*Riemann's analytic continuations and derivation of the functional equations for $\zeta$ and $\xi$ seem quite natural and intuitive from the perspective of basic complex analysis.*

Riemann in the second equation of his classic paper On the Number of Prime Numbers less than a Given Quantity (1859) writes down the Laplace (1749-1827) transform

$$\int_{0}^{+\infty}e^{-nx}x^{s-1}dx=\frac{(s-1)!}{n^s},$$ valid for $real(s)>0.$

With $n=1$ this is the iconic Euler (1707-1783) integral representation of the gamma function, and noting that

$(s-1)!=\frac{\pi}{sin(\pi s)}\frac{1}{(-s)!}$ from the symmetric relation $\frac{sin(\pi s)}{\pi s}=\frac{1}{s!(-s)!},$

this can be rewritten as

$$\frac{sin(\pi s)}{\pi}\int_{0}^{\infty}e^{-x}x^{s-1}dx=\frac{1}{(-s)!},$$

suggesting quite naturally to someone as familiar with analytic continuation as Riemann that

$$\frac{-1}{2\pi i}\int_{+\infty}^{+\infty}e^{-x}(-x)^{s-1}dx=\frac{1}{(-s)!},$$

valid for all $s$, where the line integral is blown up around the positive real axis into the complex plane to sandwich it with a branch cut for $x>0$ and to loop the origin in the positive sense from positive infinity to positive infinity. Deflating the contour back to the real axis introduces a $-exp(i\pi s)+exp(-i\pi s)=-2isin(\pi s)$. (This special contour is now called the Hankel contour after Hermann Hankel (1839-1873), who became a student of Riemann in 1860 and published this integral for the reciprocal gamma fct. in his habilitation of 1863. Most likely Riemann introduced him to this maneuver.)

Riemann in his third equation observes that

$$(s-1)!\zeta(s)=(s-1)!\sum_{n=1}^{\infty }\frac{1}{n^s}=\int_{0}^{+\infty}\sum_{n=1}^{\infty }e^{-nx}x^{s-1}dx=\int_{0}^{+\infty}\frac{1}{e^x-1}x^{s-1}dx$$

and then immediately writes down as his fourth equality the analytic continuation

$$2sin(\pi s)(s-1)!\zeta(s)=i\int_{+\infty}^{+\infty}\frac{(-x)^{s-1}}{e^x-1}dx,$$

valid for all $s$, which can be rewritten as

$$\frac{\zeta(s)}{(-s)!}=\frac{-1}{2\pi i}\int_{+\infty}^{+\infty}\frac{(-x)^{s-1}}{e^x-1}dx$$

as naturally suggested by the analytic continuation of the reciprocal of gamma above.

For $m=0,1,2, ...,$ this gives

$$\zeta(-m)=\frac{(-1)^{m}}{2\pi i}\oint_{|z|=1}\frac{m!}{z^{m+1}}\frac{1}{e^z-1}dz=\frac{(-1)^{m}}{m+1}\frac{1}{2\pi i}\oint_{|z|=1}\frac{(m+1)!}{z^{m+2}}\frac{z}{e^z-1}dz$$

from which you can see, if you are familiar with the exponential generating fct. (e.g.f.) for the Bernoulli numbers, that the integral vanishes for even $m$. Euler published the e.g.f. in 1740 (MSE-Q144436), and Riemann certainly was familiar with these numbers and states that his fourth equality implies the vanishing of $\zeta(s)$ for $m$ even (but gives no explicit proof). He certainly was also aware of Euler's heuristic functional eqn. for integer values of the zeta fct., and Edwards in **Riemann's Zeta Function** (pg. 12, Dover ed.) even speculates that ".. it may well have been this problem of deriving (2) [Euler's formula for $\zeta(2n)$ for positive $n$] anew which led Riemann to the discovery of the functional equation ...."

Riemann then proceeds to derive the functional eqn. for zeta from his equality by using the singularities of $\frac{1}{e^z-1}$ to obtain basically

$$\zeta(s)=2(2\pi)^{s-1}\Gamma(1-s)\sin(\tfrac12\pi s)\zeta(1-s),$$

and says three lines later essentially that it may be expressed symmetrically about $s=1/2$ as

$$\xi(s) = \pi^{-s/2}\ \Gamma\left(\frac{s}{2}\right)\ \zeta(s)=\xi(1-s).$$

Riemann then says, "This property of the function [$\xi(s)=\xi(1-s)$] induced me to introduce, in place of $(s-1)!$, the integral $(s/2-1)!$ into the general term of the series $\sum \frac{1}{n^s}$, whereby one obtains a very convenient expression for the function $\zeta(s)$." And then he proceeds to introduce what Edwards calls a second proof of the functional eqn. using the Jacobi theta function.

Edwards wonders:

"Since the second proof renders the first proof wholly unnecessary, one may ask why Riemann included the first proof at all. Perhaps the first proof shows the argument by which he originally discovered the functional equation or perhaps it exhibits some properties which were important in his understanding of it."

I wonder whether, as his ideas evolved *before he wrote the paper*, he first constructed $\xi(s)$ by noticing that multiplying $\zeta(s)$ by $\Gamma(\frac{s}{2})$ introduces a simple pole at $s=0$ thereby reflecting the pole of $\zeta(s)$ at $s=1$ through the line $s=1/2$ and that the other simple poles of $\Gamma(\frac{s}{2})$ are removed by the zeros on the real line of the zeta function. The $\pi^{-s/2}$ can easily be determined as a normalization by an entire function $c^s$ where $c$ is a constant, using the complex conjugate symmetry of the gamma and zeta fct. about the real axis. Riemann had fine physical intuition and would have thought holistically in terms of the the zeros of a function (see Euler's proof of the Basel problem) and its poles, the importance of which he certainly stressed.

*Let's extend the reasoning above for the Jacobi theta function*

$$\vartheta (0;ix^2)=\sum_{n=-\infty,}^{\infty }exp(-\pi n^{2}x^2).$$

Viewing a modified Mellin transform as an interpolation of Taylor series coefficients (MO-Q79868), it's easy to guess (note the zeros of the coefficients) that

$$\int^{\infty}_{0}\exp(-x^2)\frac{x^{s-1}}{(s-1)!} dx = \cos(\pi\frac{ s}{2})\frac{(-s)!}{(-\frac{s}{2})!} = \frac{1}{2}\frac{(\frac{s}{2}-1)!}{(s-1)!},$$

and, therefore,

$$\int^{\infty}_{0}\exp(-\pi (n x)^2)x^{s-1} dx = \frac{1}{2}\pi^{-s/2}(\frac{s}{2}-1)! \frac{1}{n^s}.$$

By now you should be able to complete the line of reasoning to obtain, for $real(s)>1,$

$$\xi(s)=\int_{0^+}^{\infty }[\vartheta (0;ix^2)-1)]x^{s-1}dx=\pi^{-s/2}\ \Gamma\left(\frac{s}{2}\right)\ \zeta(s).$$

Do an analytic continuation as done for the gamma function in MSE-Q13956 to obtain, for `0<real(s)<1,`

$$\xi(s)=\int_{0^+}^{\infty }[\vartheta (0;ix^2)-(1+\frac{1}{x})]x^{s-1}dx.$$

Then use symmetries of the Mellin transform and the fact that $\xi(s)=\xi(1-s)$ (as explained in MSE-Q28737) to obtain the functional equation

$$\vartheta (0;ix^2)=\frac{1}{x}\vartheta (0;\frac{i}{x^{2}}).$$

Update (Jan. 5, 2021):

This perhaps answers Edwards' question and provides a simple path to the functional identity.

Euler initially acquired a reputation with solving the Basel problem to establish the value of $\zeta(2)$ and furthermore the identities

$$\frac{2}{(2\pi)^{2n}}\:(2n-1)!\:\zeta(2n)=(-1)^{n+1}\frac{B_{2n}}{2n}=(-1)^{n}\zeta(1-2n).$$

As noted above, Riemann incorporated the e.g.f. for the Bernoulli numbers in his normalized Mellin/Laplace transform for the zeta function. It's not a great leap of faith to believe that Riemann (of the eponymous surfaces and derivative) grasped the interpolating property of the Mellin transform. Given that assumption, he could have easily noted and perhaps initially formulated the Mellin integral representation from

$$b_n = D^m_{z=0} \; e^{b.z} = D^m_{z=0} \; \frac{z}{e^z-1}$$

$$ = (-1)^{m} \; \frac{1}{2\pi i}\oint_{|z|=1}\frac{m!}{z^{m+1}} \; \frac{z}{e^z-1} \; dz $$

$$= (-1)^{m} \; \frac{1}{2\pi i}\oint_{|z|=1}\frac{m!}{z^{m+1}} \; e^{b.z} \; dz$$

$$ =(-1)^{m} \frac{1}{2\pi i} \; \oint_{|z|=1}\frac{m!}{z^{m+1}} \; [1 - \frac{z}{2}+ \sum_{n \geq 2} \; \cos(\frac{\pi n}{2}) (-2) \; (2\pi)^{-n} \; n! \; \zeta(n) \; \frac{z^n}{n!}] \; dz $$

$$ =(-1)^{m} \frac{1}{2\pi i} \; \oint_{|z|=1}\frac{m!}{z^{m+1}} \; [1 - \frac{z}{2}+ \sum_{n \geq 2} \; (-n) \;\zeta(1-n) \; \frac{z^n}{n!}] \; dz . $$

The Hankel deformation maneuvers above could be reversed to obtain the Mellin transforms equivalent to these contour integrals (see this MO_A), but this would not change the equivalence of the coefficients in the two e.g.f.s for the Bernoulli numbers, so from the interpolating property of the Mellin transform (and therefore the Cauchy integrals), Riemann could have simply analytically continued $n$ to $1-s$ to surmise the target identity

$$ \cos(\frac{\pi n}{2}) \; 2 \; (2\pi)^{-n} \; n! \; \zeta(n)] \; |_{n \to 1-s} = n \;\zeta(1-n) \; |_{n \to 1-s},$$

giving the functional reflection identity

$$\cos(\frac{\pi (1-s)}{2}) \; 2 \; (2\pi)^{s-1} \; (1-s)! \; \zeta(1-s) = (1-s) \;\zeta(s) , $$

or

$$2 \; (2\pi)^{s-1} \; \sin(\frac{\pi s}{2}) \; (-s)! \; \zeta(1-s) = \zeta(s) . $$

Edit 1/24/21: A fairly simple fourth way to derive the functional equation involving Fourier series and the core Poisson summation distribution, yet avoiding the theta function, is given in my answer to this recent MO-Q.

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