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Multiple answers and comments have already pointed out that the conceptual role of $\pi^{-s/2}\Gamma(s/2)$ comes from the viewpoint of Tate's thesis, which for $\text{Re}(s) > 1$ creates this function as $\int_{\mathbf R^\times} e^{-\pi x^2}|x|^s\,dx/|x|$, an integral over the multiplicative group $\mathbf R^\times$ of the function $e^{-\pi x^2}$ that is self-dual for the Fourier transform on the additive group $\mathbf R$ relative to the self-duality $\langle x,y\rangle = e^{2\pi ixy}$ or $\langle x,y\rangle = e^{-2\pi ixy}$ on $\mathbf R$. (If we use another self-duality of $\mathbf R$ then $e^{-ax^2}$ would be self-dual for some $a \not= \pi$ instead.)

All of what I wrote so far has appeared explicitly or implicitly in some of the other comments or answers. Since there are many self-dual even Schwartz functions $f$ on $\mathbf R$, what is it about the choice $f(x) = e^{-\pi x^2}$, leading to $\Gamma_f(s) = (1/2)\pi^{-s/2}\Gamma(s/2)$ (an extra $1/2$ on both sides of the functional equation can be cancelled) that is so nice? I have not seen the following property pointed out yet: with this choice of $f$ and familiarity with the $\Gamma$-function we know $\Gamma_f(s) \not= 0$ for $\text{Re}(s) > 1$ (in fact for $\text{Re}(s) > 0$), so therefore $\Gamma_f(s)\zeta(s) \not= 0$ for $\text{Re}(s) > 1$ from $\zeta(s)$ being nonvanishing there, and then by the functional equation $\Gamma_f(s)\zeta(s) \not= 0$ for $\text{Re}(s) < 0$, which means all zeros of $\Gamma_f(s)\zeta(s)$ have $0 \leq \text{Re}(s) \leq 1$. If you want to use a totally random even Schwartz function for $f$ in order to complete the Riemann zeta-function as in Tate's thesis, you get the nice-looking nontrivial functional equation displayed above, but how are you going to use $\Gamma_f(s)\zeta(s)$ to analyze the location of zeros of $\zeta(s)$ (including discovering its trivial zeros, whether or not you consider those important) if you do not know where $\Gamma_f(s)$ has its zeros and poles?

Multiple answers and comments have already pointed out that the conceptual role of $\pi^{-s/2}\Gamma(s/2)$ comes from the viewpoint of Iwasawa and Tate, which for $\text{Re}(s) > 1$ creates this function as $\int_{\mathbf R^\times} e^{-\pi x^2}|x|^s\,dx/|x|$, an integral over the multiplicative group $\mathbf R^\times$ of the function $e^{-\pi x^2}$ that is self-dual for the Fourier transform on the additive group $\mathbf R$ relative to the self-duality $\langle x,y\rangle = e^{2\pi ixy}$ or $\langle x,y\rangle = e^{-2\pi ixy}$ on $\mathbf R$. (If we use another self-duality of $\mathbf R$ then $e^{-ax^2}$ would be self-dual for some $a \not= \pi$ instead.)

All of what I wrote so far has appeared explicitly or implicitly in some of the other comments or answers. Since there are many self-dual even Schwartz functions $f$ on $\mathbf R$, what is it about the choice $f(x) = e^{-\pi x^2}$, leading to $\Gamma_f(s) = (1/2)\pi^{-s/2}\Gamma(s/2)$ (an extra $1/2$ on both sides of the functional equation can be cancelled) that is so nice? I have not seen the following property pointed out yet: with this choice of $f$ and familiarity with the $\Gamma$-function we know $\Gamma_f(s) \not= 0$ for $\text{Re}(s) > 1$ (in fact for $\text{Re}(s) > 0$), so therefore $\Gamma_f(s)\zeta(s) \not= 0$ for $\text{Re}(s) > 1$ from $\zeta(s)$ being nonvanishing there, and then by the functional equation $\Gamma_f(s)\zeta(s) \not= 0$ for $\text{Re}(s) < 0$, which means all zeros of $\Gamma_f(s)\zeta(s)$ have $0 \leq \text{Re}(s) \leq 1$. If you want to use a totally random even Schwartz function for $f$ in order to define a factor $\Gamma_f(s)$ that completes the Riemann zeta-function, you will get the nice-looking nontrivial functional equation displayed above but how are you going to use $\Gamma_f(s)\zeta(s)$ to analyze the location of zeros of $\zeta(s)$ (including discovering its trivial zeros, whether or not you consider those important) if you do not know where $\Gamma_f(s)$ has its zeros and poles?

So although there are many even Schwartz functions $f$ on $\mathbf R$ besides $e^{-\pi x^2}$ that you could use to get a nice functional equation by multiplying $\zeta(s)$ by $\Gamma_f(s)$, the reason that the choice $f(x) = e^{-\pi x^2}$ is so convenient is that we actually know the zeros and poles of $\Gamma_f(s) = (1/2)\pi^{-s/2}\Gamma(s/2)$: it has no zeros in $\mathbf C$ and it has simple poles at $0, -2, -4, \ldots$. For even self-dual Schwartz $f$ on $\mathbf R$ that are not simple modifications of $e^{-\pi x^2}$, how feasible is it to determine whether or not $\Gamma_f(s) \not= 0$ for $\text{Re}(s) > 1$ (or $\text{Re}(s) > 0$)? The method of meromorphically continuing $\Gamma_f(s)$ from the half-plane $\text{Re}(s) > 0$ where it is analytic to all of $\mathbf C$ shows that its only possible poles are at $0, -1, -2, -3, \ldots$ with orders at most $1$ and the residue at $s = -n$ is $(-1/n!)\int_0^\infty f^{(n+1)}(x)\,dx$, which by the Fundamental Theorem of Calculus is $(-1/n!)(f^{(n)}(\infty) - f^{(n)}(0)) = f^{(n)}(0)/n!$. Therefore you could determine the poles of $\Gamma_f$ by seeing when $f^{(n)}(0)$ is 0 and not 0, but how are you going to determine where the zeros of $\Gamma_f$ are or that there are no zeros? (EDIT: for even $f$, its odd-order derivatives vanish at $0$, so the residue at $-n$ vanishes when $n$ is odd, which means the poles of $\Gamma_f(s)$ can only be at $n = 0, -2, -4, -6, \ldots$. Those are all simple poles of $\pi^{-s/2}\Gamma(s/2)$, which has no zeros, so $\Gamma_f(s)/(\pi^{-s/2}\Gamma(s/2)$ is an entire function, which makes $\pi^{-s/2}\Gamma(s/2)$ a "holomorphic gcd" of all $\Gamma_f(s)$ for even Schwartz functions $f$ on $\mathbf R$. This addresses a comment below by Venkataramana.)

So although there are many even Schwartz functions $f$ on $\mathbf R$ besides $e^{-\pi x^2}$ that you could use to get a nice functional equation by multiplying $\zeta(s)$ by $\Gamma_f(s)$, the reason that the choice $f(x) = e^{-\pi x^2}$ is so convenient is that we actually know the zeros and poles of $\Gamma_f(s) = (1/2)\pi^{-s/2}\Gamma(s/2)$: it has no zeros in $\mathbf C$ and it has simple poles at $0, -2, -4, \ldots$. For even self-dual Schwartz $f$ on $\mathbf R$ that are not simple modifications of $e^{-\pi x^2}$, how feasible is it to determine whether or not $\Gamma_f(s) \not= 0$ for $\text{Re}(s) > 1$ (or $\text{Re}(s) > 0$)? The method of meromorphically continuing $\Gamma_f(s)$ from the half-plane $\text{Re}(s) > 0$ where it is analytic to all of $\mathbf C$ shows that its only possible poles are at $0, -1, -2, -3, \ldots$ with orders at most $1$ and the residue at $s = -n$ is $(-1/n!)\int_0^\infty f^{(n+1)}(x)\,dx$, which by the Fundamental Theorem of Calculus is $(-1/n!)(f^{(n)}(\infty) - f^{(n)}(0)) = f^{(n)}(0)/n!$. Therefore you could determine the poles of $\Gamma_f$ by seeing when $f^{(n)}(0)$ is 0 and not 0, but how are you going to determine where the zeros of $\Gamma_f$ are or that there are no zeros? (EDIT: for even $f$, its odd-order derivatives vanish at $0$, so the residue at $-n$ vanishes when $n$ is odd, which means the poles of $\Gamma_f(s)$ can only be at $n = 0, -2, -4, -6, \ldots$. Those are all simple poles of $\pi^{-s/2}\Gamma(s/2)$, which has no zeros, so $G(s) := \Gamma_f(s)/(\pi^{-s/2}\Gamma(s/2))$ is an entire function. Thus $\Gamma_f(s) = G(s)\pi^{-s/2}\Gamma(s/2)$ with $G$ entire, so $\pi^{-s/2}\Gamma(s/2)$ a "holomorphic gcd" of all $\Gamma_f(s)$ for even Schwartz functions $f$ on $\mathbf R$. The exponential factor $\pi^{-s/2}$ was kind of irrelevant to drag through the calculation since it has no zeros or poles, but it's traditionally seen alongside $\Gamma(s/2)$ so I used it. This addresses comments below by Will Sawin and Venkataramana.)

So although there are many even Schwartz functions $f$ on $\mathbf R$ besides $e^{-\pi x^2}$ that you could use to get a nice functional equation by multiplying $\zeta(s)$ by $\Gamma_f(s)$, the reason that the choice $f(x) = e^{-\pi x^2}$ is so convenient is that we actually know the zeros and poles of $\Gamma_f(s) = (1/2)\pi^{-s/2}\Gamma(s/2)$: it has no zeros in $\mathbf C$ and it has simple poles at $0, -2, -4, \ldots$. For even self-dual Schwartz $f$ on $\mathbf R$ that are not simple modifications of $e^{-\pi x^2}$, how feasible is it to determine whether or not $\Gamma_f(s) \not= 0$ for $\text{Re}(s) > 1$ (or $\text{Re}(s) > 0$)? The method of meromorphically continuing $\Gamma_f(s)$ from the half-plane $\text{Re}(s) > 0$ where it is analytic to all of $\mathbf C$ shows that its only possible poles are at $0, -1, -2, -3, \ldots$ with orders at most 1 and the residue at $s = -n$ is $(-1/n!)\int_0^\infty f^{(n+1)}(x)\,dx$, so in principle you could determine all the poles of $\Gamma_f$ by seeing which of those residues are 0 and not 0, but how are you going to determine where the zeros of $\Gamma_f$ are or that there are no zeros?

So although there are many even Schwartz functions $f$ on $\mathbf R$ besides $e^{-\pi x^2}$ that you could use to get a nice functional equation by multiplying $\zeta(s)$ by $\Gamma_f(s)$, the reason that the choice $f(x) = e^{-\pi x^2}$ is so convenient is that we actually know the zeros and poles of $\Gamma_f(s) = (1/2)\pi^{-s/2}\Gamma(s/2)$: it has no zeros in $\mathbf C$ and it has simple poles at $0, -2, -4, \ldots$. For even self-dual Schwartz $f$ on $\mathbf R$ that are not simple modifications of $e^{-\pi x^2}$, how feasible is it to determine whether or not $\Gamma_f(s) \not= 0$ for $\text{Re}(s) > 1$ (or $\text{Re}(s) > 0$)? The method of meromorphically continuing $\Gamma_f(s)$ from the half-plane $\text{Re}(s) > 0$ where it is analytic to all of $\mathbf C$ shows that its only possible poles are at $0, -1, -2, -3, \ldots$ with orders at most $1$ and the residue at $s = -n$ is $(-1/n!)\int_0^\infty f^{(n+1)}(x)\,dx$, which by the Fundamental Theorem of Calculus is $(-1/n!)(f^{(n)}(\infty) - f^{(n)}(0)) = f^{(n)}(0)/n!$. Therefore you could determine the poles of $\Gamma_f$ by seeing when $f^{(n)}(0)$ is 0 and not 0, but how are you going to determine where the zeros of $\Gamma_f$ are or that there are no zeros? (EDIT: for even $f$, its odd-order derivatives vanish at $0$, so the residue at $-n$ vanishes when $n$ is odd, which means the poles of $\Gamma_f(s)$ can only be at $n = 0, -2, -4, -6, \ldots$. Those are all simple poles of $\pi^{-s/2}\Gamma(s/2)$, which has no zeros, so $\Gamma_f(s)/(\pi^{-s/2}\Gamma(s/2)$ is an entire function, which makes $\pi^{-s/2}\Gamma(s/2)$ a "holomorphic gcd" of all $\Gamma_f(s)$ for even Schwartz functions $f$ on $\mathbf R$. This addresses a comment below by Venkataramana.)