A Series of Integrals like $\sin (x^2)/x^2$

Question

Is there a general solution or pattern in the following series of integrals?

$$S(n)=\idotsint_{-\infty}^{\infty} \frac{\sin(x_1^2+x_2^2+...+x_n^2)}{x_1^2+x_2^2+...+x_n^2}\, dx_1 \dots dx_n$$

I can solve the first integral $$S(1)=\int_{-\infty}^\infty\frac{\sin(x^2)}{x^2} dx$$

$$=2\int_{0}^\infty\frac{\sin(x^2)}{x^2} dx$$ Subbing $t$ for $x^2$: $$=\int_{0}^\infty\frac{\sin(t)}{t^{3/2}} dt$$

Using the Laplace Transform functions, $\mathcal{L}(\sin t)=\frac{1}{1+p^2}$ and $\mathcal{L}^{-1}(t^{-3/2})=\frac{2}{\sqrt\pi}\sqrt{p}$:

$$=\frac{2}{\sqrt{\pi}}\int_{0}^{\infty}\frac{\sqrt{p}}{1+p^2} dp =\frac{4}{\sqrt{\pi}}\int_{0}^{\infty}\frac{x^2}{1+x^4} dx$$

And the last integral has been solved here through various methods to be $\frac{\pi}{2\sqrt{2}}$, meaning $S(1)=\sqrt{2\pi}$.

As another quick example, the second integral

$$S(2)=\iint_{-\infty}^\infty\frac{\sin(x^2+y^2)}{x^2+y^2} dx\, dy$$

can be solved by converting to polar coordinates where $r^2=x^2+y^2$.

$$=\int_0^{\infty}\int_0^{2\pi}\frac{\sin(r^2)}{r} d\theta\, dr =2\pi\int_0^{\infty}\frac{\sin(r^2)}{r} dr =\pi\int_0^{\infty}\frac{\sin(x)}{x} dx =\frac{\pi^2}{2}$$

Obviously, these techniques don't work as more variables of integration are added. What are the higher values of $S(n)$ and is there a general formula?

Answer 1

You have already computed $S(1)$ and $S(2)$. So let's assume that $n\geq 3$. Define: $$f_n(a, b) = \int_{\mathbb R}\dots \int_{\mathbb R} \frac{e^{-(a-ib)(x_1^2+\dots+x_n^2)}}{x_1^2+\dots+x_n^2}dx_1\dots dx_n$$ Clearly, by evaluating at $a=0$ and $b=1$ and taking the imaginary part: $$S(n)=\Im f_n(0, 1)$$ Now $$\begin{split} \frac{\partial f_n}{\partial a}(a,0) &= -\int_{\mathbb R}\dots \int_{\mathbb R} e^{-a(x_1^2+\dots+x_n^2)}dx_1\dots dx_n\\ &=-\left( \int_{\mathbb R}e^{-ax^2}dx\right)^n\\ &=-\left(\frac \pi a\right)^{\frac n 2} \end{split}$$ Thus, noting that $f(a,0)\rightarrow 0$ as $a\rightarrow +\infty$, $$f(a,0)=\int_a^{+\infty}\left(\frac \pi u\right)^{\frac n 2}du=-\frac{\pi^{\frac n 2}}{1-\frac n 2}\cdot\frac {1}{a^{\frac n 2 -1}}\tag{1}$$ Also $$\begin{split} \frac{\partial f_n}{\partial b}(a,b) &= i\int_{\mathbb R}\dots \int_{\mathbb R} e^{-(a-ib)(x_1^2+\dots+x_n^2)}dx_1\dots dx_n\\ &=i\left( \int_{\mathbb R}e^{-(a-ib)x^2}dx\right)^n\\ &=i\left(\frac{\pi}{a-ib}\right)^{\frac n 2} \end{split}$$ Thus $$ f(a,b)-f(a,0)=\int_0^b i\left(\frac{\pi}{a-iv}\right)^{\frac n 2}dv\\ =\frac {\pi^{\frac n 2}} {\frac n 2-1}\cdot \left(\frac 1 {(a-ib)^{\frac n 2 -1}} - \frac 1 {a^{\frac n 2 -1}}\right) $$ Combining with $(1)$ gives $$f(a,b)=\frac {\pi^{\frac n 2}} {\frac n 2-1}\cdot \frac 1 {(a-ib)^{\frac n 2 -1}} $$ Thus $$f(0,1) = \frac {\pi^{\frac n 2}} {\frac n 2-1}\cdot \frac 1 {(-i)^{\frac n 2 -1}} = -\frac {i\pi^{\frac n 2}} {\frac n 2-1}e^{\frac{i\pi n}{4}}$$ and $$\boxed{S(n)=-\frac{\pi^{\frac n 2}}{\frac n 2 -1}\cos\left(\frac {n\pi}4\right)}$$

Edit: As noted by @Ninad Munshi, this formula works for all $n\geq 1$ (not just $n\geq 3$). For $n=2$, you need to take the limit $n\rightarrow 2$.