In a paper I am reading, the author is able to obtain the following inequality for $k\neq 0$,
$$\left|\int_{[-1,1]^d}e^{ik\cdot x}dx\right|\lesssim\frac{1}{\prod_{j=1}^d(1+|k_j|)}.$$
I am not sure how they were able to obtain these terms, I can only manage as far as:
$$\left|\int_{[-1,1]^d}e^{ik\cdot x}dx\right|\lesssim\frac{1}{||k||},$$ where $||k||=\max_{1\leq j\leq d}|k_j|$.
Could someone point me to some hints? Thank you.
Best Answer
Compute $$\int_{[-1,1]^d} e^{ik\cdot x}\, dx = \prod_{j = 1}^d \int_{-1}^1 e^{ik_jx_j}\, dx_j = \prod_{j = 1}^d \frac{2\sin k_j}{k_j} = 2^d \prod_{j = 1}^d \frac{\sin k_j}{k_j}$$ and note $$\prod_{j = 1}^d \left\lvert \frac{\sin k_j}{k_j}\right\rvert = \prod_{j = 1}^d \left(\frac{\lvert \sin k_j\rvert}{\lvert k_j\rvert} + \lvert \sin k_j\rvert\right)\frac{1}{1 + \lvert k_j\rvert}\le \prod_{j = 1}^d \left(\frac{\lvert \sin k_j\rvert}{\lvert k_j\rvert} + 1\right)\frac{1}{1 + \lvert k_j\rvert}$$By the inequality $\lvert \sin t\rvert \le \lvert t\rvert, t\in \Bbb R$, the latter product is bounded above by $$\prod_{j = 1}^d \frac{2}{1 + \lvert k_j\rvert} = \frac{2^d}{\prod_{j = 1}^d (1 + \lvert k_j\rvert)}$$Hence $$\left\lvert \int_{[-1,1]^d} e^{ik\cdot x}\, dx\right\rvert \lesssim \frac{1}{\prod_{j = 1}^d (1 + \lvert k_j\rvert)}$$