Search found 56 matches
Re: A series
This one can be done pretty slick by using the residues. Consider: $$f(z)=\frac{\pi \cot(\pi z) z^{2}}{(4z^{2}-1)^{2}}$$ Note that $$\sum_{-\infty}^{\infty}f(n)=2\sum_{n=1}^{\infty}f(n)$$ $$\lim_{z\to \pm 1/2}\frac{d}{dz}\left[\frac{\pi \cot(\pi z)z^{2}}{4(2z\pm 1)^{2}}\right]=\frac{-\pi^{2}}{64}$$ ...
Re: Series
Good job with digamma-related approach, R. I will give it a go another way with residues. . Well, it still uses $\pi \cot(\pi z)$ like RD done. Using the $$\sum_{-\infty}^{\infty}f(z)=-\text{sum of residues of} \cot(\pi z) \text{at the poles of f(z)}$$ $$\sum_{k=-\infty}^{\infty}f(k)=\sum_{k=-\infty...
- Wed Dec 23, 2015 9:18 am
- Forum: Calculus
- Topic: tough Gaussian/Poisson-like integral
- Replies: 2
- Views: 2273
Re: tough Gaussian/Poisson-like integral
Good start, T.
Let's stew on it and see what we come up with.
But, I think it'll involve $erf$.
Let's stew on it and see what we come up with.
But, I think it'll involve $erf$.
- Tue Dec 22, 2015 4:12 pm
- Forum: Calculus
- Topic: tough Gaussian/Poisson-like integral
- Replies: 2
- Views: 2273
tough Gaussian/Poisson-like integral
Here is a Poisson-like integral I ran across that has gotten no bites on other forums. I reckon this means it is not doable?. $$\int_{0}^{\infty}x^{2}e^{-x^{2}+x}\log(1-2a\cos(x)+a^{2})dx$$ I ran it through Maple and Wolfram, but neither would return anything. Maple returned numerical answers for sp...
- Sat Dec 19, 2015 7:14 pm
- Forum: Calculus
- Topic: my first post and a cool and challenging integral
- Replies: 3
- Views: 2956
Re: my first post and a cool and challenging integral
Clever and consise, RD.
- Thu Dec 17, 2015 4:38 pm
- Forum: Calculus
- Topic: my first post and a cool and challenging integral
- Replies: 3
- Views: 2956
my first post and a cool and challenging integral
Evaluate:
$$\int_{0}^{\infty}\frac{(1-\sin(ax))(1-\cos(bx))}{x^{2}}dx$$
$$\int_{0}^{\infty}\frac{(1-\sin(ax))(1-\cos(bx))}{x^{2}}dx$$