# Fourier transform of $e^{-\lvert \lambda x\rvert}$

I'm working on an exercise from Rudin's Real and Complex Analysis: Chapter 9, Exercise 14. The problem is to prove the theorems about the Fourier transform on $\mathbf{R}^{1}$ discussed in the chapter in $\mathbf{R}^{k}$ instead. I'm trying to generalize Section 9.7 to $\mathbf{R}^{k}$. Let $H$ be defined in $\mathbf{R}^{k}$ by $$H(x) = e^{-\lvert x\rvert},$$ and for every positive number $\lambda$, let $h_{\lambda}$ be defined by $$h_{\lambda}(x) = \int_{\mathbf{R}^{k}} H(\lambda y)e^{ix\cdot y} dm(y).$$

I was able to show that $H$ is integrable by using polar coordinates: $$\int_{\mathbf{R}^{k}} H(x) \:dm(x) = \int_{0}^{\infty} r^{k-1} \:dr \int_{\mathbf{S}_{k-1}} e^{-\lvert ru\rvert} d\sigma(u) = k\cdot m(\mathbf{B}_{k})\int_{0}^{\infty} e^{-r}r^{k-1}\: dr = k\Gamma(k)m(\mathbf{B}_{k}),$$ where $\mathbf{S}_{k-1}$ is the unit $k$-sphere, $\sigma$ is the Borel measure defined in $\mathbf{S}_{k-1}$ by $$\sigma(E) = k\cdot m\{ru~:~0<r<1,~u\in E\},$$ and $\mathbf{B}_{k}$ is the unit $k$-ball. But I'm having trouble computing $h_{\lambda}$ in a closed form expression as is done in the section: when $k=1$, it follows from symmetry and integration by parts that $$h_{\lambda}(x) = \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} e^{-\lambda t}\cos(tx) \:dt = \sqrt{\frac{2}{\pi}} \frac{\lambda}{\lambda^{2}+x^{2}}$$ (here ordinary Lebesgue measure is scaled by $1/\sqrt{2\pi}$). I'm not sure how to do the calculation for $k>1$.

• If you use $$\lvert x\rvert_1 = \sum_{\kappa = 1}^k \lvert x_{\kappa}\rvert$$ rather than the Euclidean norm, you get an easy-to-handle function. – Daniel Fischer Aug 7 '17 at 21:00