Answer to Question #97505 in Calculus for Rachel

Question

Let f(x) = x^a, xER^n, where a is a multi-index. Compute f-hat.

Answer 1

$\hat{f}(\omega)=\frac{1}{(2\pi)^{\frac{n}{2}}}\int\limits_{\mathbb R^n}f(x)e^{-i(x,\omega)}dx$

$\hat{f}(\omega)=\frac{1}{(2\pi)^{\frac{n}{2}}}\int\limits_{\mathbb R^n}x^{\alpha}e^{-i(x,\omega)}dx=\frac{1}{(2\pi)^{\frac{n}{2}}}\int\limits_{\mathbb R^n}\prod\limits_{k=1}^n x_k^{\alpha_k}e^{-ix_k\omega_k}dx=$

$=\prod\limits_{k=1}^n\bigl(\frac{1}{\sqrt{2\pi}}\int\limits_{\mathbb R}x_k^{\alpha_k}e^{-ix_k\omega_k}dx_k\bigr)$

Consider $\frac{1}{\sqrt{2\pi}}\int\limits_{\mathbb R}x^{\alpha}e^{-ix\omega}dx$. Prove by induction that $\frac{1}{\sqrt{2\pi}}\int\limits_{\mathbb R}x^{\alpha}e^{-ix\omega}dx=i^{\alpha}\sqrt{2\pi}\delta^{(\alpha)}(\omega)$.

Since $\delta(\omega)=\overline{\delta(\omega)}$ , for $\alpha=0$ we have $\frac{1}{\sqrt{2\pi}}\int\limits_{\mathbb R}e^{-ix\omega}dx=\overline{\frac{1}{\sqrt{2\pi}}\int\limits_{\mathbb R}e^{ix\omega}dx}=$

$=\sqrt{2\pi}\cdot\overline{\delta(\omega)}=\sqrt{2\pi}\delta(\omega)$.

Suppose that $\frac{1}{\sqrt{2\pi}}\int\limits_{\mathbb R}x^ke^{-ix\omega}dx=i^k\sqrt{2\pi}\delta^{(k)}(\omega)$, where $k\ge 0$.

Then $i^{k+1}\sqrt{2\pi}\delta^{(k+1)}(\omega)=i\frac{d}{d\omega}\bigl(i^k\sqrt{2\pi}\delta^{(k)}(\omega)\bigr)$. By the induction hypothesis we obtain $i^{k+1}\sqrt{2\pi}\delta^{(k+1)}(\omega)=i\frac{d}{d\omega}\bigl(\frac{1}{\sqrt{2\pi}}\int\limits_{\mathbb R}x^ke^{-ix\omega}dx)\bigr)=$

$=\frac{1}{\sqrt{2\pi}}\int\limits_{\mathbb R}ix^k\frac{de^{-ix\omega}}{d\omega}dx=\frac{1}{\sqrt{2\pi}}\int\limits_{\mathbb R}x^{k+1}e^{-ix\omega}dx$

So we prove this statement for $\alpha=k+1$.

By the principle of mathematical induction we obtain that $\frac{1}{\sqrt{2\pi}}\int\limits_{\mathbb R}x^{\alpha}e^{-ix\omega}dx=i^{\alpha}\sqrt{2\pi}\delta^{(\alpha)}(\omega)$ for all $\alpha\ge 0$.

So $\hat{f}(\omega)=\prod\limits_{k=1}^n\bigl(\frac{1}{\sqrt{2\pi}}\int\limits_{\mathbb R}x_k^{\alpha_k}e^{-ix_k\omega_k}dx_k\bigr)=$

$=\prod\limits_{k=1}^n\bigl(i^{\alpha_k}\sqrt{2\pi}\delta^{(\alpha_k)}(\omega_k)\bigr)=i^{|\alpha|}(2\pi)^\frac{n}{2}\prod\limits_{k=1}^n\delta^{(\alpha_k)}(\omega_k)$

Answer: $i^{|\alpha|}(2\pi)^\frac{n}{2}\prod\limits_{k=1}^n\delta^{(\alpha_k)}(\omega_k)$