# $L(1+it,\chi)\neq 0$ whenever $t \neq 0 \in \mathbb{R}$

I understand that the proof of the assertion in the title uses the same method which proves that zeta function satisfies $\zeta(1+it)\neq 0$, where the above $L$ is Dirichlet L-function.

I.e, you write down the function $D(\sigma)=\zeta^3(\sigma) \zeta^4(\sigma+it_0) \zeta(\sigma+2it_0)$, where $t_0\neq 0$ s.t $\zeta(1+it_0)=0$ then you get on the one hand that $D \rightarrow 0$ when $\sigma \rightarrow 1$, but on the other hand $|D|\geq 1$.

But how do I mimic this proof with L function, I mean I am not sure does does L function has a simple pole in s=1?

Thanks in advance.

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There is no pole at $s=1$. This is somewhat of a duplicate question with mathoverflow.net/questions/25794/…. That MO question is about $L(1,\chi)$ being nonzero, but some proofs in the answers there carry over to show $L(1+it,\chi)$ is nonzero for nontrivial $\chi$ and $t \not= 0$. –  KCd May 13 '12 at 6:26
I didn't see there a rigorous proof of this claim, and anyways I was told the proof for L should use the same method as the proof for zeta, with the same D, just replacing zeta with L appropiately, the problem is that I don't see how to show that D converges to zero when sigma approaches 1?! –  MathematicalPhysicist May 13 '12 at 8:00
$D\to0$ yet $|D|\ge1$? Are you sure that's accurate? –  anon May 13 '12 at 10:52
I want to show a contradiction by assuming there's $t_0 \neq 0$ s.t $L(1+it_0,\chi)=0$ and thus arriving that $D$ converges to 0 but yet its modulus is at least 1. –  MathematicalPhysicist May 13 '12 at 14:36
The link in my previous comment provides you with several proofs that $L(1,\chi) \not= 0$. When you say there is no "rigorous proof" there, what exactly are you expecting? If you want full-blown details then look in books on analytic number theory. Some of the answers in the link I gave before do sketch how to bootstrap a proof of $L(1,\chi)$ being nonzero to $L(1+it,\chi)$ being nonzero. The proofs usually exploit the fact that $\zeta(s)$ is known to have a simple pole at $s = 1$. –  KCd May 13 '12 at 23:41
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