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Questions tagged [analytic-continuation]

For questions related to analytic continuation

4
votes
2answers
680 views

Alternating series test for complex series

I want to show that we can continue Riemann's zeta function to Re$(s)>0$, $s\neq 1$ by the following formula \begin{align} (1-2^{1-s})\zeta(s)&=\left(1-2\frac{1}{2^s}\right)\left(\frac1{1^s}+\...
4
votes
2answers
283 views

Alternative analytic continuation to zeta, not giving $-\frac{1}{12}$ for sum of integers

Apologies if this has been asked already. Inspired partly by this answer where an $n e^{-\epsilon n}$ rather than $n^s$ regularization was made in the 'evaluation' of $\sum\limits_{n=1}^{\infty}n$ and ...
2
votes
1answer
180 views

Does the Abel sum 1 - 1 + 1 - 1 + … = 1/2 imply $\eta(0)=1/2$?

If $\sum_{n=1}^\infty a_n$ is Abel summable to $A$, then necessarily $\sum_{n=1}^\infty a_n n^{-s}$ has a finite abscissa of convergence and can be analytically continued to a function $F(s)$ on a ...
5
votes
0answers
125 views

Is $\int_0^\infty \frac{dt}{e^t-xt}$ analytic continuation of $\sum_{k=1}^\infty \frac{(k-1)!}{k^k} x^{k-1}$?

The following power series apparently converges only for $-e \leq x <e$: $$f(x)=\sum_{k=1}^\infty \frac{(k-1)!}{k^k} x^{k-1}$$ We can use it to define a real function $f(x)$, analytic in that ...
4
votes
3answers
495 views

Gamma & Zeta Summation $\sum_{n=0}^{\infty}\frac{\Gamma(n+s)\zeta(n+s)}{(n+1)!}=0$

According to Gamma Summation & Zeta Summation: $$ \sum_{n=0}^{\infty} {(-1)^n \frac{\Gamma(n+s) \zeta(n+s)}{(n+1)!}}=\Gamma(s-1) \quad : \space Re\{s\}<2 $$ Show that: $$ \sum_{n=0}^{\...
2
votes
3answers
144 views

Gamma Infinite Summation $\sum_{n=0}^{\infty}\frac{\Gamma(n+s)}{n!}=0$

Avoiding the analytic continuation of extended binomial theorem, $$ \sum_{n=0}^{\infty}\frac{\Gamma(n+z)}{n!}\,x^n = \frac{\Gamma(z)}{(1-x)^z} \quad\colon\space |x|\lt1 $$ How to prove: $$ \...
2
votes
2answers
119 views

Evaluating $\zeta\left(\frac{1}{2}\right)$ as an integral $ \zeta\left(\frac{1}{2}\right) = \frac{1}{2} \int_0^\infty \frac{[x]-x}{x^{3/2}} \, dx$

I am reading the second chapter of Titchmarsh's book on the Riemann Zeta Function. I would have written: $$ \zeta\left(\frac{1}{2}\right) = 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots = \...
1
vote
7answers
720 views

Why do $f(x)= \frac {{x²-1}}{{x-1} } $ and $g(x)=x+1$ not have the same domain of definition, but are the same? [closed]

Let $f(x)= \frac {{x²-1}}{{x-1} } $ and $g(x)=x+1$. It is clear that $f(x)=g(x)$ for all real numbers $x$. My question: Why do $f$ and $g$ not have the same domain of definition, however $f(x)=g(x)$?...
10
votes
7answers
3k views

Intuition behind $\zeta(-1)$ = $\frac{-1}{12}$ [duplicate]

When I first watched numberphile's 1+2+3+... = $\frac{-1}{12}$ I thought the sum actually equalled $\frac{-1}{12}$ without really understanding it. Recently I read some wolframalpha pages and watched ...
11
votes
2answers
943 views

On every simply connected domain, there exists a holomorphic function with no analytic continuation.

I am working on a question that requires me to prove that on every simply connected open subset of $\mathbb{C}$, there exists a holomorphic function that cannot be extended to a holomorphic function ...
7
votes
3answers
417 views

Can we use analytic continuation to obtain $\sum_{n=1}^\infty n = b, b\neq -\frac{1}{12}$

Intuitive question It is a popular math fact that the sum definition of the Riemann zeta function: $$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} $$ can be extended to the whole complex plane (except ...
1
vote
1answer
428 views

Analytical continuation of moment generating function

Let's say some distribution $F(t)$ has finite moment generating function on an open ball (-R, R). $M(x) = \sum m_n x^n /n!$ Let's extend $M(x)$ to $M(z)$ on a complex strip $S = \{z| |Re(z)| <R\}$...
6
votes
2answers
371 views

Analytic continuation of harmonic series

Is there an accepted analytic continuation of $\sum_{n=1}^m \frac{1}{n}$? Even a continuation to positive reals would be of interested, though negative and complex arguments would also be interesting. ...
2
votes
1answer
150 views

Weierstrass factorization theorem and primality function

I'm interested in application of the Weierstrass factorization theorem to the primality function. Let $np(x)\colon \mathbb N\to \mathbb N$ is a "not-prime" function: $$ np(x) = \begin{cases}1, & \...
1
vote
1answer
935 views

real part of analytic function

Assume $f(x) \in D$ is analytic, where $x \in R$ and $D$ denotes the unit circle, which means $x$ is real and $f(x)$ is complex. Then what kind of function is the real part of $f(x)$? harmonic or ...
0
votes
0answers
51 views

Maximal analytically continued domain

Can a power series or a Laurent series always be analytically continued into a domain strictly larger than its convergent annulus of finite radii? Is there a way to find its maximal domain that it can ...
4
votes
2answers
223 views

Analytic continuation for $\zeta(s)$ using finite sums?

$\zeta(s)$ converges for $\sigma >1$ but not for $\sigma =1/2.$ But for some reason for $s = 1/2 + i t $ and fixed finite $N,~$ $\zeta_N(s) =\sum_{n=1}^N\frac{1}{n^s}$ is very close to $\zeta(s)$ ...
3
votes
1answer
602 views

Analytical continuation of complete elliptic integral of the first kind

I am dealing with a problem involving the complete elliptical function of the first kind, which is defined as: $K(k)=\int_0^{\pi/2} d\theta \frac{1}{\sqrt{1-k^2\sin^2(\theta)}}=\int_0^1 dt \frac{1}{\...
2
votes
0answers
45 views

For a solution to an ODE, when do we apply analytic continuation at a point where ODE is undefined?

As a continuation of this question, the given answer states that at a point where the ODE is undefined, we apply "analytic continuation" (if this is the correct terminology) to the solution so that ...
1
vote
1answer
472 views

Accumulation points of a holomorphic function.

I am reviewing some complex analysis and I have just gotten to the portion on analytic continuation. My question is about the proof of the following theorem: Theorem. Suppose $f$ is a holomorphic ...
1
vote
1answer
268 views

Given complex series, the natural boundary is the unit circle

How to prove that the series $$\sum_{n=1}^{\infty} z^{n !}=z^{1!}+z^{2!}+z^{3!}+ \cdots $$ has the natural boundary $|z|=1$. I did a similar problem to this where I was able to get an iterative ...
0
votes
1answer
164 views

Every uniformly convergent sequence of polynomials converges on a simply connected domain

Here is a really weird problem. Suppose $P_i$ is a sequence of polynomials which converges to the analytic $f$ on a domain $U$ in $\mathbb{C}$, uniformly on compact subsets. Then there is a simply ...