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I have the function: $$ G_k(x)_=\frac{1}{(4\pi)^{k/2}\Gamma(k/2)}\int_0^\infty e^{-\pi|x|^2/\delta}e^{-\delta/4\pi}\delta^{-(n-k)/2}\,\frac{d\delta}{\delta}, $$ for all $x\in\mathbb{R}^n$ and $k>0$. I know that $G_k\in L^1(\mathbb{R}^n)$ and: $$ \mathcal{F}(G_k)(\xi)=(1+4\pi^2|\xi|^2)^{-k/2},\quad\forall\xi\in\mathbb{R}^n,k>0,$$ i want to prove that: $$ \mathcal{F}(G_{2k})(\xi/2\pi)=(2\pi)^k\mathcal{F}(G_{2k})(2\pi\xi),\quad\forall\xi\in\mathbb{R}^n,k>0.$$ I don't know how to go on. Any help would be appreciated.

$\mathcal{F}$ is the Fourier transform: for all $f\in L^1(\mathbb{R}^n)$, we set: $$\mathcal{F}(f)(\xi)=\int_{\mathbb{R}^n} f(x)e^{-2\pi i x\cdot \xi}\,dx\quad\forall\xi\in\mathbb{R}^n.$$

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    $\begingroup$ The third display contradicts the second display. Just plug in and you will see. $\endgroup$
    – GH from MO
    Oct 28, 2020 at 10:07
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    $\begingroup$ The display on the bottom of page 269 is incorrect (the first equation is OK, the second is not). I think the author meant $\mathcal{F}[G_{2k}(2\pi x)](\xi)$ in place of $\mathcal{F}[G_{2k}](2\pi\xi)$. At any rate, on the top of page 270, $2\pi\xi$ should be $\xi/2\pi$ and $(2\pi)^k$ should be omitted in the first line, and then the next line is correct by Plancherel. Note also that the Fourier transform is defined differently in this paper than in your post. See the display below (2.1). $\endgroup$
    – GH from MO
    Oct 28, 2020 at 10:23
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    $\begingroup$ Yes, the factor $(2\pi)^k$ should be deleted in the first line of the first display on page 270, and $2\pi\xi$ should be $\xi/2\pi$ in that line. $\endgroup$
    – GH from MO
    Oct 28, 2020 at 10:29
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    $\begingroup$ If i set: $$ G_{2k}^{(2\pi)}(x):=G_{2k}(2\pi x),\quad\forall x\in\mathbb{R}^n, $$ then, is true that; $$ \mathcal{F}(G_{2k}^{(2\pi)})(\xi)=(2\pi)^{-n}\mathcal{F}(G_{2k})(\xi/2\pi) =(2\pi)^{-n}(1+|\xi|^2)^{-k},\quad\forall\xi\in\mathbb{R}^n. $$ I think the author means this. $\endgroup$
    – inoc
    Oct 28, 2020 at 10:41
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    $\begingroup$ Yes, and this is what I wrote in my previous comments. $\endgroup$
    – GH from MO
    Oct 28, 2020 at 10:45

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