Double checked this - did it by converting it to an integral of an infinite sum, converted that to an infinite sum of integrals, integrated by parts three times to get the integral resolved, then found the result of the remaining infinte sum (there was a prize given by a professor who challenged attendees at a seminar to prove the result given in the integral table - and shared the prize with my math major friend who could make the proof rigorous
If you have a result in an integral table, surely the easy way to prove it is to just differentiate what’s in the table?
As I should have said, the table had only the definite integral from 0 to infinity.
I believe that the modern description of Rodrigues’ formula is the following. Start by representing every point on the 2-sphere centered at the origin by a pure quaternion (no real part) of norm 1. So the point (a,b,c) is
(ai+bj+ck)/\sqrt{a^2+b^2+c^2}. Then to rotate by an angle \theta in degrees with axis at z, you conjugage by z^{\theta/180}. So for example, if you want to rotate 90 degrees around the vertical axis, represented by k, you conjugate by \sqrt{k}=(1+k)/\sqrt{2} whose inverse is (1-k)/\sqrt{2}. I leave to you to calculate that \frac{1+k}{\sqrt{2}}i\frac{1-k}{\sqrt{2}}=j and \frac{1+k}{\sqrt{2}}j\frac{1-k}{\sqrt{2}}=-i.
I definitely maybe remember some Rodrigues-attributed formulae related to recursively calculating orthogonal polynomials. For example, for Legendre-type polynomials you get P_n(x) = 1/(2^nn!){\bigl(\frac{d}{dx}\bigr)}^n(x^2-1)^n.
If you don’t want your answer to this to have explicit infinite sums in it, don’t you need to write it in terms of polylogarithms or something? Maybe there is a reason why the table of integrals only included the definite integral.
The final answer was the satisfying simple (pi^4)/15. If the integral hadn’t been from zero to infinity, there probably wouldn’t be a closed form solution
Similarly to how the erf integral can be calculated from -infinity to infinity, but there’s no closed form for erf(x) in general.
I just ran across this Numberphile video which discusses the expression
(1+9^{{-4}^{6\cdot 7}})^{{3^{2}}^{85}}
It’s interesting because, first, it uses every digit between 1 and 9 exactly once, which isn’t that remarkable, but second, it is an incredibly accurate approximation of e. It is correct to 1.8 \times 10^{25} decimal places!
The reason isn’t hard to understand though, once you realize that 9^{{4}^{6\cdot 7}} is huge and is actually equal to {3^{2}}^{85}. Since e = \lim_{n \to \infty} (1+\frac{1}{n})^n, taking (1+n^{-1})^n and substituting a huge number for n produces a value very close to e.
I didn’t watch the video but I did see the formula elsewhere. My first reaction was “wait, that’s crazy”. And then thinking on it more, I realized the same thing you did, and thought “ok, not so crazy, actually kind of mundane really.” But then I noticed the fact that it had the digits 1-9 once each and was back to thinking it was clever.
Here’s an easy proof of a way to simplify a wide swath of integrals:
We don’t need no steenking denominators. Only condition is that g(x) is even and f() is odd.
Oops. I initially read the (6*7) term as (6-7). Too small a font size I guess.
You get a very different and nonsensical outcome if you think that’s a subtraction, not a multiplication. Which at first, ref the PE(r)MDAS thread, sent me off in the direction of an order of operations error.
I’m happy with 2.718281828459045. Easy to recall and 16 digits. There is no such easy sequence for pi. Too bad there is no e day since no month has 71 days.
Was the concept of e (regardless of the name then used) known on Feb 7th, 1828? Or on July 2nd 1828?
We certainly could have chosen to celebrate both dates 7 years ago in '18.
At some point I memorized pi up to 3.14159265358979. Not quite enough for 64-bit float precision, but close. There are some patterns that make it easier:
3. 14-15 92 65-35 89-79
The next digits are 32-38. Not too bad, pattern-wise.
Now, I have a rhyme assisting my feeble brain its tasks resisting.
Only 12 digits, but it helps.
There’s always: “How I want a drink, alcoholic of course, after the heavy lectures involving quantum mechanics”.
Yes, the number itself was known to be significant by the late 1600s. Euler gave it the name e around 1730. So it was well known by 1828. I don’t know if the modern practice of identifying special days with the concatenation of the digits of the date existed back then; I kind of doubt it.
I know pi to the exact same number of digits as Dr.Strangelove mentioned, and I know e to the exact same number of digits as Hari Seldon mentioned. I guess in both cases it’s an convenient stopping point.