The Fourier transform of a Lebesgue integrable function $f \in L^{1}(\mathbb{R})$ is given by

$$\hat{f}(u)=\int_{\mathbb{R}} f(x) e^{i x u} d \mu(x)$$

where $\mu$ is Lebesgue measure on the real line. For $f(x)=e^{-a x^{2}}, x \in \mathbb{R}, a>0$, prove that

$$\hat{f}(u)=\sqrt{\frac{\pi}{a}} e^{-\frac{u^{2}}{4 a}}$$

[You may use properties of derivatives of Fourier transforms without proof provided they are clearly stated, as well as the fact that $\phi(x)=(2 \pi)^{-1 / 2} e^{-x^{2} / 2}$ is a probability density function.]

State and prove the almost everywhere Fourier inversion theorem for Lebesgue integrable functions on the real line. [You may use standard results from the course, such as the dominated convergence and Fubini's theorem. You may also use that $g_{t} * f(x):=\int_{\mathbb{R}} g_{t}(x-y) f(y) d y$ where $g_{t}(z)=t^{-1} \phi(z / t), t>0$, converges to $f$ in $L^{1}(\mathbb{R})$ as $t \rightarrow 0$ whenever $\left.f \in L^{1}(\mathbb{R}) .\right]$

The probability density function of a Gamma distribution with scalar parameters $\lambda>0, \alpha>0$ is given by

$$f_{\alpha, \lambda}(x)=\lambda e^{-\lambda x}(\lambda x)^{\alpha-1} 1_{[0, \infty)}(x)$$

Let $0<\alpha<1, \lambda>0$. Is $\widehat{f_{\alpha, \lambda}}$ integrable?