# Does this oscillatory integral exist?

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 ?

Motivation:

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 ? ) $$\int_{\mathbb{R}^{3}}\frac{e^{\dot{\imath}|x-y|^2}}{1+|y|}dy= \int_{\mathbb{S}^{2}}\int_{0}^{\infty} \frac{e^{\dot{\imath}|\rho\omega-x|^2}\rho^2}{1+\rho}d\rho d\omega\\ =e^{i|x|^{2}}\int_{\mathbb{S}^{2}}\int_{0}^{\infty} \frac{e^{\dot{\imath} (\rho^2-2x\cdot \omega\,\rho)}\rho^2}{1+\rho}d\rho d\omega.$$

Observations:

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

## 1 Answer

Does the existence of this limit imply the existence of the integral ?

No. Take $$n=1$$. Then $$\lim_{R\to\infty}\int_{-R}^{R}x\,dx$$ exists and equals $$0$$ because postive parts are exactly canceling negative parts as $$R$$ grows. But I would not say $$\int_{\mathbb{R}}x\,dx$$ exists.

Maybe there is a question yet to answer if you require the function to be positive.

• I think the most common definition of $\int_0^{\infty}f(x)\,dx$ is $\lim_{R\to\infty}\int_{0}^{R}f(x)\,dx$. In that case, you of course cannot find such an $f$. So what definition of $\int_0^{\infty}f(x)\,dx$ are you using? – alex.jordan Nov 14 '18 at 23:01
• Exactly my question: How define the improper integral $\int_{\mathbb{R}^n} F(x)dx$ when $n\geq2$ and $F$ is a continuous function. – Medo Nov 14 '18 at 23:04