Let us compute
$$\lim_{s\to 1}\sum_{(m,n)\neq(0,0)}\frac{(-1)^{m+n}}{(m^2+n^2)^s}$$
since that's what you want to know anyway. It's mostly about arithmetics in $\mathbb{Z}[i]$, as $m^2+n^2=(m+in)(m-in)=:N(m+in)$. Also $(-1)^{m+n}=(-1)^{m^2+n^2}$, so we're after
$$f(s)=\sum_{\alpha\in\mathbb{Z}[i]...
Can we express $\sin 1^\circ$ in a closed, not repetitive radical forms? Any radical forms mean you can use any roots but without constants $\pi$, $e$ or other trigonometry functions.
@JohanLarsson hehe, yup, may be, may be its because now that the threshold for publishing things is so low, everyone and their dogs and cats can publish stuff.
It's true, my cats co-authored a paper last month, entitled "Hegemony and the Privilege of Pooping Outside: The Tyranny of Co-habitation with Canines."
@robjohn if $\Phi_{p}(t) = \frac{t^{p-1}}{\Gamma(p)} \chi(t)$ then $\Phi_{p}(t) * \Phi_{q}(t-a) = \Phi_{p+q}(t-a)$. I've done it only for $\Re p >0$, $\Re q > 0$. I have to obtain it for any $p,q$ but I mistaked when I tried to use uniqueness theorem
@robjohn by the context $\Phi_{p}(t)$ is a generalized function and $\Phi_{q}(t-a)$ is test function but it doesn't decrease at infinity, so I don't know what to do
@robjohn are you speaking about convolution with $\Phi_{q}(t-a)$?
@robjohn because if $f(t)$ decreases sufficiently fast at infinity convolution $\Phi_{p}(t) * f(t)$ is defined well via analytical continuation for any $p \in \mathbb{C}$, for $p = -3$ for example it is $f^{(3)}(t)$
@robjohn $\int\limits_{0}^{\infty} t^{p-1} f(t) dt = - \frac{1}{p}\int\limits_{0}^{\infty} t^{p} f'(t) dt$ if $f$ decreases sufficiently fast e.t.c., so $\langle t^{p-1} \chi(t), f(t) \rangle$ is defined for any noninteger $p$, right? Only if $f$ decreases sufficiently fast (I mean it analytical continuation)
@robjohn hm, for $\Re p > 0$ this is ok (term is killed by $t^p$), for general $p$ we define $\langle t^{p-1} \chi(t), f(t) \rangle$ as this integral on the right, forgetting about terms, isn't true?
@robjohn $ t^{p-1}\chi(t) * f(t) = \int\limits_{0}^{\infty} \tau^{p-1} f(t - \tau) d\tau$ It is sufficient to say that $\mathrm{supp}\; f \subseteq [a,+\infty)$ in order to solve problems, right?
@Nimza that is what I was saying. $f$ needs to decay fast on the domain you are looking at, so to define the convolution on $[0,a]$ you'd need $f(x)=0$ on $[0,a]$
@robjohn oh, why we are interested in behavior of $f$ at it's support if integral is taken only over finite segment (when $f(x) = 0$ for any $x<a$, where $a$ is a fixed const) and replacing $\tau^{p-1}$ by $\tau^{p+k}$ we can reduce it to integral of continuous function?
@Nimza So if you completely avoid the singularity, by having your function be zero over a certain range and only looking at the convolution over a certain range, the integration by parts works.
@robjohn hence convolution $\frac{t^{p-1}}{\Gamma(p)} \chi(t) * f(t)$ is defined for any smooth function $f$ with support in some inverval $[a,\infty)$ for any complex $p$ and for any real $t$
@robjohn I mean that $h(p,t) = \int\limits_{0}^{\infty} \Phi_{p}(\tau) f(t-\tau) d\tau$ is originally defined for any real $t$ for $\Re p > 0$ right? Here $f$ is with support in $[a,\infty)$
@Nimza usually one shows some contour integral that equals this convolution and then you can use integration by parts on that. But I am not sure how you are defining this for general $f$.
@robjohn that's defined for general smooth $f$ with support in some $[a,\infty)$. Then $\Phi_{p}(t)*f(t)$ is just $f$'s fractional derivative of order $-p$ in sense of distributions (by definition in a paper i'm reading)
@Nimza if you have $\mathrm{Re}(p)>0$, you don't need to limit the support of $f$, for $\mathrm{Re}(p)\le0$, we need to restrict the support of $f$ and then $\Phi_p*f$ is only defined where $f\equiv0$
@Nimza I have not figured out how to analytically continue it yet. I need to think on that.
For those whose an idle brain, could you evaluate $$ \frac{y(x)}{3}=e^{-x}\sin(2x)+\int_0^x\left[e^{-\lambda}\sin(2\lambda) e^{-(x-\lambda)}\sin(x-\lambda)\right]\textrm{d}\lambda $$
What is the limit of U_n = \frac{2U_n + 3}{U_n + 2} and U_0 = 1? I need a the detail, and another way than using the solution of f(x)=x, as f(x) = (2x+3)/(x+2) because I can't found that f(I) C I as I = ]-infini; -2[U]-2; +infini[
Where Beta function may be meromorphically continued? From $B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ I see that it has problems at $x + y = -k$. So if the maximal domain is $\mathbb{C}^2$ without $x+y = -k$, $k=0,1,2,\ldots$?
Can we express $\sin 1^\circ$ in a closed, not repetitive radical forms? Any radical forms mean you can use any roots but without constants $\pi$, $e$ or other trigonometry functions.
What is the limit of U_n = \frac{2U_n + 3}{U_n + 2} and U_0 = 1? I need a the detail, and another way than using the solution of f(x)=x, as f(x) = (2x+3)/(x+2) because I can't found that f(I) C I as I = ]-infini; -2[U]-2; +infini[
