Bessel function and upper bound I'm stuck on this following problem: Let $G$ a function such that $0\leq G(t)\leq 1$, and $G(t)=1$ if  $B^2\leq t\leq 4B^2$, with $\operatorname{supp}G\subset [\frac{1}{4}B^2, 9B^2]$ and $G^{(j)}\ll B^{-2j                    }$ then:
$\pi\int_{0}^{\infty}J_{0}(2\pi\sqrt{st})G(t)dt\ll B^2(1+sB^2)^{-\frac{3}{2}}$
where $J_{0}$ is the bessel function. I'm quite sure that I must use integration by parts to use the condition on $G$ but my attempts have failed miserably. Thank you in advance.
 A: By repeated integration by parts, we have
$$\int_0^\infty J_0(2\pi \sqrt{st}) G(t) \,dt = - \int_0^\infty \frac{1}{\pi}\left(\frac{t}{s}\right)^{1/2}J_1(2\pi \sqrt{st}) G'(t) \,dt \\= \int_0^\infty \frac{1}{\pi^2} \left(\frac{t}{s}\right)J_2(2\pi \sqrt{st}) G''(t) \,dt = -\int_0^\infty \frac{1}{\pi^3} \left(\frac{t}{s}\right)^{3/2} J_3(2\pi \sqrt{st}) G'''(t) \,dt.$$
You could keep going, but this is far enough.  Since the Bessel functions $J_\nu(x)$ are bounded on the positive reals, we have
$$\int_0^\infty J_0(2\pi \sqrt{st}) G(t) \,dt = -\int_0^\infty \frac{1}{\pi^3} \left(\frac{t}{s}\right)^{3/2} J_3(2\pi \sqrt{st}) G'''(t) \,dt = O\left(\frac{1}{Bs^{3/2}}\right).$$
Since also $J_0$ is bounded, we also have
$$\int_0^\infty J_0(2\pi \sqrt{st}) G(t) \,dt = O(B^2).$$
Put both those bounds together to get
$$\int_0^\infty J_0(2\pi \sqrt{st}) G(t) \,dt = O\left(\frac{B^2}{(1+sB^2)^{3/2}}\right).$$
REMARK: This technique relies heavily on the smoothness of the function $G(t)$.  If we just had a sharp cut-off for $G$, then (for fixed $s$) the best bound we could gotten would have been $O(B^{1/2})$ vs $O(B^{-1})$ with the smoothing.
ADDED LATER (at Bear's request):
All the integration by parts steps are justified by the following identity:

$$\int x^{n+1} J_n(x) \, dx = x^{n+1} J_{n+1}(x) + c$$ for $n \geq 0$.

