# Holomorphic extension of Dirichlet Series

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• April 7th 2010, 03:52 AM
EinStone
Holomorphic extension of Dirichlet Series
Define $L : \{s \in \mathbb{C}| Re(s) > 0\} \rightarrow \mathbb{C}$ by $L(s) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}$.

Note that $L(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{t^{s-1}}{e^t+e^{-t}}dt$

Show that L can be extended to an entire function $L : \mathbb{C} \rightarrow \mathbb{C}$
• April 7th 2010, 11:13 AM
chiph588@
Quote:

Originally Posted by EinStone
Define $L : \{s \in \mathbb{C}| Re(s) > 0\} \rightarrow \mathbb{C}$ by $L(s) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}$.

Note that $L(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{t^{s-1}}{e^t+e^{-t}}dt$

Show that L can be extended to an entire function $L : \mathbb{C} \rightarrow \mathbb{C}$

Tell me more i.e. what else do you know?

Are you familiar with the derivation of the analytic continuation of $\zeta(s)$? Because this series has a similar continuation.

If not, that's ok because I don't think you're asking for the functional equation itself, just that it exists.

Before I make any head way on this problem, what are your thoughts on how to approach it?
• April 8th 2010, 12:12 AM
EinStone
What I know is how to show that the Gamma Function is meromorphic on $\mathbb{C}$. Then by evaluating $\Gamma(s) * \zeta(s)$ in a specific way, one can show that $\Gamma(s)*\zeta(s)$ is also meromorphic on $\mathbb{C}$, and therefore $\zeta(s)$ has to be meromorphic as well.

I actually thought that $\int_0^\infty \frac{t^{s-1}}{e^t+e^{-t}}dt$ can somehow be extended holomorphically to $\mathbb{C}$ similarly to the Gamma function. But thats all I can think of.
• April 8th 2010, 01:12 AM
Laurent
Quote:

Originally Posted by EinStone
Note that $L(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{t^{s-1}}{e^t+e^{-t}}dt$

Show that L can be extended to an entire function $L : \mathbb{C} \rightarrow \mathbb{C}$

I guess you know that $\frac{1}{\Gamma}$ is an entire function. Then all you have to prove is that the integral defines an entire function as well. There is a general theorem for proving that functions of the form $s\mapsto\int f(s,t)dt$ are analytic. If you know it (which is probably the case), then you know what left you have to do... and this shouldn't be too hard (very similar to Gamma). What did you already try this way?
• April 8th 2010, 02:24 AM
EinStone
Actually I don't remember the proof exactly, it was very strange. Also I don't know the theorem you are talking about :(. What can I do?
• April 8th 2010, 01:08 PM
EinStone
I still need help!

Meanwhile I asked myself why the integral converges in the first place for $Re(s) > 0$ ?
• April 8th 2010, 01:39 PM
Laurent
Quote:

Originally Posted by EinStone
I still need help!

Actually I was wrong: the integral is only defined when ${\rm Re}(s)>0$... It is not simple to extend it to $\mathbb{C}$. Like Chiph588@ said, you should repeat arguments used for zeta.

Quote:

Meanwhile I asked myself why the integral converges in the first place for $Re(s) > 0$ ?
We have $\frac{|t^{s-1}|}{e^t+e^{-t}}\leq |t^{s-1}e^{-t}|$, so if you know that the integral defining Gamma converges, then so does this one.
• April 8th 2010, 02:53 PM
EinStone
Ok, So to prove that Zeta is holomorphic everywhere, we used $\frac{t}{e^t-1} = \sum_{k=0}^\infty \frac{B_k}{k!} t^k$ where $B_k$ are the Bernoulli numbers.

So I think I need a similar expression for $\frac{t}{e^t+e^{-t}}$, but I don't see it.
• April 8th 2010, 03:08 PM
chiph588@
Quote:

Originally Posted by EinStone
Ok, So to prove that Zeta is holomorphic everywhere, we used $\frac{t}{e^t-1} = \sum_{k=0}^\infty \frac{B_k}{k!} t^k$ where $B_k$ are the Bernoulli numbers.

So I think I need a similar expression for $\frac{t}{e^t+e^{-t}}$, but I don't see it.

$\frac{t}{e^t+e^{-t}} = \frac12 t \,\text{sech}(t) = \sum_{n=0}^\infty \frac{E_{2n}t^{2n+1}}{2(2n)!} \;\; |t|<\frac\pi2$, where $E_k$ are the Euler numbers.

I don't know if this will help at all though, since $|t|<\frac\pi2$...
• April 8th 2010, 04:35 PM
chiph588@
Quote:

Originally Posted by EinStone
Ok, So to prove that Zeta is holomorphic everywhere, we used $\frac{t}{e^t-1} = \sum_{k=0}^\infty \frac{B_k}{k!} t^k$ where $B_k$ are the Bernoulli numbers.

What all was said to show $\zeta(s)$ is meromorphic everywhere?
• April 9th 2010, 02:33 AM
EinStone
Ok here is the proof for Zeta being meromorphic everywhere.

$\frac{t}{e^t-1} = \sum_{k=0}^\infty \frac{B_k}{k!} t^k = 1 - \frac{t}{2} + \frac{t^2}{12} + 0t^3 - \frac{t^4}{720} + \cdots$ where $B_k$ are the Bernoulli numbers.

Fix $n > 0$, $f_n(t):= \sum_{k=0}^n (-1)^k \frac{B_k}{k!} t^k = 1 + \frac{t}{2} + \frac{B_2}{2!}t^2 + \frac{B_4}{4!}t^4 + \cdots + \frac{B_n}{n!}t^n$ (all terms even)

Now for Re(s) > 1:
$\Gamma(s) * \zeta(s) = \int_0^\infty \frac{t e^t}{e^t-1} e^{-t} t^{s-2}dt = \int_0^\infty (\frac{t e^t}{e^t-1} - f_n(t))e^{-t} t^{s-2}dt$ $+ \int_0^\infty f_n(t) e^{-t} t^{s-2}dt = I_1(s) + I_2(s)$

$\frac{t e^t}{e^t-1}$ is holomorphic at t = 0, has Taylor expansion: $1 + \frac{t}{2} + \frac{B_2}{2!}t^2 + \frac{B_4}{4!}t^4 + \cdots = \frac{t}{e^t-1} +t$
hence $\frac{t e^t}{e^t-1}-f_n(t) = O(t^{n+1})$ near t = 0.

$\rightsquigarrow$ Integrand of $I_1$ near t = 0 is $\approx O(t^{n+1}t^{s-2}) = O(t^{n+s-1})$. So $I_1$ converges if Re(n+s-1) > -1, i.e. if Re(s) > -n.

For $I_2$: $I_2(s) = \int_0^\infty (1 + \frac{t}{2} + \sum_{k=2}^n \frac{B_k}{k!}t^{k}) e^{-t} t^{s-2}dt = \Gamma(s-1) + \frac{1}{2}\Gamma(s) + \sum_{k=2}^n \frac{B_k}{k!} \Gamma(s+k-1)$ , a finite sum of meromorphic functions on $\mathbb{C}$.

$\rightsquigarrow \zeta$ is meromorphic for Re(s) > -n, n arbitrary, hence $\zeta$ is meromorphic on $\mathbb{C}$.
• April 9th 2010, 04:34 AM
Laurent
I didn't know this proof; it is rather simple, and adaptable to many situations.

Notice that you only use the fact that there exists an expansion $\frac{t e^t}{e^t-1}=a_0+a_1t+\cdots+a_n t^n+O(t^{n+1})=f_n(t)+O(t^{n+1})$ when $t\to 0$, regardless of the value of $a_n$ (or of the convergence of the series expansion).

Since $t\mapsto\frac{t e^t}{e^t+e^{-t}}$ is $\mathcal{C}^\infty$ (even analytic), there is also such an expansion $\frac{t e^t}{e^t+e^{-t}}=a_0+a_1t+\cdots+a_n t^n+O(t^{n+1})$ when $t\to0$ (this is Taylor-Young's theorem). And you can transpose the proof seamlessly.

It will show you that $L$ is meromorphic on $\mathbb{C}$, so you also have to justify it has no poles... That may require the values of the coefficients $a_n$ ; I'll think about it.

Sidenote: there is a justification missing in your proof, but maybe it was an implicit reference to a proof you did for Gamma. An argument is indeed necessary to justify that $I_1$ is analytic; it is of the form $\int_0^\infty f(t,s)dt$ where, for all $t>0$, $s\mapsto f(t,s)$ is analytic, but this is not sufficient to conclude that the integral itself is analytic. A "domination" is needed, like for the continuity or differentiability theorems under the integral sign. But it is simple to apply here, so I guess you were told it is just like for Gamma.

----

Addendum: in fact, the same method proves the analyticity on $\mathbb{C}$.

Indeed, we get an expression like $L(s)=\frac{1}{\Gamma(s)}{I_1(s)}+\sum_{k=0}^n a_k\frac{\Gamma(s+k-1)}{\Gamma(s)}$, for ${\rm Re}(s)>-n$, where $I_1(s)$ is analytic. We know that $\Gamma(s)\neq 0$, hence the first term $\frac{I_1(s)}{\Gamma(s)}$ defines an analytic function.
Furthermore, for $k\geq 1$, $\Gamma(s+k-1)=(s+k-2)(s+k-3)\cdots s \Gamma(s)$ (usual functional equation of $\Gamma$ used several times) hence $\frac{\Gamma(s+k-1)}{\Gamma(s)}=(s+k-2)(s+k-3)\cdots s$ is just a polynomial, hence it is analytic.

The only piece that is not analytic for $\zeta$ is the term $\frac{\Gamma(s-1)}{\Gamma(s)}=\frac{1}{s-1}$, which comes from $k=0$. (This proves that 1 is the only pole for $\zeta$; it is simple with residue 1).

However, for $L(s)$, the term $a_0$ is 0 because $\frac{te^t}{e^t+e^{-t}}$ is 0 at 0. Therefore, all terms are analytic. Thus $L$ is analytic on ${\rm Re}(s)>-n$ for all n, hence on $\mathbb{C}$. qed.
• April 9th 2010, 05:26 AM
EinStone
So I want to apply the same proof, but how do I find $I_1$ for L? Also to show that $I_2$ is meromorphic, don't I need to know the $a_n$?
• April 9th 2010, 05:29 AM
Laurent
Quote:

Originally Posted by EinStone
So I want to apply the same proof, but how do I find $I_1$ for L? Also to show that $I_2$ is meromorphic, don't I need to know the $a_n$?

All you have to do is replace $\frac{t e^t}{e^t-1}$ by $\frac{t e^t}{e^t+e^{-t}}$, and of course $\zeta(s)$ by $L(s)$. Everything is the same, except that $a_0=0$ for $L$ (hence the analyticity), but you don't need the other values (neither for zeta of for L), just the fact that they exist. They are constants, so they matter in no way.
• April 9th 2010, 05:36 AM
EinStone
Now I get it, its just the existence (even for Zeta) that is needed, which follows from analyticity.
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