What's the hardest integral you've ever encountered, Veeky Forums?

Pro-tip: if an automatic solver like Matlab's or Mathematica's can do it, it's not a hard integral.

>being this retarded
if you think that Mathematica can give you a closed form solution to that integral you are brain dead.
also, why the fuck are you even comparing Matlab's solver to Mathematica's?
are you fucking 2?

jesus christ this is autistic

But sempai maple beats them all

I once did a triple integral. I'm planning to major in that.

can you guys solve this?

is this one of the cleo ones

Why is physics notation so absolutely abhorrent?

>why the fuck are you even comparing Matlab's solver to Mathematica's?
I am?

4pi arccot sqrt(phi)

Prove it. In a different way to Ron Gordon.

The function is continuous so it exists. I shall call it [math] \vartheta [/math]

The answer is [math] \vartheta [/math].

brilliant

the integral function
you can solve this for any abstract f(x)

Obviously the answer is

[eqn]2\frac{1}{x} \sqrt{\frac{1+x}{1-x}} \ln\left( \frac{2x^2+2x+1}{2x^2-2x+1} \right)[/eqn]

Norm is a blessing and he should be writing the syllabus. Kids would be finding doing complex geometry before the end of primary school.

why must all endeavors serve a practical purpose?
why can't people partake in mathematics simply because they're interested? I mean, you started your shit reply with "Other than pumping up your brain", as if making yourself smarter is a pointless exercise.

let me guess, liberal arts major?

you raging autist

When I took Calculus 2 I had to solve both of these:

[math]\int x \tan x \cdot dx[/math]
[math]\int \tan^{-1} x \cdot dx[/math]

No you can't. f has to be a locally integrable function.

wait till you do renormalization in qcd, those calculations are fucking hard as hell

dimensional regularization is only regularization that doesnt shit all over lorentz symmetry

good job man

Isn't zeta function and dimensional regularisation equivilent?

yeah well come back after uve finished primary school mathematics

>manipulating equations is intrinsic math/physics beauty
i want undergrads to leave

How though? How can you have lorentz symmetry without an integer number of dimensions

i found funny that this question usually not being asked for sports, games or whatever useless in general hobby

Why is the dx written right after the integral and not after whatever the fuck is being integrated?

standard practice for physicists

>computing integrals
>hard

wtf is wrong with you guys?

In physics it is common functions of multiple variables and with many parameters being integrated in only some of them, so it makes sense to know right away what you are integrating.

It's okay, the world needs HR reps and sales guys too.

Why the fuck are you even here?

because physicists like cocks in their ass and this is how they signal it to each other

>any
Tried a non-measurable function, brainlet?
Or a nowhere continuous function, you fucking brainlet?

Are you a brainlet?

Most (Riemann)integrals have no analytic solution and besides numerically solving them there is nothing you can do.

If your notion of integral depends on measure theory, then it isn't the most general notion of integral.

no, i am the opposite of jelly

there is nothing more retarded than people who think big integrals=more smart

Proving that if an operator which maps functions to non-negative functions is weak (1,1) and weak (infinity, infinity), then it is strong (p,p) for p>1. I'll skip the set up and just post the integral computation:
[math]
\|Mf\|_p^p = \int Mf^p\,d\mu = \int_0^\infty \mu(\{x: Mf(x)^p > y)\}dy = \int_0^\infty pt^{p-1}\mu(\{x: Mf(x) > t\})\,dt \leq \int_0^\infty pt^{p-1} \int_{x: |f(x)|>t/2} 2c|f(x)|/t \,dx\,dt
[/math]
[math]
= 2cp \int_0^\infty \int_{x: |f(x)|>t/2} |f(x)|t^{p-2}\,dx\,dt = 2cp \int_{\mathbb{R}} \int_0^{2|f(x)|} |f(x)|t^{p-2}\,dt\,dx = \int_{\mathbb{R}} |f(x)| \frac{2^{p-1}|f(x)|^{p-1}}{p-1}\,dx = \frac{2^p cp}{p-1} \int_{\mathbb{R}} |f(x)|^p \,dx = \frac{2^p cp}{p-1}\|f\|_p^p.
[/math]
So taking the p-th root of both sides gives the strong (p,p) bound.

>why do something if your employer doesn't require it