[proofplan] We express $\pi^{-s/2}\Gamma(s/2)\zeta(s)$ as the Mellin transform of the nonconstant part of the [Jacobi theta function](/page/Jacobi%20Theta%20Function). The [Poisson summation formula](/page/Poisson%20Summation%20Formula) gives the theta transformation law $\theta(t)=t^{-1/2}\theta(1/t)$, and this law converts the Mellin integral over $(0,1)$ into an integral over $(1,\infty)$ with the exponent $s$ replaced by $1-s$. The only nonsymmetric-looking terms come from the constant part of $\theta$; they combine into \begin{align*} \frac{1}{s(s-1)}. \end{align*} Multiplication by \begin{align*} \frac12 s(s-1) \end{align*} removes the poles and leaves a symmetric entire function. [/proofplan] [step:Represent the completed zeta factor as a theta Mellin transform] Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. Define the [Jacobi theta function](/page/Jacobi%20Theta%20Function) $\theta:(0,\infty)\to\mathbb{R}$ by \begin{align*} \theta(t):=\sum_{n\in\mathbb{Z}} e^{-\pi n^2 t}. \end{align*} For $\operatorname{Re}(s)>1$, define the completed zeta factor $\Lambda:\{s\in\mathbb{C}:\operatorname{Re}(s)>1\}\to\mathbb{C}$ by \begin{align*} \Lambda(s):=\pi^{-s/2}\Gamma\!\left(\frac{s}{2}\right)\zeta(s), \end{align*} where $\Gamma$ is the [Gamma function](/page/Gamma%20Function) and $\zeta$ is the [Riemann zeta function](/page/Riemann%20Zeta%20Function). Using the defining Dirichlet series for $\zeta$ and the [Gamma integral formula](/page/Gamma%20Function), we compute \begin{align*} \Lambda(s) &=\sum_{n=1}^{\infty}\pi^{-s/2}n^{-s}\Gamma\!\left(\frac{s}{2}\right)\\ &=\sum_{n=1}^{\infty}\int_0^\infty e^{-\pi n^2 t}t^{s/2-1}\,d\mathcal{L}^1(t)\\ &=\int_0^\infty \sum_{n=1}^{\infty}e^{-\pi n^2 t}t^{s/2-1}\,d\mathcal{L}^1(t)\\ &=\frac12\int_0^\infty\bigl(\theta(t)-1\bigr)t^{s/2-1}\,d\mathcal{L}^1(t). \end{align*} The interchange of sum and integral is justified by [Tonelli's theorem](/page/Tonelli%27s%20Theorem), since the absolute-value summands $e^{-\pi n^2t}t^{\operatorname{Re}(s)/2-1}$ are nonnegative and their integral sum is finite for $\operatorname{Re}(s)>1$. [guided] The goal is to put the zeta function into a form where the substitution $t\mapsto 1/t$ can act on it. Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. Define \begin{align*} \theta:(0,\infty)&\to\mathbb{R}\\ t&\mapsto\sum_{n\in\mathbb{Z}}e^{-\pi n^2t}. \end{align*} Since the summand is even in $n$, we have \begin{align*} \theta(t)-1=2\sum_{n=1}^{\infty}e^{-\pi n^2t}. \end{align*} For $\operatorname{Re}(s)>1$, define the completed zeta factor $\Lambda:\{s\in\mathbb{C}:\operatorname{Re}(s)>1\}\to\mathbb{C}$ by \begin{align*} \Lambda(s):=\pi^{-s/2}\Gamma\!\left(\frac{s}{2}\right)\zeta(s). \end{align*} Using the [Gamma integral formula](/page/Gamma%20Function) with the change of variables $u=\pi n^2t$, so that $d\mathcal{L}^1(u)=\pi n^2\,d\mathcal{L}^1(t)$, gives \begin{align*} \int_0^\infty e^{-\pi n^2t}t^{s/2-1}\,d\mathcal{L}^1(t) &=(\pi n^2)^{-s/2}\int_0^\infty e^{-u}u^{s/2-1}\,d\mathcal{L}^1(u)\\ &=\pi^{-s/2}n^{-s}\Gamma\!\left(\frac{s}{2}\right). \end{align*} Therefore \begin{align*} \Lambda(s) &=\sum_{n=1}^{\infty}\int_0^\infty e^{-\pi n^2t}t^{s/2-1}\,d\mathcal{L}^1(t)\\ &=\int_0^\infty\sum_{n=1}^{\infty}e^{-\pi n^2t}t^{s/2-1}\,d\mathcal{L}^1(t)\\ &=\frac12\int_0^\infty\bigl(\theta(t)-1\bigr)t^{s/2-1}\,d\mathcal{L}^1(t). \end{align*} [Tonelli's theorem](/page/Tonelli%27s%20Theorem) applies because the relevant absolute-value integrand $e^{-\pi n^2t}t^{\operatorname{Re}(s)/2-1}$ is nonnegative and the resulting series of integrals is finite when $\operatorname{Re}(s)>1$. [/guided] [/step] [step:Use the theta transformation law on the interval $(0,1)$] We use the theta transformation law \begin{align*} \theta(t)=t^{-1/2}\theta(1/t), \qquad t>0. \end{align*} This follows from the [Poisson summation formula](/page/Poisson%20Summation%20Formula) applied to the Gaussian $g_t:\mathbb{R}\to\mathbb{R}$, $x\mapsto e^{-\pi t x^2}$. For each fixed $t>0$, the function $g_t$ is a [Schwartz function](/page/Schwartz%20Space), since every derivative is a polynomial in $x$ times $e^{-\pi t x^2}$ and hence decays faster than every inverse power of $|x|$. We use the [Fourier transform](/page/Fourier%20Transform) convention \begin{align*} \hat f(\xi):=\int_{\mathbb{R}} f(x)e^{-2\pi i x\xi}\,d\mathcal{L}^1(x). \end{align*} With this normalization, the Fourier transform of $g_t$ is $\hat g_t(\xi)=t^{-1/2}e^{-\pi \xi^2/t}$, so Poisson summation gives \begin{align*} \sum_{n\in\mathbb{Z}}e^{-\pi n^2t}=t^{-1/2}\sum_{m\in\mathbb{Z}}e^{-\pi m^2/t}. \end{align*} This is exactly the displayed theta transformation law. Split the Mellin integral at $1$: \begin{align*} \Lambda(s) &=\frac12\int_1^\infty(\theta(t)-1)t^{s/2-1}\,d\mathcal{L}^1(t) +\frac12\int_0^1(\theta(t)-1)t^{s/2-1}\,d\mathcal{L}^1(t). \end{align*} For the integral over $(0,1)$, the theta transformation gives \begin{align*} \theta(t)-1=t^{-1/2}\bigl(\theta(1/t)-1\bigr)+t^{-1/2}-1. \end{align*} Hence \begin{align*} \frac12\int_0^1(\theta(t)-1)t^{s/2-1}\,d\mathcal{L}^1(t) &=\frac12\int_0^1 t^{-1/2}\bigl(\theta(1/t)-1\bigr)t^{s/2-1}\,d\mathcal{L}^1(t)\\ &\quad+\frac12\int_0^1\left(t^{(s-1)/2-1}-t^{s/2-1}\right)\,d\mathcal{L}^1(t). \end{align*} The last integral is explicit: \begin{align*} \frac12\int_0^1\left(t^{(s-1)/2-1}-t^{s/2-1}\right)\,d\mathcal{L}^1(t) =\frac{1}{s-1}-\frac{1}{s} =\frac{1}{s(s-1)}. \end{align*} [guided] We first prove the theta transformation law used in the split integral. For $t>0$, define $g_t:\mathbb{R}\to\mathbb{R}$ by $g_t(x)=e^{-\pi t x^2}$. This function lies in the [Schwartz space](/page/Schwartz%20Space), because each derivative is a polynomial in $x$ multiplied by $e^{-\pi t x^2}$, and the Gaussian factor dominates every polynomial as $|x|\to\infty$. We use the [Fourier transform](/page/Fourier%20Transform) normalization \begin{align*} \hat f(\xi):=\int_{\mathbb{R}} f(x)e^{-2\pi i x\xi}\,d\mathcal{L}^1(x). \end{align*} With this normalization, completing the square in the [Gaussian integral](/theorems/1140) gives \begin{align*} \hat g_t(\xi)=t^{-1/2}e^{-\pi \xi^2/t}. \end{align*} The [Poisson summation formula](/page/Poisson%20Summation%20Formula) applies to $g_t$ because $g_t$ is Schwartz, and it yields \begin{align*} \sum_{n\in\mathbb{Z}}e^{-\pi n^2t}=\sum_{m\in\mathbb{Z}}\hat g_t(m)=t^{-1/2}\sum_{m\in\mathbb{Z}}e^{-\pi m^2/t}. \end{align*} By the definition of $\theta$, this is exactly \begin{align*} \theta(t)=t^{-1/2}\theta(1/t), \qquad t>0. \end{align*} Now split the Mellin integral at $1$: \begin{align*} \Lambda(s) &=\frac12\int_1^\infty(\theta(t)-1)t^{s/2-1}\,d\mathcal{L}^1(t) +\frac12\int_0^1(\theta(t)-1)t^{s/2-1}\,d\mathcal{L}^1(t). \end{align*} On $(0,1)$, the transformation law gives \begin{align*} \theta(t)-1=t^{-1/2}\bigl(\theta(1/t)-1\bigr)+t^{-1/2}-1. \end{align*} Substituting this identity into the integral over $(0,1)$ separates the part involving $\theta(1/t)-1$ from the constant correction: \begin{align*} \frac12\int_0^1(\theta(t)-1)t^{s/2-1}\,d\mathcal{L}^1(t) &=\frac12\int_0^1 t^{-1/2}\bigl(\theta(1/t)-1\bigr)t^{s/2-1}\,d\mathcal{L}^1(t)\\ &\quad+\frac12\int_0^1\left(t^{(s-1)/2-1}-t^{s/2-1}\right)\,d\mathcal{L}^1(t). \end{align*} For $\operatorname{Re}(s)>1$, both elementary integrals converge, and direct evaluation gives \begin{align*} \frac12\int_0^1\left(t^{(s-1)/2-1}-t^{s/2-1}\right)\,d\mathcal{L}^1(t) =\frac{1}{s-1}-\frac{1}{s} =\frac{1}{s(s-1)}. \end{align*} This is the only nonsymmetric rational term that appears in the argument. [/guided] [/step] [step:Change variables to reveal the symmetry $s\leftrightarrow 1-s$] In the remaining integral over $(0,1)$, use the substitution $u=1/t$. Then $t=1/u$, the interval $(0,1)$ becomes $(1,\infty)$, and $d\mathcal{L}^1(t)=u^{-2}\,d\mathcal{L}^1(u)$ after reversing the limits. Therefore \begin{align*} \frac12\int_0^1 t^{-1/2}\bigl(\theta(1/t)-1\bigr)t^{s/2-1}\,d\mathcal{L}^1(t) &=\frac12\int_1^\infty(\theta(u)-1)u^{(1-s)/2-1}\,d\mathcal{L}^1(u). \end{align*} Combining this with the integral over $(1,\infty)$ yields, for $\operatorname{Re}(s)>1$, \begin{align*} \Lambda(s) = \frac{1}{s(s-1)} +\frac12\int_1^\infty(\theta(t)-1)\left(t^{s/2-1}+t^{(1-s)/2-1}\right)\,d\mathcal{L}^1(t). \end{align*} [guided] The substitution $u=1/t$ is the point where the symmetry appears. In the term \begin{align*} \frac12\int_0^1 t^{-1/2}\bigl(\theta(1/t)-1\bigr)t^{s/2-1}\,d\mathcal{L}^1(t), \end{align*} set $u=1/t$. Then $t=1/u$, the domain $0<t<1$ becomes $1<u<\infty$, and the one-dimensional Lebesgue measure transforms as \begin{align*} d\mathcal{L}^1(t)=u^{-2}\,d\mathcal{L}^1(u) \end{align*} after reversing the orientation of the limits. The power of $u$ is \begin{align*} t^{-1/2}t^{s/2-1}d\mathcal{L}^1(t) &=t^{s/2-3/2}d\mathcal{L}^1(t)\\ &=u^{-s/2+3/2}u^{-2}\,d\mathcal{L}^1(u)\\ &=u^{(1-s)/2-1}\,d\mathcal{L}^1(u). \end{align*} Thus \begin{align*} \frac12\int_0^1 t^{-1/2}\bigl(\theta(1/t)-1\bigr)t^{s/2-1}\,d\mathcal{L}^1(t) &=\frac12\int_1^\infty(\theta(u)-1)u^{(1-s)/2-1}\,d\mathcal{L}^1(u). \end{align*} Adding the original integral over $(1,\infty)$ and the explicitly computed constant term gives \begin{align*} \Lambda(s) = \frac{1}{s(s-1)} +\frac12\int_1^\infty(\theta(t)-1)\left(t^{s/2-1}+t^{(1-s)/2-1}\right)\,d\mathcal{L}^1(t). \end{align*} This formula has separated the obstruction to holomorphy, namely $1/(s(s-1))$, from an integral whose two terms are exchanged by $s\mapsto 1-s$. [/guided] [/step] [step:Extend the symmetric formula to the whole complex plane] For $t\geq 1$, \begin{align*} 0\leq \theta(t)-1=2\sum_{n=1}^{\infty}e^{-\pi n^2t} \leq 2\sum_{n=1}^{\infty}e^{-\pi nt} =\frac{2e^{-\pi t}}{1-e^{-\pi t}} \leq \frac{2e^{-\pi t}}{1-e^{-\pi}}. \end{align*} Thus, for every compact set $K\subset\mathbb{C}$, the functions \begin{align*} s\mapsto (\theta(t)-1)t^{s/2-1}, \qquad s\mapsto (\theta(t)-1)t^{(1-s)/2-1} \end{align*} are dominated on $K$ by an exponentially decaying function of $t$ times a fixed power of $t$. Therefore the map $H:\mathbb{C}\to\mathbb{C}$ defined by \begin{align*} H(s):=\frac12\int_1^\infty(\theta(t)-1)\left(t^{s/2-1}+t^{(1-s)/2-1}\right)\,d\mathcal{L}^1(t) \end{align*} is entire by the [holomorphic parameter integral theorem](/page/Holomorphic%20Parameter%20Integral): for each fixed $t\geq1$ the integrand is entire in $s$, and on every compact set $K\subset\mathbb{C}$ all $s$-derivatives are dominated by the same exponential bound multiplied by a fixed power of $t$ and a fixed power of $\log t$, which is integrable on $(1,\infty)$. The formula \begin{align*} \Lambda(s)=\frac{1}{s(s-1)}+H(s) \end{align*} gives the meromorphic continuation of $\Lambda$ to $\mathbb{C}$ with at most simple poles at $0$ and $1$. Moreover the definition of $H$ is invariant under $s\mapsto 1-s$, since that substitution interchanges the two powers of $t$. Hence \begin{align*} H(s)=H(1-s) \end{align*} for all $s\in\mathbb{C}$. [guided] The estimate on $\theta(t)-1$ is what permits [analytic continuation](/page/Analytic%20Continuation). For $t\geq1$, we use $n^2\geq n$ for $n\geq1$ and the geometric series formula to obtain \begin{align*} 0\leq \theta(t)-1=2\sum_{n=1}^{\infty}e^{-\pi n^2t} \leq 2\sum_{n=1}^{\infty}e^{-\pi nt} =\frac{2e^{-\pi t}}{1-e^{-\pi t}} \leq \frac{2e^{-\pi t}}{1-e^{-\pi}}. \end{align*} Let $K\subset\mathbb{C}$ be compact. Since $\operatorname{Re}(s)$ is bounded on $K$, there is a constant $A_K>0$ such that the powers $t^{s/2-1}$ and $t^{(1-s)/2-1}$ are bounded in absolute value by a fixed power $t^{A_K}$ for all $s\in K$ and $t\geq1$. The preceding exponential estimate then gives an integrable majorant on $(1,\infty)$ with respect to $\mathcal{L}^1$. Define \begin{align*} H:\mathbb{C}&\to\mathbb{C}\\ s&\mapsto\frac12\int_1^\infty(\theta(t)-1)\left(t^{s/2-1}+t^{(1-s)/2-1}\right)\,d\mathcal{L}^1(t). \end{align*} For each fixed $t\geq1$, the integrand is an entire function of $s$. The same exponential estimate, with additional powers of $\log t$ after differentiating with respect to $s$, still gives an integrable majorant on compact subsets of $\mathbb{C}$. Hence the [holomorphic parameter integral theorem](/page/Holomorphic%20Parameter%20Integral) implies that $H$ is entire. The identity already proved for $\operatorname{Re}(s)>1$ therefore gives the meromorphic continuation \begin{align*} \Lambda(s)=\frac{1}{s(s-1)}+H(s) \end{align*} to all of $\mathbb{C}$, with at most simple poles at $0$ and $1$. Finally, the definition of $H$ is invariant under $s\mapsto1-s$, because this substitution only interchanges the two summands $t^{s/2-1}$ and $t^{(1-s)/2-1}$. Therefore \begin{align*} H(s)=H(1-s) \end{align*} for every $s\in\mathbb{C}$. [/guided] [/step] [step:Multiply by $s(s-1)/2$ to obtain the xi symmetry] For $s\in\mathbb{C}\setminus\{0,1\}$, the continued expression gives \begin{align*} \xi(s) &=\frac12s(s-1)\Lambda(s)\\ &=\frac12s(s-1)\left(\frac{1}{s(s-1)}+H(s)\right)\\ &=\frac12+\frac12s(s-1)H(s). \end{align*} Since $H$ is entire, the right-hand side extends holomorphically to $s=0$ and $s=1$; this is the entire continuation of $\xi$. Using $H(s)=H(1-s)$ and \begin{align*} (1-s)((1-s)-1)=s(s-1), \end{align*} we obtain \begin{align*} \xi(1-s) &=\frac12+\frac12(1-s)((1-s)-1)H(1-s)\\ &=\frac12+\frac12s(s-1)H(s)\\ &=\xi(s). \end{align*} Therefore $\xi(s)=\xi(1-s)$ for every $s\in\mathbb{C}$. [guided] Define the completed zeta function $\xi:\mathbb{C}\to\mathbb{C}$ by the holomorphic continuation of \begin{align*} \xi(s):=\frac12s(s-1)\Lambda(s). \end{align*} For $s\in\mathbb{C}\setminus\{0,1\}$, the meromorphic continuation formula for $\Lambda$ gives \begin{align*} \xi(s) &=\frac12s(s-1)\Lambda(s)\\ &=\frac12s(s-1)\left(\frac{1}{s(s-1)}+H(s)\right)\\ &=\frac12+\frac12s(s-1)H(s). \end{align*} The right-hand side is holomorphic at $s=0$ and $s=1$ because $H$ is entire and $s(s-1)$ is a polynomial. Thus this formula supplies the entire continuation of $\xi$. Now apply the symmetry already proved for $H$. Since \begin{align*} (1-s)((1-s)-1)=s(s-1), \end{align*} and since $H(1-s)=H(s)$, we have \begin{align*} \xi(1-s) &=\frac12+\frac12(1-s)((1-s)-1)H(1-s)\\ &=\frac12+\frac12s(s-1)H(s)\\ &=\xi(s). \end{align*} Therefore the completed zeta function satisfies $\xi(s)=\xi(1-s)$ for every $s\in\mathbb{C}$, which is the asserted functional equation. [/guided] [/step]