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9 votes
2 answers
2k views

Why does this theta function value yield such a good Riemann sum approximation?

Let $T(q)$ denote the third Jacobi theta function at $z=0$; i.e., $$T(q) := \sum_{n\in\mathbb{Z}} q^{n^2}.$$ Then for any $x>0$, $T(\exp(-1/x)) \approx \sqrt{\pi x}$. For example, as the entry for ...
Timothy Chow's user avatar
  • 82.7k
1 vote
0 answers
102 views

Conjectured closed form of $\int_0^1\frac{\ln^3(1+x)\,\ln^3x}x\mathrm{d}x$

I posted this question on Math Stack Exchange, but there were no helpful comments or answers https://math.stackexchange.com/q/4874446/1298448 How to integrate $${\displaystyle \int_0^1\frac{\ln^3(1+x)\...
Martin.s's user avatar
  • 224
7 votes
0 answers
299 views

An integral for the tribonacci constant and the general case

When I asked for integrals involving the tribonacci constant $T$, user @nospoon gave the nice answer, $$\int_0^{\infty} \eta( i \, x)\,\eta(i \,11 x) \,dx = \frac{ \ln T}{\sqrt{11}} $$ However, the ...
Tito Piezas III's user avatar
13 votes
1 answer
812 views

Summation of series involving $\sinh$ of a square root

Consider the following series: $$ S = \sum_{\text{odd } n} \sum_{\text{odd } m} \frac{(-1)^{(n+m)/2}}{nm} \frac{\sinh( \pi \sqrt{n^2 + m^2}/2)}{\sinh( \pi \sqrt{n^2 + m^2})} $$ From the physical ...
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