An implication is that conditional density functions are not invariant under coordinate transformation of the conditioning variable. Well, duh!

Well, I was not aware of that.

@HenningMakholm Oops, sorry, I didn't mean to be so awful :-(

@robjohn - what's coördinate transformation?

I suppose one has to keep up the meanness from time to time. :-)

@HenningMakholm I have just had many problems with that very thing, that I guess it seems very obvious to me.

Things tend to feel obvious when one gets sufficiently used to them.

Henning looks so manly.

I spend a lot of time getting the densities right because I have seen how messed up the probabilities get when transformed.

Hmm, I drank too much.

Sorry, I don't swing that way.

what a thread killer ;-)

Have to point this one out: chat.stackexchange.com/transcript/36?m=2351906#2351906

@Srivatsan See my link.

@HenningMakholm If QED weren't ignoring you, you'd hear the following quote by JvN: "Young man, in mathematics you don't understand things. You just get used to them."

Well, I didn't mean that...

Neumann!

Perhaps the alcohol has damaged more of Jonas' Y chromosomes than X...

@JonasTeuwen It wasn't at all clear what you meant, so we're sort of guessing. =)

I'm not sure what it meant but it certainly didn't mean that!

@tb Is he now? My experience from Usenet is the a very small fraction of killfiling announcements actually correspond to actual killfiling.

Hello, are questions about definite integrals on-topic?

Here or on the main site?

@JonasTeuwen It didn't by any chance mean that an older sister of yours momentarily forced you away from the keyboard, did it?

The burnt rubber has been replaced by apricots.

@HenningMakholm I don't have any sisters.

Damn.

@JonasTeuwen alcohol tastes better with consumption.

All tag descriptions are indefinite integrals .. Questions on the evaluation of definite and indefinite integrals

@HenningMakholm I don't know.

@Srivatsan On the main site

@Gigili Sure, go ahead!

Uhum, thank you.

@HenningMakholm Yes?

@tb I see, but "Questions on the evaluation of definite and indefinite integrals"

Oh, I'm blind, just saw the definite word

@Gigili Sorry, I overlooked that you already saw it. But out of curiosity, what's the integral you wanted to ask about?

@Gigili So we were both blind :)

:D

@tb First I have to find out how to type an integral here

Just like this: int

For limits: $ \int_{a}^{b} f(x) dx $

$ int x dx / sinx^2

What's up with the unnecessary brackets?

Um, like that? @JonasTeuwen

@JonasTeuwen STFU.

@JonasTeuwen Don't worry, that'll be taken care of. =)

@Gigili: Substitute u=x^2?

STFU? Hmm.

We're talking LaTeX here.

(Shut the \cdots up)

Are we solving the question here or Gigili going to post it on the main site?

@HenningMakholm The whole sinx raised to the power of 2

@HenningMakholm F\ast\ast\ast looks good also :)

@robjohn Exactly

@Srivatsan e\cho\cho\cho :)

echo? What does that mean, tb?

@Srivatsan what it means

@tb Oh, cool :-)

tb: I got something, I'll pretend that's what you meant. :)

@Gigili Did you try plain integration by parts? You probably know how to integrate 1/(sin x)^2?

@tb - cot(x) ?

@Gigili Exactly. And how does the integral look after integrating by parts?

I don't get what do you mean by "integrating by parts", do you mean something like substitution, udv?

@Gigili I mean this

Aha, I knew it, right .. Let me try to solve it that way then

@robjohn Fantastic, and the second part is ln (sinx) ?

I used to write it as ln |sin x|, but yes, you're right.

Yes, you're right.

Is it taught as C-B-S in Europe? math.stackexchange.com/questions/84361/… // As opposed to just C-S.

Oops, forgot some backslashes

I learned it as simply C-S.

Okay, I gotta go. Night everyone.

Good night, Henning.

@HenningMakholm nighty-night

@Srivatsan I learned it as Cauchy-Schwarz, too.

@tb without the 't' ;-)

And I think robjohn said the same thing as well: about C-S vs. C-B-S.

Problem solved, you mathematicians are totally great. I was about to bang my head against the wall.

@Gigili Don't do that, please, the wall's too precious :)

Hih =)

@Srivatsan Indeed, I learned it as Cauchy-Schwarz

Ok. Me too. Except for one stray Russian problem text I used to read.

That's how I know the existence of the B.

Having a math problem is something, typing a question is another issue you need to overcome or so, which makes your math problem a side issue.

@Srivatsan Well, of course, they are going to insert a Russian name in the middle and hope no one notices until it is too late :-D

@Srivatsan Actually, in a later course it was called C-S-B by one lecturer and he elaborated on the reasons, but I don't remember why, exactly.

One story is that C proved it for finite sums, B extended it to integrals, and S generalized C's proof to cover general inner product spaces. In particular, Schwarz (I am going to switch to writing full names =)) gave the proof using the quadratic form (x - lambda y)^T (x - lambda y).

How can I check if I type it correctly?

$\int_{\pi/3}^{\pi/4}\frac{\x}{\sin^2 x}dx$?

Sorry, that should read <x-lambda y, x-lambda y>?

I am considering the real case only. Complex case involves a tweak.

@Srivatsan Yeah, but there was some strange reason for the order of letters. Anyway, it's certainly explained in detail in Steele's book.

@tb Oh that I am not aware of.

@Srivatsan I think it was an idiosyncrasy.

I used to know it as CSB too. For a long time, I didn't know what B meant. In my head, I expanded it to "Badger" just so that I had some pronunciation. =)

Oh, \exp^{tx}, thats news to me :)

@tb clumsy and defeats the purpose of exp. Nice :-)

@robjohn I'm sorry =\ , I don't know what to substitute x for? pi/3 or pi/4?

@Gigili in the numerator there's a backslash too much

Wikipedia article has no definite integral exampel

try $\int_{\pi/3}^{\pi/4}\frac{x}{\sin^2 x}dx$

@Gigili You sub pi/3 then subtract the same at pi/4

@tb Where can I see the result?

@robjohn I know, same for the first part?

-xcot(x) has no integral

@tb Found it, thanks a lot

Plug in x = \pi/4 then subtract the same with x=\pi/3

For both parts?

the whole function

@tb He writes +infinity as ${}^+ \infty$. :)

(Phew! Done with that comment, finally... =))

ok, dinner time. Bye all.

one minute left to today on MSE.

@Srivatsan enjoy dinner.

Bye Srivatsan