# Proving an integral is finite

I have the following integral:

$$\displaystyle \int_{\mathbb{R}^2} \left( \int_{\mathbb{R}^2} \frac{J_{1}(|\alpha|)J_{1}(|k- \alpha|)}{|\alpha||k-\alpha|} \ \mathrm{d}\alpha \right)^2 \ \mathrm{d}k,$$

where both $\alpha$ and $k$ are vectors in $\mathbb{R}^2$, with $k \neq 0$, and $J_{\nu}$ denotes the Bessel function of the first kind. I'm having some trouble with the best way to approach this integral. If we focus on the inner integral first, then using the fact that for sufficiently large, positive $z$ we have $|J_{\nu}(z)| \leqslant C|z|^{-1/2},$ then the inner integral can be reduced to

$$\displaystyle \int_{\mathbb{R}^2} |\alpha|^{-3/2}|k-\alpha|^{-3/2} \ \mathrm{d}\alpha.$$

However, as can be seen in this answer, this integral is $O(|k|^{-1})$, which, after squaring, is clearly not integrable over all $|k| \geqslant 1$ after switching to polar co-ordinates (obviously not including $0$ in the lower limit of the outer integral). We would need an estimate of at least $O(|k|^{-1 - \epsilon})$ for any $\epsilon > 0$ to guarantee convergence of the outer integral.

One idea might be to try to bring the outer integral inside (though one would need to justify interchanging the order of integration). Using the asymptotics for the Bessel functions gives a product of cosines, and then one can use polar co-ordinates (taking $r = |\alpha|$). This would cancel out the $|\alpha|$ in the denominator, but then the $|k-\alpha|$ terms get very messy, which seems to make things worse. The Bessel functions appear to cause the most trouble. Does anyone have any ideas on how to proceed?

One idea is to notice (as someone suggested in the comments) that the above is the $L^2$ norm of a convolution, and also that (up to a constant) we have

$$\displaystyle f(\xi) = \frac{J_{d/2}(|\xi|)}{|\xi|^{d/2}} = \mathcal{F}(\chi)(\xi),$$

where $\chi$ is the characteristic function of the unit ball in $\mathbb{R}^d$. It can be shown that $f \in L^2(\mathbb{R}^d)$ and even $L^{p}(\mathbb{R}^d)$ for any $p \geqslant 2$. This lets us write the entire expression as

$$\|f \ast f\|_2^2 = \|\mathcal{F}(\chi) \ast \mathcal{F}(\chi)\|_2^2,$$

but unfortunately there is no kind of convolution theorem for functions on $L^2(\mathbb{R}^d)$ that I am aware of, without going into the theory of distributions. Moreover, $f$ does not even belong to $S(\mathbb{R}^d)$ for any $d$, so we cannot say much about the convolution. Thus, the problem is equivalent to asserting the finiteness of the above norm, $\|f \ast f\|_2$. If anyone has any ideas on how else the problem could be approached, then I would be very keen to hear about them.

• I don't know if this can help but you may write your expression as an L_2 norm of a convolution. – GGG Feb 7 '17 at 12:24
• @GGG It is interesting that you mention that, because that is mostly how this integral was derived -- it began as a discrete convolution which we then approximated by this integral. Roughly speaking this expression corresponds to the $L^4$ norm of the difference between the number of integer points inside a ball and its volume. – user363087 Feb 7 '17 at 16:01
• Interesting! I was mentioning that because when you have the norm of a convolution, then you may use the Fourier transform. I'm not an expert of Bessel functions, so I don't know if their explicit Fourier transform is known. – GGG Feb 7 '17 at 23:09
• Furthermore to give you some mathematical input: It looks like Cauchy-Schwarz could be very helpful here – tired Feb 9 '17 at 15:41
• having a quick look the cauchy schwarz approach should yield (applying your inequality for bessel's functions) an upper bound for the inner integrals like $Const/(\rho^2|k|^4)$ which yield a finite value if integrated over $|k|\geq1$. Correct? – tired Feb 9 '17 at 16:31

## 1 Answer

I think this problem can be tackled by bounding $\left|\frac{J_1(z)}{z}\right|$ with $\frac{1}{2}e^{-z^2/8}$ for $z\in[0,2\pi]$ and with $\sqrt{\frac{2}{\pi z^3}}$ for $z\geq 2\pi$. These bounds come from the Taylor series at the origin and Laplace method. Let $f(z)=\frac{J_1(z)}{z}$ for simplicity. We have:

$$\int_{\mathbb{R}^2} f(\left|\alpha\right|)\,f(\left|k-\alpha\right|)\,d\mu\\ \leq \frac{1}{2}\int_{\left|k-\alpha\right|\leq 2\pi}f(\left|\alpha\right|)\,e^{-|k-\alpha|^2/8}\,d\mu+\sqrt{\frac{2}{\pi}}\int_{\left|k-\alpha\right|\geq 2\pi}f(\left|\alpha\right|)\frac{d\mu}{|k-\alpha|^{3/2}}$$ and the original integral should be simple to approximate by splitting the integration range in four parts:

1. $\alpha$ and $k$ being close to each other and close to the origin;
2. $\alpha$ and $k$ being close to each other, far from origin;
3. $\alpha$ and $k$ being far from each other, both far from the origin;
4. $\alpha$ and $k$ being far from each other, one of them being close to the origin.

For instance, let we approximate $$I_1 = \int_{\mathbb{R}^2}\left(\int_{\mathbb{R}^2}\exp\left(-\frac{1}{8}(\left|\alpha\right|^2+\left|k-\alpha\right|^2\right)\,d\alpha\right)^2\,dk .$$ By assuming $k=\rho_1(\cos\theta_1,\sin\theta_1)$ and switching to polar coordinates, the innermost integral equals $$\int_{0}^{2\pi}\int_{0}^{+\infty}\rho\,\exp\left(-\frac{1}{8}\left(2\rho^2+\rho_1^2-2\rho\rho_1\cos(\theta-\theta_1)\right)\right)\,d\rho\,d\theta$$ that is positive and bounded by $$2\pi\int_{0}^{+\infty}\exp\left(-\frac{1}{8}\left(2\rho^2-\rho_1^2-2\rho\rho_1\right)\right)\,d\rho\leq 16\pi \exp\left(-\frac{(\rho_1-4)^2}{8}\right)$$ that is a Schwartz function in $\rho_1$. It follows that $I_1$ is finite.

On the other hand, most of the mass of the integral is concentrated on the region over which $\left|\alpha\right|,\left|k\right|,\left|\alpha-k\right|$ are large, and by this previous answer (that ultimately boils down to the triangle inequality) $I_3$ is not finite. This is how the world ends: not with a bang, but with a whimper.

• This looks very promising, but I have a few questions -- firstly, what do you mean when you write $\mathrm{d}\mu$? Secondly, since the inner integral is being squared, this seems to make it very difficult to then integrate the (square) of the two integrals you have bounded the inner integral by -- and despite partitioning the outer integral over $\mathbb{R}^d$ into 4 parts, I am not sure how we can deal with the squares of those integrals. How can this issue be resolved? – user363087 Feb 20 '17 at 17:53
• @user363087: with $\mu$ I denoted the $2$-dimensional Lebesgue measure. I do not like to use $d\alpha$ since $\alpha$ is a vector, but that is just a matter of personal taste. About your second question: if $\alpha$ or $k-\alpha$ are small, a gaussian function, i.e. a Schwartz function appears in the integral. It follows that the only problematic case to deal with is the case in which both $\alpha$ and $k-\alpha$ are large. But in such a case we just have to approximate the integral of a rational function, and the exponent $\frac{3}{2}$ should be enough to grant convergence. – Jack D'Aurizio Feb 20 '17 at 17:57
• Apologies if this is a silly question, but I still can't see how to do any of the integrals. For instance, if $\alpha$ is small, then one of the integrals we deal with is $\displaystyle \int \left( \int e^{-(|\alpha|^2 + |k - \alpha|^2)/8} \ \mathrm{d}\alpha \right)^{2} \ \mathrm{d}k$. I have no idea how to do this. Could you by any chance show me how to do one of the cases so I can try to do the others? – user363087 Feb 20 '17 at 19:02
• @user363087: all right, updating. – Jack D'Aurizio Feb 20 '17 at 19:21
• Thanks. In the last line, how were you able to get rid of the factor of $\rho$ outside the exponential, and how was the inequality deduced? – user363087 Feb 20 '17 at 20:07