# Asymptotic behavior and of an integral on a d-dimensional torus

I am trying to evaluate the asymptotic behavior of the following integral as $$t \to \infty$$:

$$I(t; \mathbf{v}) = \int_{[-\pi, \pi]^d} \frac{\sin(t f(\mathbf{k}))}{\sin(f(\mathbf{k}))} e^{i t \mathbf{k} \cdot \mathbf{v}} \, d\mathbf{k}$$

where $$\mathbf{v}$$ is a $$d$$-dimensional constant vector and

$$f(\mathbf{k}) = \arcsin\left(\sqrt{1 - \left(\frac{1}{d} \sum_{j=1}^d \cos(k_j)\right)^2}\right).$$

### Key points and progress

I think the factor $$\frac{\sin(t f(\mathbf{k}))}{\sin(f(\mathbf{k}))}$$ becomes a kind of $$\delta$$ function as $$t \to \infty$$. However I am not sure how to handle it.

### Question

How can I rigorously evaluate the asymptotic behavior of the integral $$I(t; \mathbf{v})$$ as $$t \to \infty$$? How can I evaluate the convergence rate with respect to $$t$$?

• have you tried the method of stationary phase/Laplace's method? Commented Jul 16 at 14:12
• @LSpice Thank you. I have corrected it. Commented Jul 16 at 21:47
• @nervxxx Thank you. I have tried the method of stationary phase. In this case, the phase function does not have zero of its gradient. Therefore I thought the method does not work for Fourier transform. Isn’t it? Commented Jul 16 at 21:47

The dominant contributions for large $$t$$ come from the two regions around $$\mathbf k_0 = \{0,\ldots,0\}$$ and $$\mathbf k_\pi = \{\pi,\ldots,\pi\}$$, where $$f(\mathbf k)\ll 1$$. Note that $$f(\mathbf k_0)=f(\mathbf k_\pi)=0$$ are the only zeroes of $$f$$ in the integration interval. $$\DeclareMathOperator{\sign}{sign} \DeclareMathOperator{\sinc}{sinc}$$

We first focus on the region around $$\mathbf k_0$$. Substituting $$\mathbf k \mapsto \mathbf q/t$$ and expanding the integrand around $$t=\infty$$ we find \begin{align} \frac{\sin(t f(\mathbf q/t))}{t\sin(f(\mathbf q/t))} = \sinc\left(\frac{|\mathbf q|}{\sqrt d}\right) + \mathcal O(t^{-2}) , \end{align} with the cardinal sine, $$\sinc(x)=\frac{\sin x}{x}.$$ For the one-dimensional case $$d=1$$ this gives \begin{align} I^{(d=1)}_{0}(\infty;v)&=\int_{-\infty}^{\infty}dq \sinc(q) \, e^{i q v} \\ &=\frac{\pi}{2}[\sign(1-v)+\sign(1+v)], \end{align} which is the well known Fourier transform of $$\sinc(q)$$. The region around $$\mathbf k_\pi$$ contributes the same, but with an additional phase factor, such that in total \begin{align} I^{(d=1)}(\infty;v)&=\frac{\pi}{2}[\sign(1-v)+\sign(1+v)][1+\cos(\pi v)]. \end{align} In higher dimensions the calculation is quite similar, \begin{align} I^{(d)}_{0}(\infty;\mathbf v)&=\int_{-\infty}^{\infty}d^d \mathbf q \sinc\left(\frac{|\mathbf q|}{\sqrt d}\right) \, e^{i \mathbf q \cdot \mathbf v}, \end{align} which now is the $$d$$-dimensional Fourier transform of the cardinal sine, with argument $$|\mathbf q|/\sqrt{d}$$. In $$d=2$$, Mathematica can evaluate this transform, with result \begin{align} I^{(d=2)}_{0}(\infty;\mathbf v)&= \Re\left(\frac{4\pi}{\sqrt{1-2 |\mathbf v|^2}}\right). \end{align} The region around $$\mathbf k_\pi$$ should again contribute a phase shift. Note that the result for $$I^{(d)}_{0}(\infty;\mathbf v)$$ must be a function of $$|\mathbf v|$$, as the Fourier transform of an isotropic function is isotropic, see the update below. Maybe it can be generalized to general $$d$$, the op should check the literature. I must go to bed now...

Update #1

See, e.g., MO:315613 and references therein for a discussion of the case $$d>2$$.

In $$d=3$$, the solution seems to be (maybe up to a constant) \begin{align} I^{(d=3)}_{0}(\infty;\mathbf v)&= \sqrt{\frac{\pi}{2}}\frac{\delta(1-\sqrt3|\mathbf{v}|)}{\sqrt3|\mathbf{v}|}. \end{align}

• Thank you for your rapid reply. I am trying to verify the calculations for the multi-dimensional case. One thing I want to know is the order of convergence of $I(t;\mathbf{v})$ with respect to $𝑡$. Is it easy? I cannot obtain this by the method of stationary phase. Commented Jul 17 at 6:02
• @KoHey See new first equation (for the convergence), and my update #1. Commented Jul 17 at 11:12
• Thank you for your update. I have calculated them for d dimensional cases. I want to evaluate the convergence rate with respect to $t$ which may depends on $d$. How can we evaluate them. Commented Jul 20 at 8:21
• I calculated them but I found that the factor $\frac{1}{t^{d-1}}$ may be missed through variable transform. @fred Commented Jul 23 at 12:19