The output of this integral seems irrational, but how would you go about proving it? I can't even integrate it without using computer assisted estimations
The output of this integral seems irrational, but how would you go about proving it...
Liam Cox
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Jonathan Rodriguez
>The output of this integral seems irrational, but how would you go about proving it?
en.wikipedia.org
Parker Murphy
Hard to do that when you can't even represent the integral exactly in reasonable terms.
Integral equals some integers p/q but at a certain point you need to concretely evaluate the integral, which is resisting all of my efforts
Josiah Hill
I don't really give a fuck about pure math so I always approx my integrals with a computer. What do mathematicians even do when then run into a function that has no closed-form anti-derivative?
Christian Nelson
No one cares about exact values for integrals. Analysts are usually happy with just knowing it's finite.
David Sanders
>The output of this integral seems irrational, but how would you go about proving it?
I don't know, but I guess that the proof isn't simple, at least on the first glance.
Maybe you had a better chance if you could express it in its closed form?
Michael Brooks
Usually you'd do it by first finding a bounds for the solution (let's call it I) that's between two fractions. Then you'd show that for every such interval there's a smaller interval between two fractions for where I lies, so I cannot be a fraction.
For example I is in (a/b, c/d) => I is in (a2/b2, c2/d2), where a2/c2 > a/c and c2/d2 < c/d.
It's been a long time since I done this, but I remember proving irrationality for integrals being somewhat easy for first year... I could be completely wrong. Good luck
Joseph Foster
Estimate it using a series expansion, then show it is close to a rational (within 1/q^2 of p/q)
Isaiah Evans
Taylor series. Approximate the function with a polynomial and then calc the area of the polynomial. Approximation gets better and better with more terms.
Brayden Lee
It's probably impossible. Just look at how unnatural that integrand is.
Mason Miller
How about setting z:=exp(ix) and play around with complex integrals?
Xavier Gonzalez
compute infinity many digits then check for repetition
Dominic James
This is actually really interesting.
Have you tried evaluation using one of the sine integral functions? Or simply naming a function to be its integral?
Something like
[Math]\int x^{cos(x)} = \phi(x) [\math] and then based on that you analyse its properties.
Then a proof by contradiction could work but im not a mathematitian.
Lincoln Jackson
Shit. I placed the wrong slash.
[Math]\int x^{cos(x)} = \phi(x) [/math]
Brandon Stewart
You wrote "Math" instead of "math".
[math]\int x^{cos(x)} = \phi(x)[/math]
Jackson Mitchell
All I can say is that the function is tough to work with and the expansion converges amazingly badly
William Ward
Ill off myself its no problem
Matthew Smith
>Approximation gets better and better with more terms.
Only if the function is analytic
Austin Watson
Why do you even care? -Physicist
Andrew Hill
it's irrational, you can tell easily from the series
Dylan Reed
it's not an algebraic antiderivative so its automatically transcendental this irrational (basic algebra)
Juan Lopez
>it's not an algebraic antiderivative so its automatically transcendental this irrational (basic algebra)
elaborate
Nicholas Bell
>it's irrational, you can tell easily from the series
elaborate
Alexander Robinson
Actually beginning from the very next decimal place it repeats itself.