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Bessel function integral

Let J_0 denote the Bessel function of the first kind. Prove that

    \[\int_0^\infty J_0(x) \, \mathrm{d}x=1\]

Solution

We recall that

    \[J_0(x)=\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{4^n (n!)^2}\]

Hence,

    \begin{align*} \int_{0}^{\infty}J_0(x)e^{-px}\,\mathrm{d}x &= \sum_{n=0}^{\infty}\frac{(-1)^n}{4^n p^{2n+1}}\binom{2n}{n}\\ &=\frac{1}{p}\sum_{n=0}^{\infty} \binom{-1/2}{n}\frac{1}{p^{2n}} \\ &= \frac{1}{\sqrt{1+p^2}} \end{align*}

Then,

    \[\lim_{p \rightarrow 0^+} \int_{0}^{\infty}J_0(x)e^{-px}\,\mathrm{d}x = \lim_{p \rightarrow 0^+} \frac{1}{\sqrt{1+p^2}} =1\]

Using the fact that the J_0(x) looks like an ‘almost periodic’ function with decreasing amplitude. If we denote by \{\alpha_k\}_{k \geq 0} the zeros of J_0 then \alpha_k \nearrow \infty as k \to \infty and furthermore

    \[\left| \int_{\alpha_k}^{\infty} J_0(x)e^{-px}\,\mathrm{d} x \right| \leq \int_{\alpha_k}^{\alpha_{k+1}} |J_0(x)|e^{-px}\,\mathrm{d} x \rightarrow 0\]

as k \rightarrow \infty for each p \geq 0. So the integral converges uniformly in this case justifying the interchange of limit and integral.

The result follows.

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