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If one starts with the Weierstrass factorisation $$ \Gamma(z) = \frac{e^{-\gamma z}}{z} \prod_{k=1}^\infty \left(1 + \frac{z}{k}\right)^{-1} e^{z/k}$$ of the Gamma function, applied to $z = -x, -\omeg …

I am recording here a "regularized" version of the standard contour integral argument that avoids detours (though it still needs the standard contour traversing the standard fundamental domain). I fi …

Just a note that Carlo's nice inequality $$ \left|\left(1+\frac{it}{n}\right)^n - 1\right| \geq |e^{it} - 1| \label{1}\tag{1}$$ in fact is valid for all real $t$ and natural numbers $n \geq 1$. For $ …

This summation method gives answers that are close to, but do not always match, traditional divergent summation methods. For instance, for constants $a,b>0$, the divergent sum $$ \sum_{n=1}^\infty (a …

We have an exact formula \begin{align*} \frac{1}{\zeta'(\rho)} &= \lim_{s \to \rho} \frac{s-\rho}{\zeta(s)} \\ &= \lim_{s \to \rho} \frac{(s-\rho) (s-1) \Gamma(1+s/2) \pi^{-s/2}}{\xi(s)} \\ &= (\rho- …

This function is (on the real line, at least) the product of $$ \exp( \mu^2 \sum_{i=1}^\infty |z_i|^2 - 2 \mu \Re(\sum_{i=1}^\infty z_i)) \quad (1)$$ and the Hadamard type product $$ \prod_{i=1}^\inft …

This follows from the work of Miller, Michael J., On Sendov’s conjecture for roots near the unit circle, J. Math. Anal. Appl. 175, No. 2, 632-639 (1993). ZBL0782.30007. and independently Vâjâitu, Vior …

Hmm, it was more difficult than I expected to leverage universality to establish the claim. But one can proceed by probabilistic reasoning instead, basically exploiting the phase transition in the li …

The problem is not with Fubini's theorem (or more precisely, Tonelli's theorem), which is valid for any non-negative function measurable in the product sigma algebra. The error instead lies in the cl …

[Reposted as an answer, as requested. -T.] While this can be done for any given $f$ (as indicated by Angelo's answer), it can't be done in a way which is stable with respect to perturbations of $f$ ( …

This is (the adjoint of) the Beurling transform (also known as the Beurling-Ahlfors transform). It's also a Calderon-Zygmund operator and a pseudodifferential operator of order 0, so all the usual th …

I guess it depends on whether the objective is (a) to introduce and motivate the Gamma function, and only the Gamma function, in as efficient a manner as possible, or (b) to present some useful mathem …

This is not a serious answer, but one can "prove" the fundamental theorem of algebra by applying the spectral theorem to the matrix $$ A := \begin{pmatrix} 0 & 1 & 0 & \ldots & 0 \\\ 0 & 0 & 1 & \ldo …

Well, real-valued analytic functions are just as rigid as their complex-valued counterparts. The true question is why complex smooth (or complex differentiable) functions are automatically complex an …