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.
 A: 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 
$$ \sum_k \beta_k  < |\alpha|. $$
