Mathematics - 2016-11-09 (page 1 of 5)

Maks

00:00

@DHMO ok, then the answer wolfram gives me when it solves the integral is the convergence point

Thz

DHMO

@saturatedexpo still no idea

you are welcome

saturatedexpo

@DHMO only one missing element would be easier imo

but still hard for me haha

DHMO

@saturatedexpo how?

saturatedexpo

f(x)=0.5 if x=0
f(x)=0.5 + 0.5x if x in Q without 0
f(x)=0.5 -0.5x if x in R\Q

DHMO

do we really have to use dirichlet?

what is f^-1(0.25)?

saturatedexpo

00:05

ouch

0.5?

mmh

too disturbing haha

@DHMO would the problem be the same with [0,1] and ]2,10[ ?

Adeek

@BalarkaSen here ?

NaCl

00:37

@Ramanujan 1729 <3

Kasmir Khaan

Hello guys i need some help

I have posted a question with no good answers so far

I really need someone to explain to me so i can finish the rest of my homework

anyone can help me ?

NaCl

can't help ya if you don't post the actual question

saturatedexpo

@NaCl hi!

NaCl

too lazy to go to your profile

Kasmir Khaan

5

Q: Double improper integral , how to see if it diverge

$$\iint_D \frac{(x+y) e^{y-x}}{x^2 y^2}dx \, dy$$ $$D= \{(x,y) ; 0\leq y+1\leq x , xy\geq 1 \}$$ Iv been stuck on this for past two hours , I need some hint . My bounds are : $\frac{1+\sqrt{5}}{2}\leq X<\infty $ $\frac 1 x \leq Y\leq x-1$ are the bounds correct ? I need some hints, ...

improper-integrals

NaCl

00:38

@saturatedexpo yo

Kasmir Khaan

do you see it ?

saturatedexpo

yes

NaCl

ok that's above my current level

Kasmir Khaan

okay thanks for trying

I have 7 other of same type if only i can crack one

NaCl

there are more people here than me ;)

Kasmir Khaan

00:40

how do i make it get more attention ?

NaCl

bounty

Kasmir Khaan

too many ppl posting its not on the top anymore

it sais i have to wait 20 hours

NaCl

thats how things work here

Kasmir Khaan

indeed thanks

you have been very helpfull thanks

saturatedexpo

if you want i bounty it lol

Kasmir Khaan

00:44

i would be very glad if you could

NaCl

@KasmirKhaan sorry

saturatedexpo

but isnt it answered already?

Kasmir Khaan

nothing to be sorry about

it is but its not correct

the double integral is divergent but he claims to be convergent

saturatedexpo

where is the bounty button? :D

Kasmir Khaan

it sais i should wait 2 days to be able to bounty

1 day few hours have passed

saturatedexpo

00:46

ah, i dont see anything, so i cant

NaCl

@saturatedexpo you can in 20 hours

Kasmir Khaan

its okay mate , ill have to wait untill tomorrow

We are not even doing real analysis and they give us such questions

NaCl

Today I heard about a cool problem. Take two integers $a, b$. Let $c=\frac{a}{b}$ such that $c\in\mathbb{Q}$. Now let n\in\mathbb{N} be fixed and $0\le d_1,d_2,d_3,\cdots,d_n\le 9$, such that $d_1,d_2,d_3,\cdots,d_n\in\mathbb{N}$ denote digits of some number $c$. Given all $d_1,d_2,d_3,\cdots,d_n$ prove that there exist infinitely many pairs $(a,b)$ such that $\frac{a}{b}$ results in a number, where the given digit sequence $d_1,d_2,d_3,\cdots,d_n$ will appear somewhere at least once.

Kasmir Khaan

we cant see the problem

well we do but in a messed up way , unreadable

NaCl

tinyurl.com/cfqcvpc

Mees de Vries

00:52

Just take $a$ some number with the given digit sequence and $b = 1$.

NaCl

and then?

I forgot: $b$ mustn't divide $a$

Mees de Vries

Maybe I don't understand the problem.

NaCl

also if you just give one pair, that's no proof that there exist infinitely many such pairs

Mees de Vries

I gave infinitely many, because $a$ was picked from an infinite set...

Kasmir Khaan

6

Q: Double improper integral , how to see if it diverge

$$\iint_D \frac{(x+y) e^{y-x}}{x^2 y^2}dx \, dy$$ $$D= \{(x,y) ; 0\leq y+1\leq x , xy\geq 1 \}$$ Iv been stuck on this for past two hours , I need some hint . My bounds are : $\frac{1+\sqrt{5}}{2}\leq X<\infty $ $\frac 1 x \leq Y\leq x-1$ are the bounds correct ? I need some hints, ...

improper-integrals

Can anyone post something here so my question goes back on the top of list ?

NaCl

00:56

@MeesdeVries ok you misunderstand.

Mees de Vries

Anyway, if you do not want $b \mid a$, then pick $a$ any number containing the given digit sequence and let $b$ be a string of the same length consisting only of 9s.

saturatedexpo

@KasmirKhaan posting doesnt "bump" i believe

Mees de Vries

Can you rephrase so I do?

NaCl

I'll try partially

Kasmir Khaan

I think i should just give up :D sorry for this

NaCl

00:57

the sequence of digits is given prior to the proof, then you're supposed to prove that there exist infinitely many $a, b$

that fulfill the condition

Kasmir Khaan

I hope tomorrow i would get help

I would give all my points as bounty

NaCl

o, good catch

Mees de Vries

I know. So fix your sequence of digits. Then take any $a$ containing the digit sequence (that's already infinitely many) and take $b$ the string of 9s such that $b$ and $a$ have the same length.

NaCl

@MeesdeVries that actually sounds good

Then you can just append for $a$ a zero and for $b$ a $9$ again

and so on and so forth

Mees de Vries

technically this does not work if the digit sequence is all 9s, in which case you require $a$ to always contain at least one non-nine digit in addition to the digit sequence.

Yes... that's a special case of what I said, yeah.

NaCl

00:59

true

Ramanujan

@NaCl 1729 means?

saturatedexpo

salt

NaCl

@Ramanujan you call yourself "Ramanujan" and don't know about 1729?

I'm disappointed

@saturatedexpo sup bro

Ramanujan

OK,iknow

Know igot

saturatedexpo

@NaCl 1729 i dont know tho, maybe the year of chemestry?

Ramanujan

01:03

Special no which can be written as sum of cubes of 2 smallest no.

NaCl

@saturatedexpo First taxicab-number

Ramanujan

In two different ways

saturatedexpo

lol, this seems arbitary^^

NaCl

and has some nice properties as well. e.g. 1729 = 19 * 91

1+7+2+9=19

Ramanujan

1^3 + 12^3 and 9^3 + 10^3

NaCl

01:04

ye

saturatedexpo

so a very saltinated taxi

yum yum

NaCl

oh and as far as I remember it always divides any natural number $x^{1729}-x$

Ramanujan

@NaCl water Molecule will disovle you,be careful

NaCl

@Ramanujan I am, thank you. :)

saturatedexpo

@NaCl year 1729: "August 1 – The Comet of 1729, possibly the largest comet, with the highest apparent magnitude, on record, is discovered by Fr. Nicolas Sarrabat, a professor of mathematics at Marseille."

"a professor of mathematics"

"comet"

brilliant :D

NaCl

01:07

wow

That is awesome as hell

Ramanujan

2^(74,207,281) − 1

Who knows?

saturatedexpo

some big number

Ramanujan

Without googling

saturatedexpo

biggest prime?

(known)

Ramanujan

Right

Kasmir Khaan

01:10

good save

:)

saturatedexpo

i think i googled that last night

but of course i cant remember the number haha

NaCl

wat

honestly

Ramanujan

I too!:D

NaCl

Wasn't it bigger?

Ramanujan

I too copied

saturatedexpo

01:11

yes something like 10 digits in the exponent? i look it haha

NaCl

No, I looked it up just now

saturatedexpo

wow

NaCl

primes.utm.edu/largest.html

saturatedexpo

that's kinda small isnt it?

NaCl

I think the guys from numberphile printed this as book

Ramanujan

01:12

Yes

2-3 book

NaCl

New World's Biggest Prime Number (PRINTED FULLY ON PAPER) - Numberphile

►

saturatedexpo

haha, one misprinted digit and its worthless^^

NaCl

that's just mad

Ramanujan

I watched before:D

saturatedexpo

i have a good one:

Ramanujan

01:13

@saturatedexpo I can't imagine computer printer make mistakes

Mees de Vries

Oh boy, have you never used a printer before?

saturatedexpo

@Ramanujan they do, ESPECIALLY with numbers

8=0, 9=0, ect

happens all the time

Ramanujan

OK,i did not used printers

Mees de Vries

I have this printout of a homework assignment somewhere, a tex-based PDF, where the printer shifted every unicode character by 1.

saturatedexpo

well, i don't know if they fixed it

but it made many many buiseness year calculations practically worthless

NaCl

01:15

depends on the type of printer you use

anyways, おやすみなさい！it's like 2AM here lmfao

need to get up at 6am

Ramanujan

不也

saturatedexpo

@NaCl i need to get up at ~9

Ramanujan

It's morning in India!

Maks

10pm here on Argentina :D

saturatedexpo

@MeesdeVries sounds fun :D (was your excercise to get exactly that?)

saturatedexpo

01:38

Does $G=\mathbb{R}/ \mathbb{Z}$ mean that G is partitioned?

Kasmir Khaan

01:51

6

Q: Double improper integral , how to see if it diverge

$$\iint_D \frac{(x+y) e^{y-x}}{x^2 y^2}dx \, dy$$ $$D= \{(x,y) ; 0\leq y+1\leq x , xy\geq 1 \}$$ Iv been stuck on this for past two hours , I need some hint . My bounds are : $\frac{1+\sqrt{5}}{2}\leq X<\infty $ $\frac 1 x \leq Y\leq x-1$ are the bounds correct ? I need some hints, ...

improper-integrals

Akiva Weinberger

02:30

A nice alternate proof of V-E+F=2, with animation: youtu.be/-9OUyo8NFZg

I like the black background aesthetic

saturatedexpo

@DHMO there?

saturatedexpo

03:32

does $G=\mathbb{R}/\mathbb{Z}$ (quotient groups) mean anything in G is lying in ]-1,1[?

Fargle

@saturatedexpo First, it would be $[0,1)$ (don't forget $0$!)--second, "kinda", in the sense that each coset of $\Bbb Z$ has a representative in $[0,1)$, but elements of $G$ are cosets, not single real numbers.

That being said

ohgodpleasegivememoremathIcan'thandlethiselection

saturatedexpo

03:48

i don't understand, why is -0.5 for example not element of G then?

or do you mean -0.5 is "the same" as 0.5?

@Fargle are you in the US?

Fargle

04:05

@saturatedexpo Yeah.

saturatedexpo

trump wins

if you believe the prematurely results

oh man

this is close as fuc*

Fargle

04:20

Time to move to Canada.

Adeek

hahaha

Fargle

@saturatedexpo The latter.

Semiclassical

hi chat

saturatedexpo

we will all die

Semiclassical

glad to see people aren't panicking, then :P

saturatedexpo

04:32

oh my god

florida trump

Semiclassical

nnnggg

ugh

538 blog now has Trump at a 77% chance to get elected

538 blog now has Trump at a 77% chance to win.

saturatedexpo

ah german moderator in tv called him even "Arbeiterführer"=Worker's Leader

Fargle

@Semiclassical You live close to Canada right? Get me into Canada

saturatedexpo

tough times are coming

Semiclassical

Me first

Fargle

04:35

That's fine, just get me there ASAP

I mean Jesus, I'm a white male and I'm this scared

Semiclassical

ugh

saturatedexpo

is the President commander in chief? or is the military strongly seperated from politics?

Semiclassical

commander in chief, yes.

saturatedexpo

ugh

Semiclassical

'what rough beast slouches towards bethlehem to be born'

saturatedexpo

04:40

at least after ww3 no land will have any debt.

the disadvantage is, there will be no land left

utah/iowa trump

Semiclassical

wisconsin, michigan, pennsylvania, nevada

that's what its resting on

saturatedexpo

the question is

would you have voted for elon musk? lol

PVAL-inactive

FN already called WI.

Semiclassical

ugh.

i'm really hoping that's just fox being fox

because if not, then...fuck.

PVAL-inactive

If Clinton wins MI, PENN,NV and NH the vote will be decided by a congressional district.

Semiclassical

04:46

jfc

PVAL-inactive

If Trump takes any of those its over.

538 gives Trump a 78% chance and Clinton a 20% chance.

Adeek

what is 538 ?

PVAL-inactive

An analytics blog.

Adeek

our algebra prof said if trump wins she will move permanently to canada haha.

user228700

Hi everyone :-)

user228700

04:56

Mathworld by Wolfram alpha defines many-one functions as follows:

user228700

> "A function f which may (but does not necessarily) associate a given member of the range of f with more than one member of the domain of f."

user228700

"May, but not necessarily"? Can anybody explain this to me?

Balarka Sen

@Kaumudi It means f : A --> B may have the property that f^-1(x) for x in B is of cardinality more than 1. But it may as well be 1.

So one-to-one functions are also considered as many-to-one.

Kasmir Khaan

7

Q: Double improper integral , how to see if it diverge

$$\iint_D \frac{(x+y) e^{y-x}}{x^2 y^2}dx \, dy$$ $$D= \{(x,y) ; 0\leq y+1\leq x , xy\geq 1 \}$$ Iv been stuck on this for past two hours , I need some hint . My bounds are : $\frac{1+\sqrt{5}}{2}\leq X<\infty $ $\frac 1 x \leq Y\leq x-1$ are the bounds correct ? I need some hints, ...

improper-integrals

I need help with this

anyone pls ?

user228700

@BalarkaSen Oh .__.

user228700

05:07

OK, thanks, I'll think about this.

saturatedexpo

@BalarkaSen so many to one functions are simply not injective?

(or can be not injective?)

Balarka Sen

can be not injective, according to mathworld's defn.

I'd define it to not be injective at least somewhere.

user228700

What is the correct definition, then?

saturatedexpo

@Kaumudi balarka's is the most precise IMO

Balarka Sen

Function which's not injective somewhere. There is a point such that more than one point gets mapped to it.

I wouldn't say it's "the correct definition", just what people usually means by many-to-one.

user228700

05:12

@BalarkaSen I don't really get this definition :|

Balarka Sen

Can you be specific?

user228700

Can u pls elaborate on the whole point?

saturatedexpo

@Kaumudi $f(x)=x^2$ is many to one. but $f(x)=x$ according to WFs def too

Balarka Sen

@Kaumudi Well, what do you mean by the whole point?

user228700

I just graduated high school and it would be helpful if u explained it in simpler terms, if possible..?

saturatedexpo

05:15

@Kaumudi examples are probably a quicker way google.de/…

in contrast look at one to one and the other terms ;)

Balarka Sen

@Kaumudi A many-to-one function is a function which is not injective. Forget mathworld's definition if that's confusing you.

saturatedexpo

@BalarkaSen would sin(x) be a good example?

Balarka Sen

Sure.

user228700

But what do they mean by "not necessarily"? Isn't that the definition? Then how can they say " not necessarily"?!

Balarka Sen

@Kaumudi They mean a many-to-one function associates a given member of the range with one or more than one member of the domain.

It's rather poorly phrased, I agree.

user228700

05:19

So all many-one functions are injective?

Balarka Sen

No, why would that imply what you said?

According to that definition, all injective functions are many-one. Not the other way around.

user228700

> "With one or more than one member of the domain"

user228700

Huh.

Balarka Sen

I mean, $f(x) = x^2$ is many-one using that definition. Do you see that?

user228700

Yes, that ^, sorry.

Balarka Sen

05:21

Right.

saturatedexpo

i think many to one is a stupid definition. but ok

(or stupid phrasing, how you like it)

Balarka Sen

nobody ever really uses the word many-to-one.

most people just say "not injective"

so, Trump's gonna win

Fargle

@BalarkaSen Yeah. Literally actually get me out of this country. I'm so done.

user228700

My textbook says "If a function is one-one, it cannot be many-one and vice versa" :/

user228700

Arfgh.

Balarka Sen

05:27

@Kaumudi So put mathworld's definition in the dustbin and move on.

Like I said 20 minutes ago.

saturatedexpo

@Kaumudi one to one, onto, many to one are IMO very rarely used anyways ;)

user228700

Yeah :| OK, thanks.

Balarka Sen

@Fargle Move to Europe for grad-school I guess.

saturatedexpo

then there is one to many which says nothing to me...

Fargle

@BalarkaSen Canada's the thought right now.

Balarka Sen

05:28

Better idea.

Adeek

@Fargle come to canada it is awesome here

Fargle

@Adeek Poutine is a natural draw. Not having a fascist leader is about to be another.

saturatedexpo

one to many function doesnt exist because of function-definition ?

user228700

@saturatedexpo If I understand what you mean by "one to many" correctly, then yeah, I think that would be correct.

user228700

I always think about the position-time graph when visualizing functions-no object can be at two different places at the same time.

PVAL-inactive

05:32

He isn't a facist. He's a sociopathic scammer and he won't be the first.

saturatedexpo

@PVAL-inactive at least he's old so he might die before the 4th year after election?

Fargle

@PVAL-inactive I dunno, calling for bans on immigration by religion is pretty fashy to me. But I do agree with your second statement.

@saturatedexpo I'm sure people hoped the same for Reagan. Didn't happen. (Not that it should have, I don't have much of a problem with Reagan, relatively speaking)

user228700

"How to move to Canada: Immigration website crashes as Donald Trump election victory looks imminent"

user228700

google.co.in/amp/www.independent.co.uk/news/world/americas/…

Balarka Sen

oops

saturatedexpo

05:42

if R gets a multiplicative inverse for 0, named c (to not fall into traps). would it then count as a group?

Kasmir Khaan

05:53

Who can help me with a question ?

it is very important for me pls anyone ?

Cbjork

Just ask

Kasmir Khaan

7

Q: Double improper integral , how to see if it diverge

$$\iint_D \frac{(x+y) e^{y-x}}{x^2 y^2}dx \, dy$$ $$D= \{(x,y) ; 0\leq y+1\leq x , xy\geq 1 \}$$ Iv been stuck on this for past two hours , I need some hint . My bounds are : $\frac{1+\sqrt{5}}{2}\leq X<\infty $ $\frac 1 x \leq Y\leq x-1$ are the bounds correct ? I need some hints, ...

improper-integrals

this is it

and thanks

Vrouvrou

hello

what is a smooth domain

Kasmir Khaan

hello sir

Cbjork

@KasmirKhaan is the answer given not good with you?

Kasmir Khaan

05:57

he claims it converge when the answer is diverge

so it cant be right

and it is from an old exam so they cant have made a mistake

and to be honest i don't know if all the steps he did are legal , like the boundings he made

saturatedexpo

mmh, that rises to me the question: is a infinitly nested integral always divergent? (not talking about 0)

Cbjork

Kasmir I really don't know how to do that integral and it's late. Maybe ask Ted?

Good luck!

Kasmir Khaan

thanks cbjork :)

user228700

I am back w/ another quick question! :-P

user228700

06:12

My textbook says "If either $f'(x)≥0$ $\forall x$ $\in$ the domain of $f$, or $f'(x)≤0$ $\forall x$ $\in$ the domain of $f$, where equality can hold at discrete point(s) only (ie. Strictly monotonic, then function is one-one, otherwise many-one".

user228700

What do they mean by "equality holds at discrete points only"?

saturatedexpo

saddlepoints

Balarka Sen

@Kaumudi $f' = 0$ holds only for a bunch of isolated points in the domain. It can't be zero along an interval, say.

user228700

@BalarkaSen Yeah, OK, u mean it can't stay zero-there cannot be any portion on the graph that is a straight line parallel to the x-axis, yeah?

Balarka Sen

Right, that's the intuition.

user228700

06:15

OK, thanks :-)

Balarka Sen

If that happens it won't be strictly monotonic. Whereas for discrete critical points (that's where $f' = 0$ happens), look at eg $x^3$.

saturatedexpo

i guess a textbook that requires explanation is rubbish?

i mean i knew the term, but the definition looks ugly

anyways

trumped

Kasmir Khaan

anyone can help me ?

i got a hard question

7

Q: Double improper integral , how to see if it diverge

$$\iint_D \frac{(x+y) e^{y-x}}{x^2 y^2}dx \, dy$$ $$D= \{(x,y) ; 0\leq y+1\leq x , xy\geq 1 \}$$ Iv been stuck on this for past two hours , I need some hint . My bounds are : $\frac{1+\sqrt{5}}{2}\leq X<\infty $ $\frac 1 x \leq Y\leq x-1$ are the bounds correct ? I need some hints, ...

improper-integrals

Adeek

06:39

hey @BalarkaSen here

Balarka Sen

yep

Adeek

I just want to check the logic of something with you.

Let $f : X \rightarrow Y$ be differentiable map between manifolds of the same dimension. If df(p)(differential map) is non-singular for all p in X, then f is an open map. Give counter example if we drop the condition df(p) is non-singular for all p in X.

So I proved the following result given $f : X \rightarrow Y$ be differentiable map between manifolds of same dimension with df(p) being non-singular. Then there is nbhds $U_p$ and $V_{f(p)}$ such that $f : U \rightarrow V$ is a diffeomorphism.

So suppose $f : X \rightarrow Y$ satisfy the conditions in the hypothesis of the problem. If U subset of X is open we which to show that f(U) is open.

for each $p \in U$ we have df(p) is open so we have that there exists diffeomorphism $f : U_p \rightarrow f(U_p)$ where $U_p \subset U$ is open, so $f : \bigcup_{p \in U}U_p \rightarrow f(\bigcup_{p \in U}U_p)$ is a diffeomorphism. Hence f(U) is open.

what do you think ?

Balarka Sen

@Adee You need to show that diffeomorphisms are open maps.

You didn't do that anywhere.

Adeek

oh I thought we get it directly but not necessarily.

your right.

Balarka Sen

I don't know what that's supposed to mean

Adeek

06:50

I mean I thought if we have a diffeomorphism $f : U \rightarrow f(U)$ then it is true that $f(U)$ must be open.

Balarka Sen

Well, prove it, then.

That's the key thing; because your $f$ is a local diffeomorphism.

Adeek

I see.

ok yes it is not that bad.

Balarka Sen

right

Adeek

@BalarkaSen Suppose that U is open and let $x \in U$. Then we have df(p) is non-singular. So There exists nbhd $U_p$ and $V_{f(p)}$ open such that $f : U_p \rightarrow V_{f{p}}$ is a diffeomorphism. $f(U) = \bigcup_{p \in U}V_{f(p)}$ i.e union of open sets hence open.

Balarka Sen

That works.

Adeek

07:05

is there an easy counter example for the second part ?

Balarka Sen

Lots

Adeek

maybe something like $S^{1}$.

Balarka Sen

You can give me easy examples $f : \Bbb R \to \Bbb R$

Adeek

$x^{3/2}$ ?

hmm let us see

Balarka Sen

I mean, that's not defined in $\Bbb R$

Danu

07:11

Oh god...

It's actually happening.

Balarka Sen

Mhm

I am not surprised, but I don't live in USA.

Thomas Andrews

Yeah, I'm up do to anxiety about our stupid stupid voters.

Anybody got a good math puzzle or contest problem to tackle?

Kasmir Khaan

7

Q: Double improper integral , how to see if it diverge

$$\iint_D \frac{(x+y) e^{y-x}}{x^2 y^2}dx \, dy$$ $$D= \{(x,y) ; 0\leq y+1\leq x , xy\geq 1 \}$$ Iv been stuck on this for past two hours , I need some hint . My bounds are : $\frac{1+\sqrt{5}}{2}\leq X<\infty $ $\frac 1 x \leq Y\leq x-1$ are the bounds correct ? I need some hints, ...

improper-integrals

Adeek

hm I am little bit confused so if we have $f : \mathbb{R} \rightarrow \mathbb{R}$ which is differentiable then it is necessarily continous right so we must have it to be a homeomorphism ?

@BalarkaSen?

Balarka Sen

@Adeek ?????!? Certainly not every continuous map is a homeomorphism

Adeek

07:18

I remember seeing a result before that if we have $g : R^n \rightarrow R^n$ which is continous bijection then it must be homeomorphism.

Balarka Sen

Where did the condition bijection come from????

Adeek

hm yeah we don't need to have bijection. I was thinking in terms of bijective function which failed in all cases.

lol

@BalarkaSen here is a trivial example.

$f : \mathbb{R} \rightarrow \mathbb{R}$ given by $x \mapsto 1$

this is differentiable map between manifolds of same dimensions. All the points have singular derivatives and this map isn't open.

lol

Balarka Sen

@Adeek Sure

There are nonconstant examples, eg $x^2$

It's not open near $0$

OK, I gotta go

Adeek

ok thanks

Kasmir Khaan

07:58

7

Q: Double improper integral , how to see if it diverge

$$\iint_D \frac{(x+y) e^{y-x}}{x^2 y^2}dx \, dy$$ $$D= \{(x,y) ; 0\leq y+1\leq x , xy\geq 1 \}$$ Iv been stuck on this for past two hours , I need some hint . My bounds are : $\frac{1+\sqrt{5}}{2}\leq X<\infty $ $\frac 1 x \leq Y\leq x-1$ are the bounds correct ? I need some hints, ...

improper-integrals

SteamyRoot

What a great day to not be American

Fargle

08:10

@SteamyRoot The sentiment is shared by this American.

Adeek

haha

Kasmir Khaan

08:29

7

Q: Double improper integral , how to see if it diverge

$$\iint_D \frac{(x+y) e^{y-x}}{x^2 y^2}dx \, dy$$ $$D= \{(x,y) ; 0\leq y+1\leq x , xy\geq 1 \}$$ Iv been stuck on this for past two hours , I need some hint . My bounds are : $\frac{1+\sqrt{5}}{2}\leq X<\infty $ $\frac 1 x \leq Y\leq x-1$ are the bounds correct ? I need some hints, ...

improper-integrals

DHMO

08:40

@saturatedexpo yes, -0.5=0.5

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$Mathematics$ Mathematics