I have asked this question three weeks ago here

https://math.stackexchange.com/questions/2998601/does-this-oscillatory-integral-exist/2998930#2998930 but received no relevant answers.

Let $n\geq 2$ and consider the improper integral $$I:=\int_{\mathbb{R}^{n}}F(x)dx$$ where $F$ is a continuous function.

If $I$ exists then

$$I=\lim_{R\rightarrow +\infty}\int_{B_{R}}F(x)dx,$$ where $B_{R}$ is a ball with radius $R$. So if this limit does not exist we know that the integral does not exist. Does the existence of this limit imply the existence of the integral ?


I am interested in the existence of the integral $$\int_{\mathbb{R}^{3}}\frac{e^{\dot{\imath}|x-y|^2}}{1+|y|}dy.$$

Using spherical coordinates (I do not even know if we are allowed to change variables here. Are we ? ) $$\int_{\mathbb{R}^{3}}\frac{e^{\dot{\imath}|x-y|^2}}{1+|y|}dy= \int_{\mathbb{S}^{2}}\int_{0}^{\infty} \frac{e^{\dot{\imath}|\rho\omega-x|^2}\rho^2}{1+\rho}d\rho d\omega\\ =e^{i|x|^{2}}\int_{\mathbb{S}^{2}}\int_{0}^{\infty} \frac{e^{\dot{\imath} (\rho^2-2x\cdot \omega\,\rho)}\rho^2}{1+\rho}d\rho d\omega.$$


1-The inner integral does not exist for any $x$ and $\omega$.

2-We can not change order of integration

3-The limit

$$\lim_{R\rightarrow \infty}\int_{\mathbb{S}^{2}}\int_{0}^{R} \frac{e^{\dot{\imath} (\rho^2-2x\cdot \omega\,\rho)}\rho^2}{1+\rho}d\rho d\omega$$ exists. Simply apply the very nice formula [Grafakos, classical Fourier analysis-Appendix D]: $$\int_{\mathbb{S}^{n-1}} F(x.\omega)d\omega=c \int_{-1}^{1}(\sqrt{1-s^2})^{n-3} F(s|x|)ds.$$ then benefit from the oscillation in both variables $\rho$ and $\omega$ and integrate by parts in both variables.

Any ideas how to handle this ?

Thank you so much

  • $\begingroup$ How do you define the improper integral? There are a number of ways to do that. $\endgroup$ Dec 4, 2018 at 13:50

1 Answer 1


I find it convenient to use cylindrical coordinates $(\phi,\rho,z)$ rather than spherical coordinates. I orient the $z$-axis along the vector $\vec{x}=x_0\hat{z}$. The desired integral is $$I=\int_{\mathbb{R}^{3}}\frac{e^{i|\vec{x}-\vec{r}|^2}}{1+|\vec{r}|}d^3 r=2\pi\int_{-\infty}^\infty dz\int_0^\infty \rho\,d\rho \,\frac{\exp[i(x_0-z)^2+i\rho^2]}{1+\sqrt{z^2+\rho^2}}.$$ I have not been able to evaluate this in closed form, but since the OP asks about the "existence" I consider instead an integral which decays even less rapidly, $$J=2\pi\int_{-\infty}^\infty dz\int_0^\infty \rho\,d\rho \,\frac{\exp[i(x_0-z)^2+i\rho^2]}{1+\rho},$$ i.e. in the denominator I replace $z^2+\rho^2$ by $\rho^2$. The integral $J$ is independent of $x_0$ and evaluates in terms of an exponential integral Ei and Fresnel sine and cosine integrals S,C: $$J=i\pi^2+ i \pi e^i \sqrt{\pi/2}\left[(1-i) \text{Ei}(-i)+2 \pi S\left(\sqrt{2/\pi}\right)+2\pi i C\left(\sqrt{2/\pi}\right)\right].$$

The improper integral here is of the type $$\int_{-\infty}^\infty e^{iz^2}dz=(1+i)\sqrt{\pi/2},$$ discussed for example in this Physics.SE question.

  • $\begingroup$ Thank you, but I am afraid there is a significant factor of $|\vec{r}|^2$ missing in your computation. $\endgroup$
    – Medo
    Jul 9, 2020 at 16:32

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