Articles of analytic continuation

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: $$ np(x) = \begin{cases}1, & \text{$x$ is not prime}\\0, & \text{$x$ is prime}\end{cases} $$ Obviously, prime numbers are zeros of $np$. Therefore it seems reasonable to use Weierstrass factorization theorem to express […]

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 $S = \{z| |Re(z)| <R\}$. I need to prove that $M(z)$ is also analytic on $S$(thus it is an analytic continuation). What is the […]

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 interval. However, we can also use an integral to define this function: $$f(x)=\int_0^\infty \frac{dt}{e^t-xt}=\sum_{k=1}^\infty \frac{(k-1)!}{k^k} x^{k-1}$$ In the interval of convergence of the series these two definitions […]

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 $\zeta(s)$ as found using analytic continuation for $t$ in some range $$f_1(N) < t < f_2(N). $$ For example, for $N= 1000$ a […]

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 one) to obtain $\zeta(-1)=-\frac{1}{12}$. The right hand side for the above equation in $-1$ becomes the sum of the natural numbers so in some […]

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: $$ \sum_{n=0}^{\infty}\frac{\Gamma(n+s)}{n!} = 0 \quad\Rightarrow\, \frac{s}{1!}+\frac{s(s+1)}{2!}+\cdots = -1 \quad\colon\space Re\{s\}\lt0 $$

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: $$ \sum_{n=0}^{\infty} {\frac{\Gamma(n+s) \zeta(n+s)}{(n+1)!}}=0 \quad : \space Re\{s\}<1 $$ In other words, the Even & Odd parts are convergent series, equaling sums, and different signs: $$ \sum_{n=0}^{\infty} {\frac{\Gamma(2n+s) \zeta(2n+s)}{(2n+1)!}}=-\sum_{n=0}^{\infty} {\frac{\Gamma(2n+1+s) \zeta(2n+1+s)}{(2n+1+1)!}}=\frac{\Gamma(s-1)}{2} \space : \space Re\{s\}<1 $$

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}+\frac1{2^s}+\ldots \right) \\ &=\frac1{1^s}+\frac1{2^s}+\ldots -2\left(\frac1{2^s}+\frac1{4^s}+\ldots \right)\\ &=\frac1{1^s}-\frac1{2^s}+\frac1{3^s}-\frac1{4^s}+\ldots \\ &=\sum_{n=1}^\infty (-1)^{n-1}\frac1{n^s}. \end{align} In order to do that, I need to show that the series converges for Re$(s)>0$, except $s=\frac{2k\pi i}{\ln 2}+1$, $k\in \mathbb{Z}$, which are removable […]

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 the number $-\frac{1}{12}$ appeared as the only constant in the answer, and partly by a conversation with a friend claiming that $-\frac{1}{12}$ is the only ‘right’ […]

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 neighborhood of $s=0$ such that $F(0)=A$. Is that true? Edit. The Dirichlet series $\sum_{n=1}^\infty a_n n^{-s}$ converges absolutely for $\Re(s)>1$, because the radius of convergence […]