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2d
comment Two integrals with bounds
The integral you give does not converge.
2d
comment Hyperbola question
cough You mean parabola
2d
comment Is there something special about 2015?
@NikolajK There is a reason I put the word 'prove' in quotation marks.
2d
comment Is there something special about 2015?
You can "prove" by contradiction that there are no uninteresting numbers using the well-ordering property. For if uninteresting numbers exist, then there must be a smallest uninteresting number. But that would be a very interesting number! Contradiction. =)
2d
comment How to calculate the two integrals?
@FDP After a great deal work, I've found that the first integral has the simple closed form, $I=\frac{5\sqrt{2}\,\pi^3}{192}-\frac{7\sqrt{2}\,\pi}{32}+\frac{5\sqrt{2}\,\pi \ln{(2)}}{16}-\frac{\sqrt{2}\,\pi\ln^2{(2)}}{16}$. But gee whiz it was unexpectedly exhausting =o
Jan
23
revised How to calculate the two integrals?
Expanded repsonse
Jan
22
answered How to calculate the two integrals?
Jan
20
awarded  Populist
Jan
18
comment Why is $(f(x))'$ shortened $f'(x)$
This certainly isn't the only example of poorly thought out basic calculus notation that nevertheless continues to be taught in introductory courses. We really ought to fix this one of these days.
Jan
18
comment Hypergeometric 2F1 with negative c
@balping Well whether or not you need it, I realized there is a much better way to evaluate the sum, and I felt compelled to post it because my first answer was really unenlightening. Cheers =)
Jan
18
revised Hypergeometric 2F1 with negative c
added 1742 characters in body
Jan
17
comment Proving sine of sum identity for all angles
What is your definition of sine and cosine?
Jan
17
comment Hypergeometric 2F1 with negative c
@balping Remember that by the symmetry of binomial coefficients, we have: $\binom{a+n-1}{n}=\binom{a+n-1}{(a+n-1)-n}=\binom{a+n-1}{a-1}$. Does that solve your problem?
Jan
17
answered Hypergeometric 2F1 with negative c
Jan
17
answered Elliptic integral evaluation
Jan
17
comment Hypergeometric 2F1 with negative c
Ah, I didn't notice that $a$ is also an integer. I see your issue now. One solution is to let $a$ be an integer plus $\epsilon$ and then take the limit $\epsilon\to 0$ at the end of the problem.
Jan
17
comment $\lim_{n \to \infty}n^2\int_{1}^{\infty} \frac{cos(x/n)-1}{x^4}dx$
Why exactly can't you compute the limit as $k\to\infty$? Have you tried distributing the limit over the sum and evaluating them individually?
Jan
17
comment Elliptic integral evaluation
Do we know that $r_b>0$?
Jan
17
comment Green's Theorem; computing a double integral
How odd! The etymological history of math and science related terminology kind of a hobby of mine. Apparently the 'astroid' derives from the ancient Greek 'ástron+eîdos', whereas 'asteroid' derives from the ancient Greek 'astḗr+eîdos'. I guess this is the words are spelled differently in English. But apparently these two Greek roots were actually synonyms in the ancient Greek language, both meaning "star". Maybe this accounts for the lack of spelling difference in other languages such as Finnish. Cheers =)
Jan
17
comment Green's Theorem; computing a double integral
I believe you meant astroid, not to be confused with asteroids which are chunks of rocks flying around in space. ;)