18.785: Analytic Number Theory, MIT, spring 2007 (K.S. Kedlaya)
The functional equations for Dirichlet L-functions
In this unit, we establish the functional equation property for Dirichlet L-functions. Muchof the work is left as exercises.

1Even charactersLet χ be a Dirichlet character of level N . We say χ is even if χ(−1) = 1 and odd if
χ(−1) = −1.
For χ even, we can derive a functional equation for L(s, χ) by imitating the argument weused for ζ. Start withZ ∞2−s/2 s/2−sχ(n)e−πn x/N xs/2−1 dxχ(n)πN Γ(s/2)n =0

and sum over n to obtain
π−s/2Ns/21
Γ(s/2)L(s, χ) =2forθ(x, χ) =∞XZ ∞xs/2−1 θ(x, χ) dx

(1)

02

χ(n)e−πn x/N .

n=−∞(Notice there is no additive constant because χ(0) = 0.)
Applying the Poisson summation formula to θ(x, χ) looks problematic, because χ(n) does
not extend nicely to a function on all of R. Fortunately we can avoid this by doing a bit
more Fourier analysis, but this time on the additive group Z/N Z: write
χ(n) =NXcχ,m e2πimn/N

m=1withN

1 Xcχ,m =χ(l)e−2πilm/N .
N l=1I’ll come back to what this quantity cχ,m actually is in a moment. In the meantime, let’s
see what happens when we use this new expression for χ(n). Or rather, I’ll let you see what
happens as an exercise; you should getθ(x, χ) = (N/x)

1/2

NXcχ,m

m=1∞
X

n=−∞

12e−π(nN +m) /(xN ) .

(2)

To get further, we need some description of the cχ,m which is somehow uniform in m. Hereit is: if m is coprime to N , thenNcχ,m ==1 X
χ(l)e−2πilm/NN
l=1NX

1χ(m)χ(lm)e−2πilm/N
N l=1= χ(m)cχ,1 .
For m not coprime to N , we must assume χ is primitive, and then againcχ,m = χ(m)cχ,1(3)but this is not so obvious; see exercises.
This gives usθ(x, χ) = (N/x)1/2 cχ,1 θ(x−1 , χ),and now we are home free: again split the integral (1) at 1 and substitute x 7→ x−1 in one
term, to obtainZZ
1 ∞ s/2−1
1 ∞ −s/2−1 −1−s/2 s/2πN Γ(s/2)L(s, χ) =x
θ(x, χ) dx +x
θ(x , χ) dx
2 12 1ZZ ∞1 ∞ s/2−1
1 1/2xθ(x, χ) dx + N cχ,1=x(1−s)/2−1 θ(x, χ) dx.
2 121Similarly,
π−s/2Ns/2

1Γ(s/2)L(s, χ) =2Z ∞1xs/2−11θ(x, χ) dx + N 1/2 cχ,12Z ∞x(1−s)/2−1 θ(x, χ) dx.

1It is elementary to check that cχ,1 cχ,1 = N −1 (see exercises); we thus obtain
π −(1−s)/2 N (1−s)/2 Γ((1 − s)/2)L(1 − s, χ) = N 1/2 cχ,1 π −s N s/2 Γ(s/2)L(s, χ).

(4)Again, the extra factors of π, N, Γ should be thought of as an “extra Euler factor” comingfrom the “prime at infinity”.
Pay close attention to the fact that unless χ = χ, the functional equation 4 relates two
different L-functions. In a few circumstances, this makes it less useful than if it related asingle L(s, χ) to itself, but so be it.
Also note that quantity cχ,1 is related to the more commonly introduced Gauss sum
associated to χ:NX
τ (χ) = N cχ,1 =χ(l)e2πil/N .
l=1

For more about Gauss sums, see the exercises.

22Odd characters

We have to do something different if χ(−1) = −1, as then the function θ(x, χ) as defined
above is identically zero. Instead we useθ1 (x, χ) =∞X

2nχ(n)e−n πx/Nn=∞and shift s around a bit. Namely,π−(s+1)/2N(s+1)/21
Γ((s + 1)/2)L(s, χ) =2Z ∞θ1 (x, χ)x(s+1)/2−1 dx.

0Again you split the integral at x = 1 and use an inversion formula; this time the right identity
is∞∞ X
Xm  −π(n+m/N )2 N/x−n2 πx/N +2πimn/N
3/2e.
(5)
ne= i(N/x)
n+N
n=−∞
n=−∞

You should end up with the functional equation

π −(2−s)/2 N (2−s)/2 Γ((2 − s)/2)L(1 − s, χ) = −iτ (χ)N −1/2 π −(1+s)/2 N (1+s)/2 Γ((1 + s)/2)L(s, χ).
(6)
Since this is now the third time through this manner of argument, I leave further details tothe exercises.

Exercises
1. Prove the following functional equations for Γ:πsin πsΓ(s)Γ(s + 1/2) = 21−2s π 1/2 Γ(2s).
Γ(s)Γ(1 − s) =Then use these to give a simplified functional equation for ζ of the form “ζ(1 − s)equals ζ(s) times some explicit function”.
2. Prove that (3) holds for χ primitive whether or not m is coprime to N .
3. Prove that for χ primitive, τ (χ)τ (χ) = N . (Warning: the value of τ (χ)τ (χ) depends onwhether χ is even or odd.) Then exhibit an example where this fails if χ is imprimitive.
4. For χ a Dirichlet character of level N , based on the functional equation, where does
L(s, χ) have zeroes and poles in the region Re(s) ≤ 0?

35. Prove that∞Xe−(n+α)2 π/x=x1/2∞X

2e−n πx+2πinα(α ∈ R, x > 0);

n=−∞

n=−∞

then prove (5) by the same method (namely Poisson summation).
6. Use the previous exercise to deduce (2).
7. Prove the functional equation (6).
8. Pick an example of a nonprincipal nonprimitive character χ, and write out the functional equation for L(s, χ).
9. (Dirichlet) For a, b ∈ Z and f : R → C a function obtained by taking a continuousfunction on [a, b] and setting its other values to 0, the Poisson summation formula stillholds if interpreted as∞X
11f (a) + f (a + 1) + · · · + f (b − 1) + f (b) =fˆ(n)22n=−∞(you don’t have to prove this). Apply this to the function(2e2πit /N t ∈ [0, N ]f (t) =0otherwise
in order to evaluate

PN

n=1 e

2πin2 /Nfor N a positive  integer. Then use this to computeG(χ) for χ the quadratic character χ(m) =a more elementary computation of G(χ)2 .)4mp. (Optional, not to be turned in: give