@Hippalectryon Poles at z = -1 and 1, with a clear sense of the constant vertical straightlines seen, clearly an immersion of the geodesics of a hyperbolic plane. Let me guess, is that a modular funtion?

I've got another quick integration question for which I think I have a solid line of reasoning. $$ \int x^2 \sqrt{9-x^2} \text{ d}x \overset{x=3\sin\theta}= \int 9^2 \sin^2 \theta |\cos \theta| \cos \theta \text{ d}\theta $$ For $\theta \in \left[ \frac{-\pi}{2} + 2n\pi, \frac{\pi}{2} + 2n\pi \right]$ our integral becomes $$ \int 9^2 \sin^2 \theta \cos \theta \cos \theta \text{ d}\theta$$

For $\theta \in \left[ \frac{\pi}{2} + 2n\pi, \frac{3\pi}{2} + 2n\pi \right]$ our integral becomes $$ -\int 9^2 \sin^2 \theta \cos \theta \cos \theta \text{ d}\theta$$

That's right, right?

I am not concerned about domains.

You're not?

No. I am not concerned about integration at all.

Also, I forgot to mention the obvious; $n \in \mathbb{Z}$.

Oh, I see. Well I am and I think it's important.

@Khallil Care to tell @TedShifrin to unignore me?

He is doing that for months.

I mean, c'mon, I am not THAT annoying!

I already have, @BalarkaSen. Me telling Ted won't really make a difference as I've already relayed the message.

Sigh.

I just realised that I could've implemented the signum function in my integral.

:O

@BalarkaSen I think @TedShifrin is annoyed at you because you pass a lot of things that other people think are interesting off as trivial and useless. I guess if you have nothing nice to say don't say it at all.

@Khallil Do you still need any integration help? I am here for you

Hi everyone

I'm practicing math as a hobby, and I'm stuck on some 'numerological' problem regarding prime number generation

Might I interest somebody for a little discussion?

I ended up getting it anyway, @Alizter!

Say, for large enough $n$ and small enough $x>0$, $\frac{(1+x)n^{3/4}}{n^{5/6}}=o(1)$, right?

The integral was only valid for $|x| \leqslant 3$, so my $\text{sgn} \left( \sqrt{ 1 - \frac{x^2}{9} } \right)$ multiplier ended up being equal to $1$. ^_^

My current algorithm generates the first 680 prime numbers in well under a second

It's about time I bowed out. Thanks for all your help everyone. Peace out!

Hello everyone. I am in need of dire help at math.stackexchange.com/questions/922843/…. I cannot show anyhow that when a function f is Darboux integrable then it is Riemann integrable somehow...

I take what I said earlier, back.

@Khallil brilliant.org?

I need to show that if $y=x^2$, then $\displaystyle \int_{0}^{1} y \text{ d}x + \int_{0}^{1} x \text{ d}y = 1$ by considering a sketch of the integrals.

@Khallil That is easy

I've got a sketch, but my main problem lies with the second integral in that inequality.

$x=\pm \sqrt y$

and you only need +

because 0 to 1

Remember that the graph of an inverse of a function is a reflection across y=x

@UfukCanBiçici Kitap yok mu?

Oh I see.

This doesn't really have much to do with an inverse function does it?

Yada vikidan oreniyormusun?

@Khallil Think x^2 has a slice of the unit square

Yea, I got that with the sketch.

and $\sqrt y$ has the other slice

I've got it now.

so join them together they are the unit square

I think I have a problem with switching the variables of integration.

@UfukCanBiçici Yarin cevap gelir artik

How would $\displaystyle \int x \text{ d}y$ look on the $x-y$ plane?

(With $y=x^2$)

turn your head sideways

it would look like a parabola fell over

Yep, but when integrating from $y=0$ to $1$, wouldn't it look like the real line has been inverted so larger numbers are on the left and smaller numbers are on the right, @Alizter?

@BalarkaSen Nope

(I turned my head 90 degrees clockwise.)

@Khallil your integrating from 0 to 1

so it would need to fluip

smaller to larger

what sarah said

So you just flip it when you can, to avoid these discrepancies?

@Khallil

Simple geometry

:)

Yea, got it.

That's what I've had all along.

@BalarkaSen $\cos(z)^{\sin(z)}$

Why is $\displaystyle \int_{-a}^{a} \sin^n x \text{ d}x = 0$, where $n$ is an odd integer?

@Hippalectryon What about it?

Is it because the integrand is an odd function?

@Alizter Fish function

@Khallil yes.

Cool!

$-\sin x= \sin -x$

so raising it to an odd power will preserve negative

@Alizter Wikide denk geldiğim spesifik bir şeydi oradaki kanıtı bir türlü kendim gerçekleştiremedim. Ya birşeyi yanlış anladım ya da birşeyi gözden kaçırıyorum ama nedir...

@UfukCanBiçici It sounds like a very specialist subject you are interested in and many people are sleeping. You may have attention by tomorrow (too tired to write turkish sorry)

Hi Professor @TedShifrin