Show that for each $\epsilon > 0$ the function $F(\xi) = \frac 1{(1+|\xi|^2)^\epsilon}$ is the Fourier transform of an $L^1$ function.
[Hint: With $K_\delta(x)=e^{-\frac{\pi|x|^2}\delta} \delta^{-\frac d2}$ consider $f(x)=\int_0^\infty K_\delta (x) e^{-\pi \delta} \delta^{\epsilon - 1} \, d\delta$. Use Fubini's theorem to prove $f \in L^1(\mathbb R^d)$, and $$\hat f(\xi)=\int_0^\infty e^{-\pi\delta |\xi|^2}e^{-\pi \delta} \delta^{\epsilon - 1} \, d\delta,$$ and evaluate the last integral as $\pi^{-\epsilon} \Gamma(\epsilon) \frac 1{(1+|\xi|^2)^\epsilon}$. Here $\Gamma(s)$ is the gamma function defined by $\Gamma(s)=\int_0^\infty e^{-t} t^{s-1} \, dt$.]
Work attempted to prove $f \in L^1(\mathbb R^d)$:
First, notice that \begin{align*} \int_{\mathbb R^d} e^{-\frac{\pi|x|^2}{\delta}} \, dx &= \prod_{i=1}^d \int_{-\infty}^\infty e^{-\frac{\pi x_i^2}{\delta}} \, dx_i = \prod_{i=1}^d \sqrt{\frac{\delta}{\pi}} \sqrt{\pi} = \prod_{i=1}^d \sqrt{\delta} = d\sqrt{\delta} = d(\delta)^{\frac 12}. \end{align*} So we have \begin{align*} \|f(x)\|_{L^1(\mathbb R^d)} &= \int_{\mathbb R^d} |f(x)| \, dx \\ &\le \int_{\mathbb R^d} \int_0^\infty |e^{-\frac{\pi|x|^2}{\delta}} \delta^{-\frac d2}||\delta^{\epsilon - 1}| \, d\delta \, dx \\ &= \int_0^\infty \left(\int_{\mathbb R^d} e^{-\frac{\pi |x|^2}{\delta}} \, dx \right) \delta^{-\frac d2 + \epsilon - 1} \, d\delta \\ &= \int_0^\infty d(\delta)^{\frac 12} \delta^{-\frac d2+\epsilon - 1} \, d\delta \\ &= d \int_0^\infty \delta^{\frac{1-d}2+\epsilon - 1} \, d\delta \\ &= d \frac 1{\frac{1-d}2+\epsilon} \delta^{\frac{1-d}2+\epsilon} \vert_0^\infty \\ &= \end{align*}
...but how do I prove from here that $f \in L^1(\mathbb R^d)$?