Motivated by the answer in this question, I wanted to perform computation in my own way in order to compute Fourier transform of $|x|^{-\alpha}$ (in a sense of tempered distributions) in $\mathbb{R}^n$ using Schwinger trick which is the following identity for $x \neq 0_{\mathbb{R}^n}$. However, I had problems understanding all of the steps.
$$ \frac{1}{|x|^{\alpha}} = \frac{1}{\Gamma(\alpha/2)}\int \limits_0^{\infty} t^{\alpha/2-1} e^{-t|x|^2} \mathrm{d}t $$
For clarity, I define distribution $1/|x|^{\alpha}$ in the following way for $0 < \alpha < n$ and $f \in S(\mathbb{R}^n)$, where $S(\mathbb{R}^n)$ is Schwartz space in $\mathbb{R}^n$. Here, $B_{\varepsilon}(0_{\mathbb{R}^n})$ is a ball centered at $0_{\mathbb{R}^n}$ with radius $\varepsilon > 0$.
$$ \frac{1}{|x|^{\alpha}} (f) := \lim_{\varepsilon \to 0^+} \int_{\mathbb{R}^n - B_{\varepsilon}(0_{\mathbb{R}^n})} \frac{f(x)}{|x|^{\alpha}} \mathrm{d}^nx$$
I understand that by definition of Fourier transform extended to distributions, I want to compute the following integral.
$$ \widehat{\frac{1}{|x|^{\alpha}}} (f) := \frac{1}{|x|^{\alpha}}(\widehat{f}) = \lim_{\varepsilon \to 0^+} \int \limits_{\mathbb{R}^n - B_{\varepsilon}(0_{\mathbb{R}^n})} \frac{\widehat{f}(k)}{|k|^{\alpha}} \mathrm{d}^nk$$
Moreover, I understand that as $\varepsilon > 0$, for each $k$ in the domain of integration we can use Schwinger's trick identity.
$$ \widehat{\frac{1}{|x|^{\alpha}}} (f) = \lim_{\varepsilon \to 0^+} \int \limits_{\mathbb{R}^n - B_{\varepsilon}(0_{\mathbb{R}^n})} \widehat{f}(k) \cdot \left( \frac{1}{\Gamma(\alpha/2)}\int \limits_0^{\infty} t^{\alpha/2-1} e^{-t|k|^2} \mathrm{d}t \right) \mathrm{d}^nk$$
However, now the trick is to apply Fubini's theorem in the following way. However, I am not able to justify it. I would like to argue the following.
$$\lim_{\varepsilon \to 0^+} \int \limits_{\mathbb{R}^n - B_{\varepsilon}(0_{\mathbb{R}^n})} \widehat{f}(k) \cdot \left( \frac{1}{\Gamma(\alpha/2)}\int \limits_0^{\infty} t^{\alpha/2-1} e^{-t|k|^2} \mathrm{d}t \right) \mathrm{d}^nk \stackrel{?}{=} \lim_{\varepsilon \to 0^+} \int \limits_0^{\infty} \frac{t^{\alpha/2-1}}{\Gamma(\alpha/2)} \cdot \left(\int \limits_{\mathbb{R}^n - B_{\varepsilon}(0_{\mathbb{R}^n})} \widehat{f}(k)e^{-t|k|^2} \mathrm{d}^nk \right) \mathrm{d}t $$
Question 1: How to justify previous equation? I think that I should argue that for all $\varepsilon > 0$ function $|t^{\alpha/2-1}\widehat{f}(k)e^{-t|k|^2}|$ is integrable on region where $(t, k) \in (0, \infty) \times (\mathbb{R}^n - B_{\varepsilon}(0_{\mathbb{R}^n})) \subset (0,\infty) \times \mathbb{R}^n$. However, I don't know how.
After that, I believe that in order to compute the integral over $k$, we want to use that for $f,g \in S(\mathbb{R}^n)$, $\langle \widehat{f}, \widehat{g} \rangle \propto \langle f, g \rangle$, where proportionality depends on definition of Fourier transform. However, I don't see how we can apply it here, as integration is over $\mathbb{R}^n - B_{\varepsilon}(0_{\mathbb{R}^n})$ instead of $\mathbb{R}^n$. So, the best I can think of is that we have to argue for the following equality.
$$ \lim_{\varepsilon \to 0^+} \int \limits_0^{\infty} \frac{t^{\alpha/2-1}}{\Gamma(\alpha/2)} \cdot \left(\int \limits_{\mathbb{R}^n - B_{\varepsilon}(0_{\mathbb{R}^n})} \widehat{f}(k)e^{-t|k|^2} \mathrm{d}^nk \right) \mathrm{d}t \stackrel{?}{=} \int \limits_0^{\infty} \frac{t^{\alpha/2-1}}{\Gamma(\alpha/2)} \cdot \left(\int \limits_{\mathbb{R}^n} \widehat{f}(k)e^{-t|k|^2} \mathrm{d}^nk \right) \mathrm{d}t $$
Question 2: Is this what we are supposed to do? If yes, how do we argue for that? I assume we want to use dominated convergence theorem but I am not sure of how to do it.
I understand how to perform $k$ integral now (assuming previous steps), where $A \in \mathbb{R}$ depends on definitions. However, it is unclear again to me how Fubini's theorem is justified.
$$\int \limits_0^{\infty} \frac{t^{\alpha/2-1}}{\Gamma(\alpha/2)} \cdot \left(\int \limits_{\mathbb{R}^n} \widehat{f}(k)e^{-t|k|^2} \mathrm{d}^nk \right) \mathrm{d}t =\int \limits_0^{\infty} \frac{t^{\alpha/2-1}}{\Gamma(\alpha/2)} \cdot \left(\int \limits_{\mathbb{R}^n} A f(k) e^{-|k|^2/4t} t^{-n/2}\mathrm{d}^nk \right) \mathrm{d}t $$
$$ \stackrel{?}{=} \int \limits_{\mathbb{R}^n} \frac{A}{\Gamma(\alpha/2)} f(k) \cdot \left( \int \limits_0^{\infty} t^{(\alpha-n)/2-1} e^{-|k|^2/4t} \mathrm{d}t \right) \mathrm{d}^nk$$
Question 3: How is Fubini's theorem justified in the last step?
I now understand that the goal is to consider change of variables $u(t) = |k|^2/4t$. But I think that such change of variables is defined only if $k \neq 0_{\mathbb{R}^n}$. So, I think I have to separate previous integral into regions $\mathbb{R}^n - B_{\varepsilon}(0_{\mathbb{R}^n})$ and $B_{\varepsilon}(0_{\mathbb{R}^n})$ as follows.
$$ \int \limits_{\mathbb{R}^n} \frac{A}{\Gamma(\alpha/2)} f(k) \cdot \left( \int \limits_0^{\infty} t^{(\alpha-n)/2-1} e^{-|k|^2/4t} \mathrm{d}t \right) \mathrm{d}^nk = \int \limits_{\mathbb{R}^n - B_{\varepsilon}(0_{\mathbb{R}^n})} \frac{A}{\Gamma(\alpha/2)} f(k) \cdot \left( \int \limits_0^{\infty} t^{(\alpha-n)/2-1} e^{-|k|^2/4t} \mathrm{d}t \right) \mathrm{d}^nk + \int \limits_{B_{\varepsilon}(0_{\mathbb{R}^n})} \frac{A}{\Gamma(\alpha/2)} f(k) \cdot \left( \int \limits_0^{\infty} t^{(\alpha-n)/2-1} e^{-|k|^2/4t} \mathrm{d}t \right) \mathrm{d}^nk$$
Question 4: Is this the correct strategy? If yes, I believe that to obtain the intended result I have to argue that the second term (integral over $B_{\varepsilon}(0_{\mathbb{R}^n})$) tends to $0$ as $\varepsilon \to 0^+$. How to show that?
