0} \\ &= \lim_{x \downarrow 0} \, x^{-n} \sqrt{\frac{\pi}{2x}} \, J_{n+\frac{1}{2}}(x) \tag{2} \\ &=\lim_{x \downarrow 0} \, x^{-n} \sqrt{\frac{\pi}{2x}}\left(\frac{1}{\Gamma\left(n + \frac{3}{2}\right)} \left(\frac{x}{2}\right)^{n+ \frac{1}{2}} +\mathcal{O}\left(x^{n+ \frac{5}{2}} \right)\right) \tag{3} \\ &= \sqrt{\pi} \, 2^{-n-1} \, \frac{1}{\Gamma \left(n+ \frac{3}{2}\right)} \\ &= \sqrt{\pi} \, 2^{-n-1} \, \frac{2^{n+1}}{\sqrt{\pi} (2n+1)!!} \tag{4}\\ &= \frac{1}{(2n+1)!!}. \end{align}$$

$(1)$: http://dlmf.nist.gov/10.51#E3

$(2)$: https://en.wikipedia.org/wiki/Bessel_function#Asymptotic_forms

$(3)$: https://en.wikipedia.org/wiki/Bessel_function#Bessel_functions_of_the_first_kind:_J.CE.B1

$(4)$: http://mathworld.wolfram.com/DoubleFactorial.html (2)

This might not be the particular approach you were seeking, but it can be evaluated using a property of the Mellin transform.

Ramanujan's master theorem states that if $f(x)$ has an expansion of the form $$ f(x) = \sum_{k=0}^{\infty} \frac{\phi(k)}{k!} (-x)^{k} ,$$ then

$$ \int_{0}^{\infty} x^{s-1} f(x) \, dx = \Gamma(s) \phi(-s) $$

for the values of $s$ for which the integral converges.

The hypergeometric representation of the Bessel function of the first kind of order $\alpha$ is $$ J_{\alpha}(x) = \frac{(\frac{x}{2})^{\alpha}}{\Gamma(\alpha +1)} \ {}_0F_{1} \left(\alpha +1; - \frac{x^{2}}{4} \right) = \frac{(\frac{x}{2})^{\alpha}}{\Gamma(\alpha +1)} \sum_{k=0}^{\infty} \frac{\Gamma(\alpha + 1)}{\Gamma(\alpha +1+k)} \frac{(-\frac{x^{2}}{4})^{k}}{k!} .$$

So for $s+n+1 >0$ and $s <2$, we have

$$ \begin{align} \int_{0}^{\infty} x^{s-1} j_{n+1}(x) \, dx &= \int_{0}^{\infty} x^{s-1} \sqrt{\frac{\pi}{2x}} \, J_{n+\frac{3}{2}} (x) \, dx \\ &= \int_{0}^{\infty} x^{s-1} \sqrt{\frac{\pi}{2x}} \frac{(\frac{x}{2})^{n + \frac{3}{2}}}{\Gamma(n +\frac{5}{2})} \, {}_0F_{1} \left(n + \frac{5}{2}; - \frac{x^{2}}{4} \right) \, dx \\ &= \frac{\sqrt{\pi}}{2^{n+2}} \frac{1}{\Gamma(n+\frac{5}{2})} \int_{0}^{\infty} x^{s+n} \, {}_0F_{1} \left(n + \frac{5}{2}; - \frac{x^{2}}{4} \right) \, dx \\ &= \frac{\sqrt{\pi}}{2^{n+2}} \frac{1}{\Gamma(n+\frac{5}{2})} \int_{0}^{\infty} (2 \sqrt{u})^{s+n} \, {}_0F_{1} \left(n + \frac{5}{2}; - u \right) \, \frac{du}{\sqrt{u}} \\ &= \frac{ \sqrt{\pi} \ 2^{s-2}}{\Gamma(n+\frac{5}{2})} \int_{0}^{\infty} u^{\frac{s+n+1}{2}-1} \, {}_0F_{1} \left(n + \frac{5}{2}; - u \right) \, du \\ &= \frac{ \sqrt{\pi} \ 2^{s-2}}{\Gamma(n+\frac{5}{2})} \, \Gamma \left(\frac{s+n+1}{2} \right) \frac{\Gamma(n+ \frac{5}{2})}{\Gamma(n + \frac{5}{2} - \frac{s+n+1}{2})} \\ &= \sqrt{\pi} \ 2^{s-2} \, \Gamma \left(\frac{s+n+1}{2} \right) \frac{1}{\Gamma(\frac{4-s+n}{2})} . \end{align}$$

Now let $s= -n+1$.

Then

$$ \begin{align} \int_{0}^{\infty} x^{-n} j_{n+1}(x) \ dx &= \sqrt{\pi} \ 2^{-n-1} \Gamma (1) \, \frac{1}{\Gamma \left(n + \frac{3}{2} \right)} \\ &= \sqrt{\pi} \ 2^{-n-1} \frac{2^{n+1}}{\sqrt{\pi} (2n+1)!!} \tag{1} \\ &= \frac{1}{(2n+1)!!} \end{align}$$

$(1)$: http://mathworld.wolfram.com/DoubleFactorial.html (2)

Random Variable
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  • Not exactly what I wanted but was very helpful. Thank you. – user152691 May 28 '14 at 20:28
  • @user152691 Thanks. I knew it wasn't exactly the approach you wanted. If you found it helpful, you can also upvote my answer as well. – Random Variable May 28 '14 at 20:36
  • @Random_Variable i tried to upvote your answer but i'm sorry my reputation is not enough, possibly 0 (because i'm very very new member :) ) – user152691 May 28 '14 at 20:38
  • @user152691 Apparently you need 15 reputation points to cast votes. And you only have 13 points. – Random Variable May 28 '14 at 20:43
  • This answer is very helpful. How were you able to perform the integral of $_0 F_1$? – jinawee Feb 11 '17 at 23:39
  • @jinawee $$ \ {}0F{1} \left(;n+\frac{5}{2};-u \right) = \sum_{k=0}^{\infty} \frac{\Gamma(n+ \frac{5}{2})}{\Gamma(n+ \frac{5}{2}+k)} \frac{(-u)^{k}}{k!} $$ Then I applied Ramanujan's master theorem. – Random Variable Feb 12 '17 at 00:34