The output of this integral seems irrational, but how would you go about...

Playboyize
Playboyize

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

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en.wikipedia.org/wiki/Proof_by_contradiction

Methshot
Methshot

The output of this integral seems irrational, but how would you go about proving it?
en.wikipedia.org/wiki/Proof_by_contradiction

Evilember
Evilember

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

SniperWish
SniperWish

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?

PurpleCharger
PurpleCharger

No one cares about exact values for integrals. Analysts are usually happy with just knowing it's finite.

Nojokur
Nojokur

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?

GoogleCat
GoogleCat

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

Evil_kitten
Evil_kitten

Estimate it using a series expansion, then show it is close to a rational (within 1/q^2 of p/q)

hairygrape
hairygrape

Taylor series. Approximate the function with a polynomial and then calc the area of the polynomial. Approximation gets better and better with more terms.

Gigastrength
Gigastrength

It's probably impossible. Just look at how unnatural that integrand is.

Techpill
Techpill

How about setting z:=exp(ix) and play around with complex integrals?

BlogWobbles
BlogWobbles

compute infinity many digits then check for repetition

kizzmybutt
kizzmybutt

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.

Skullbone
Skullbone

Shit. I placed the wrong slash.

[Math]\int x^{cos(x)} = \phi(x) [/math]

Carnalpleasure
Carnalpleasure

You wrote "Math" instead of "math".
[math]\int x^{cos(x)} = \phi(x)[/math]

Gigastrength
Gigastrength

All I can say is that the function is tough to work with and the expansion converges amazingly badly

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happy_sad
happy_sad

Ill off myself its no problem

Deadlyinx
Deadlyinx

Approximation gets better and better with more terms.
Only if the function is analytic

FastChef
FastChef

Why do you even care? -Physicist

viagrandad
viagrandad

it's irrational, you can tell easily from the series

viagrandad
viagrandad

it's not an algebraic antiderivative so its automatically transcendental this irrational (basic algebra)

SniperWish
SniperWish

it's not an algebraic antiderivative so its automatically transcendental this irrational (basic algebra)
elaborate

Bidwell
Bidwell

it's irrational, you can tell easily from the series
elaborate

eGremlin
eGremlin

Actually beginning from the very next decimal place it repeats itself.

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