# Determining when $\int_{0}^{\infty} \cos(\alpha x) \prod_{m=1}^{n} J_{0}(\beta_{m} x) \, \mathrm dx =0$ without using contour integration

Let $$J_{0}(z)$$ be the Bessel function of the first kind of order zero, and assume that $$\alpha$$ and $$\beta_{m}$$ are positive real parameters.

$$J_{0}(z)$$ is an even function that is real-valued along the real axis.

And when $$z$$ approaches infinity at a constant phase angle, $$J_{0}(z)$$ has the asymptotic form $$J_{0}(z) \sim \sqrt{\frac{2}{\pi z}} \cos \left(z-\frac{\pi}{4} \right), \quad |\arg(z)| < \pi.$$

So by integrating the entire function $$e^{i \alpha z} \prod_{m=1}^{n} J_{0}(\beta_{m}z) , \quad \sum_{m=1}^{n} \beta_{m} < \alpha,$$ around a contour that consists of the real axis and the infinitely large semicircle above it, it would seem to follow that $$\int_{0}^{\infty} \cos(\alpha x) \prod_{m=1}^{n} J_{0}(\beta_{m} x) \, \mathrm dx =0 \, , \quad \sum_{m=1}^{n} \beta_{m} < \alpha. \tag{1}$$

(For the cases $$n=1$$ and $$n=2$$, you would need to appeal to Jordan's lemma.)

Is there way to prove $$(1)$$ that doesn't involve contour integration?

EDIT:

A similar approach also shows that $$\int_{0}^{\infty} \frac{\cos(\alpha x)}{1+x^{2}} \prod_{m=1}^{n} J_{0}(\beta_{m} x) \, \mathrm dx = \frac{\pi e^{-a} }{2} \prod_{m=1}^{n}I_{0}(\beta_{m}), \quad \sum_{m=1}^{n} \beta_{m} \le \alpha,$$ where $$I_{0}(z)$$ is the modified Bessel function of the first kind of order zero.

• The integration extends only over the positive reals. How do you intend to deform the contour so that it enclosed the upper half plane? – Mark Viola Feb 12 '16 at 22:41
• @Dr.MV $J_{0}(x)$ is an even function – Random Variable Feb 12 '16 at 22:43
• So, are you starting with $\frac12 \text{Re}\left(\int_{-\infty}^\infty e^{i\alpha x}\prod_{m=1}^{n}J_0(\beta_m x)\,dx\right)$? – Mark Viola Feb 12 '16 at 22:54
• Oh sorry ... just saw the aside ... ;-)) – Mark Viola Feb 12 '16 at 23:01
• @Dr.MV Yes. If $\int_{-\infty}^{\infty} e^{i\alpha x} \prod_{m=1}^{n} J_{0}(\beta_{m} x) \, dx =0$ under the conditions on the parameters stated, then we also know that $\int_{0}^{\infty} \cos(\alpha x)\prod_{m=1}^{n} J_{0}(\beta_{m} x) \, dx =0$ under the same conditions since $J_{0}(x)$ is an even function that is real-valued along the real axis. – Random Variable Feb 12 '16 at 23:15

It's because the Fourier transform of $\mathrm{J}_0$ vanishes outside $[-1,1]$.
Let $I$ be the integral $$\def\J{{\mathrm{J}_0}}\def\dd{{\,\mathrm{d}}}\def\ii{{\mathrm{i}}} \def\ee{{\mathrm{e}}} I(\alpha) = \int_0^\infty \cos\alpha x\prod_k \J(\beta_k x)\,\dd x.$$ I will use the integral representation $$\J(x) = \int_0^\pi \cos(x\sin\theta) \frac{\dd\theta}{\pi} = \int_0^1 \cos(x u)\frac{2\dd u}{\pi\sqrt{1-u^2}}$$ together with the Fourier transform of the Heaviside sign function in the form $$\int_0^\infty e^{\ii ax}\,\dd x = \text{P.V.}\frac{\ii}{a} + \pi\delta(a).$$
Expanding each Bessel function, we get $$I(\alpha) = \int_0^\infty\dd x\int_0^1 \Big( \prod_k \frac{2\dd u_k}{\pi\sqrt{1-u_k^2}}\Big) \cos\alpha x \prod_k \cos(\beta_k x u_k).$$
Now expand each cosine as $\cos x = \frac12(\ee^{\ii x} + \ee^{-\ii x})$: $$\cdots = \int_0^\infty \dd x\int_0^1 \Big( \prod_k \frac{2\dd u_k}{\pi\sqrt{1-u_k^2}} \Big) \sum_{s\in\{\pm1\}^{n+1}} 2^{-n-1} \exp\Big( \ii s_0\alpha x + \sum_k \ii s_k \beta_k u_k x \Big),$$ where the sum is taken over all $2^{n+1}$ choices of signs $s_0,\ldots,s_n = \pm1$ that come from expanding the cosines in exponentials.
The integral over $x$ now can be done directly, as above: $$\cdots = \frac{1}{2\pi^{n-1}} \int_0^1 \Big( \prod_k \frac{\dd u_k}{\sqrt{1-u_k^2}} \Big) \sum_{s\in\{\pm1\}^{n+1}} \delta\Big(s_0\alpha + \sum_k s_k \beta_k u_k \Big).$$ (The imaginary part has to vanish so only the $\delta$ term remains.)
This makes it clear why the integral vanishes: the integral representation of the $n$ Bessel functions integrates over the $n$-cube $[0,1]^n$, but the $2^{n+1}$ hyperplanes $$s_0\alpha + \sum_k s_k \beta_k u_k = 0$$ do not intersect this cube at all when $$\sum_k \beta_k < |\alpha|.$$