To extend on Matt's comment about Euler, here is something I wrote up some years ago about Euler's discovery of the functional equation only at integral points. I hope there are no typos.
Although Euler never found a convergent analytic expression for
$\zeta(s)$ at negative numbers, in 1749 he publishedpresented a method of
computing
values of the zeta function at negative integers by a precursor of
Abel's Theorem applied to a divergent series. The(The paper appeared in 1768: see here, Sections 3, 9, and 10.) The
computation led him to the asymmetric functional equation of
$\zeta(s)$.
The technique uses the function
$$
\zeta_{2}(s) = \sum_{n \geq 1} \frac{(-1)^{n-1}}{n^s} =
1 - \frac{1}{2^s} + \frac{1}{3^s} - \frac{1}{4^s} + \dots.
$$
This looks not too different from $\zeta(s)$, but has the advantage
as an alternating series of
converging for all positive $s$. For $s > 1$,
$\zeta_2(s) = (1 - 2^{1-s})\zeta(s)$.
Of course this is true for complex $s$, but
Euler only worked with real $s$, so we shall as well.
Disregarding convergence issues, Euler wrote
$$
\zeta_{2}(-m) = \sum_{n \geq 1} (-1)^{n-1}n^m = 1 - 2^m + 3^m - 4^m +
\dots,
$$
which he proceeded to evaluate as follows. Differentiate
the equation
$$
\sum_{n \geq 0} X^n = \frac{1}{1-X}
$$
to get
$$
\sum_{n \geq 1} nX^{n-1} = \frac{1}{(1-X)^2}.
$$
Setting $X = -1$,
$$
\zeta_{2}(-1) = \frac{1}{4}.
$$
Since $\zeta_{2}(-1) = (1-2^2)\zeta(-1)$, $\zeta(-1) = -1/12$. Notice we can't set $X = 1$ in the second power series and compute $\sum n = \zeta(-1)$ directly. So $\zeta_2(s)$ is nicer than $\zeta(s)$ in this Eulerian way.
Multiplying the second power series by $X$ and then differentiating, we get
$$
\sum_{n \geq 1} n^2X^{n-1} = \frac{1+X}{(1-X)^3}.
$$
Setting $X = -1$,
$$
\zeta_{2}(-2) = 0.
$$
By more successive multiplications by $X$ and differentiations, we get
$$
\sum_{n \geq 1} n^3X^{n-1} = \frac{X^2+4X+1}{(1-X)^4},
$$
and
$$
\sum_{n \geq 1} n^4X^{n-1} = \frac{(X+1)(X^2+10X+1)}{(1-X)^5}.
$$
Setting $X = -1$, we find $\zeta_{2}(-3) = -1/8$ and
$\zeta_{2}(-4) = 0$. Continuing further, with the recursion
$$
\frac{d}{dx} \frac{P(x)}{(1-x)^n} = \frac{(1-x)P'(x) + nP(x)}{(1-x)^{n+1}},
$$
we get
$$
\sum_{n \geq 1} n^5X^{n-1} = \frac{X^4+26X^3+66X^2 + 26X +1}{(1-X)^6},
$$
$$
\sum_{n \geq 1} n^6X^{n-1} = \frac{(X+1)(X^4 + 56X^3 + 246X^2 + 56X+1)}
{(1-X)^7},
$$
$$
\sum_{n \geq 1} n^7X^{n-1} = \frac{X^6 + 120X^5 + 1191X^4 + 2416X^3 +
1191X^2 + 120X + 1}{(1-X)^8}.
$$
Setting $X = -1$, we get $\zeta_{2}(-5) =
1/4, \ \zeta_{2}(-6) = 0, \ \zeta_{2}(-7) = -17/16$.
Apparently $\zeta_{2}$ vanishes at the negative even integers, while
$$
\frac{\zeta_{2}(-1)}{\zeta_{2}(2)} = \frac{1}{4}\cdot\frac{6\cdot 2}{\pi^2} =
\frac{3\cdot 1!}{1\cdot \pi^2}, \ \ \ \
\frac{\zeta_{2}(-3)}{\zeta_{2}(4)} =
-\frac{1}{8}\cdot\frac{30\cdot24}{7\pi^4} = -\frac{15\cdot 3!}{7\cdot \pi^4},
$$
$$
\frac{\zeta_{2}(-5)}{\zeta_{2}(6)} = \frac{1}{4}\cdot
\frac{42\cdot 6!}{31\pi^6} =
\frac{63 \cdot 5!}{31\cdot \pi^6}, \ \ \ \
\frac{\zeta_{2}(-7)}{\zeta_{2}(8)} = -\frac{17}{16}\cdot
\frac{30\cdot 8!}{127\cdot \pi^8}
= -\frac{255\cdot 7!}{127\pi^8}.
$$
The numbers $1,
3, 7, 15, 31, 63, 127, 255$ are all one less than a power of 2,
so Euler was led to the observation that for $n \geq 2$,
$$
\frac{\zeta_{2}(1-n)}{\zeta_{2}(n)} =
\frac{(-1)^{n/2+1}(2^n-1)(n-1)!}{(2^{n-1}-1)\pi^n}
$$
if $n$ is even and
$$
\frac{\zeta_{2}(1-n)}{\zeta_{2}(n)} =
0
$$
if $n$ is odd. Notice how
the vanishing of $\zeta_{2}(s)$
at negative even integers nicely compensates for
the lack of knowledge of $\zeta_2(s)$ at positive odd integers $> 1$ (which
is the same as not knowing $\zeta(s)$ at positive odd integers $> 1$).
Euler interpreted the $\pm$ sign at even $n$ and the vanishing
at odd $n$ as the single factor $-\cos(\pi n/2)$, and with
$(n-1)!$ written as $\Gamma(n)$ we get
$$
\frac{\zeta_{2}(1-n)}{\zeta_{2}(n)} = -\Gamma(n)\frac{2^n-1}{(2^{n-1}-1)\pi^n}
\cos\left(\frac{\pi n}{2}\right).
$$
Writing $\zeta_{2}(n)$ as $(1 - 2^{1-n})\zeta(n)$ gives the
asymmetric functional
equation
$$
\frac{\zeta(1-n)}{\zeta(n)} = \frac{2}{(2\pi)^n}
\Gamma(n)\cos\left(\frac{\pi n}{2}\right).
$$
Euler applied similar ideas to $L(s,\chi_4)$ and found its
functional equation. You can work this out yourself in
Exercise 2 below.
Exercises
Show that Euler's computation of zeta values at negative
integers can be put in the form
$$
(1 - 2^{n+1})\zeta(-n) =
\left.\left(u\frac{d}{du}\right)^{n}\right\vert_{u=1}\left(\frac{u}{1+u}
\right) = \left.\left(\frac{d}{dx}\right)^{n}\right\vert_{x=0}
\left(\frac{e^x}{1+e^x}\right).
$$
To compute the divergent series
$$
L(-n,\chi_4) = \sum_{j \geq 0} (-1)^{j}(2j+1)^n =
1 - 3^n + 5^n - 7^n - 9^n + 11^n - \dots
$$
for nonnegative integers $n$, begin with the formal identity
$$
\sum_{j \geq 0} X^{2j} = \frac{1}{1-X^2}.
$$
Differentiate and set $X = i$ to show $L(0,\chi_4) = 1/2$.
Repeatedly multiply by $X$, differentiate, and set
$X = i$ in order to compute $L(-n,\chi_4)$ for $0 \leq n \leq 10$. This computational technique is not rigorous, but the answers are correct. Compare with the values of $L(n,\chi_4)$ for positive $n$, if you know those, to get a formula for $L(1-n,\chi_4)/L(n,\chi_4)$. Treat alternating signs like special values of a suitable trigonometric function.
