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It is well known that, if $J_{\nu}$ denotes the Bessel function of the first kind, then we have

$$\displaystyle \int_{0}^{\infty} J_{0}(x) \ \mathrm{d}x = 1.$$

Moreover, since $\int J_{1}(x) = -J_{0}(x)$, then we also have

$$\displaystyle \int_{0}^{\infty} J_{1}(x) \ \mathrm{d}x = 1,$$

since $J_{0}(0) = 1$ and $J_{0}$ tends to zero at infinity. Can we say anything similar about $\displaystyle \int_{0}^{\infty}J_{\nu}(x) \ \mathrm{d}x,$ where $\nu > 0$ is either an integer or a half-integer? Do all of these integrals converge? Are they listed somewhere in a table of integrals? Gradshteyn-Ryzhik lists some examples, but they only seem applicable to $\nu = 0$ or $1$.

user363087
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1 Answers1

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Assuming $\nu\in\mathbb{N}$, we have: $$ \mathcal{L}\left(J_\nu(x)\right) = \frac{1}{\sqrt{1+s^2}\left(s+\sqrt{1+s^2}\right)^\nu},\qquad \mathcal{L}^{-1}(1)=\delta(s) \tag{1}$$ hence: $$ \int_{0}^{+\infty}J_\nu(x)\,dx = \lim_{s\to 0^+}\frac{1}{\sqrt{1+s^2}\left(s+\sqrt{1+s^2}\right)^\nu} = \color{red}{1}.\tag{2}$$ I exploited a useful property of the Laplace transform. The first identity can be easily derived from the generating function for Bessel functions of the first kind: $$ \exp\left(\frac{z}{2}\left(t-\frac{1}{t}\right)\right)=\sum_{n\in\mathbb{Z}}t^n J_n(z).\tag{3}$$ $(1)$ (then $(2)$) holds also if $\nu\in\mathbb{R}^+$ since $(1)$ follows from Bessel's differential equation, too.

Clarification: we have $$ \int_{0}^{+\infty}J_\nu(x)\,dx = \lim_{s\to 0^+}\int_{0}^{+\infty}J_\nu(x)e^{-sx}\,dx $$ since $J_\nu(x)$ is an improperly Riemann-integrable function, because $J_\nu(x)$ is continuous (much more: entire) in a right neighbourhood of the origin and $$J_\nu(x)\approx\sqrt{\frac{2}{\pi x}}\cos(x-\theta_\nu)$$ for large $x$ (equation $(58)$ here), so $\int_{0}^{+\infty}J_\nu(x)\,dx$ is convergent by Dirichlet's test, since $\cos(x-\theta_\nu)$ is a function with mean zero.

Jack D'Aurizio
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  • The formula you give for the Laplace transform is usually under the condition that the real part of $s$ should be positive. What is the argument that you are allowed to take $s\to 0$? – mickep Feb 04 '17 at 19:12
  • @mickep: that should be a $s\to 0^+$, now fixing. – Jack D'Aurizio Feb 04 '17 at 19:13
  • May I ask the reason for the downvote? – Jack D'Aurizio Feb 04 '17 at 19:23
  • Let me clarify that I am not the down voter. I still don't see how the validation of the limit $s\to0^+$ follows (I by no means doubt it, but when I in a discussion with a colleague some time ago about this very problem, we (he) ended up with a rather complicated argument.) – mickep Feb 04 '17 at 19:26
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    @mickep: in general, $$\int_{0}^{+\infty}f(x),dx = \lim_{s\to 0^+}(\mathcal{L}f)(s)$$ as soon as the dominated convergence theorem ensures $$\int_{0}^{+\infty}f(x),dx = \lim_{s\to 0^+}\int_{0}^{+\infty}f(x) e^{-sx},dx.$$ – Jack D'Aurizio Feb 04 '17 at 19:28
  • Sorry, I'm probably slow right now. But $J_\nu\not\in L^1(\mathbb R)$, so what is the dominated function in this case? – mickep Feb 04 '17 at 19:56
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    @mickep Since $\int_{0}^{\infty} J_{\nu}(x) , dx, \ v >-1,$ does not converge absolutely, justification here comes from the fact that the integral $\int_{0}^{\infty} f(x) e^{-p x} , dx$ converges uniformly on $[0, \infty)$ if $\int_{0}^{\infty} f(x) , dx$ exists as an improper Riemann integral. See Daniel Fischer's answer to this question. – Random Variable Feb 04 '17 at 20:07
  • @RandomVariable: thank you, I was going to tell exactly the same. – Jack D'Aurizio Feb 04 '17 at 20:31
  • @RandomVariable Thank you very much, it is now clear. I don't want to put words into the mouth of OP, but I actually got the feeling that it was this convergence that was the main problem. Jack, now when this is sorted out, you certainly get my vote. – mickep Feb 04 '17 at 21:19
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    @mickep: I added a clarification, I hope it is clear enough :) – Jack D'Aurizio Feb 04 '17 at 22:18