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-1 votes
1 answer
236 views

Natural candidates for sub-half-exponential which limit to half-exponential function from below

There are no closed form candidates for half-exponential functions "Closed-form" functions with half-exponential growth. However sub-half-exponentials (functions whose composition grows ...
VS.'s user avatar
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-1 votes
1 answer
208 views

Does this function belong to $L^2(\mathbb{D})$?

Edit: After the answer of Prof. Eremenko to the previous version, I realized that a weaker assumption works for the main motivation of this post. so I revise the question. The unit ...
Ali Taghavi's user avatar
-1 votes
1 answer
369 views

Would this go to 0 [closed]

Let $t_{m}$ be the sup of the sum of the pairwise distances between any $2m$ points in the unit disk. Does $t_{m}/m^{2}$ go to $0$ as $m\rightarrow\infty$?
u51245's user avatar
  • 1
-1 votes
1 answer
204 views

Cauchy reduction formula with measure (a variation)

The Cauchy reduction formula conveniently compresses $n$ integrations of a function $F(x)$ into a single integral. Here I am interested in reducing the following "curved-space" ...
Math2024's user avatar
  • 141
-2 votes
1 answer
314 views

Series representation for $\log(|\zeta(\frac{1}{2}+it)|)$

(Question is short and straight-forward. ) What is/are "nice and non-trivial" series representation/s of $\log(|\zeta(\frac{1}{2}+it)|)$ ?? By "nice and non-trivial" I mean contains no ...
bambi's user avatar
  • 375
-6 votes
1 answer
614 views

Proof of formula for $\pi$ [closed]

The number $\pi$ can be expressed as $\pi=\lim_{n\to\infty} \frac{n\sqrt[n]{-1}-n}{\sqrt{-1}}$ or more poetically $\pi=\frac{\infty\sqrt[\infty]{-1}-\infty}{\sqrt{-1}}$. Here we choose the principal ...
Mikhail Katz's user avatar
  • 16.6k
-6 votes
2 answers
2k views

Is there a transformation or a proof for these integrals?

Here are certain weighted Gaussian integrals I have encountered for which numerical computation reassures equality. Question. Is this true? If so, is there an underlying transformation or just a ...
T. Amdeberhan's user avatar

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