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00:00 - 16:0016:00 - 00:00

00:00
So $\ker h \subset \ker f$.
@BalarkaSen I actually added time for you
missed the geometric analysis talk
@LeakyNun thank. i will get back to it as soon as possible
@ÉricoMeloSilva I'm Taylor expanding the length of a closing edge of a geodesic hinge rn
the what of a what of a what
00:04
it's exactly what it sounds like
lol
@Erico You're still stuck in Princeton campus?
00:21
@BalarkaSen btw Nepo won a game against Magnus
Nice, rekt
00:33
@Erico: Yes, if you have some good books to recommend to @JingeonAn, you certainly are a better source.
00:45
ah i missed enjoys /:
@Enjoys I admire your persistance
@geocalc33 did you upvote my post yet?
:P
@BalarkaSen Magnus Carlsen messed up the opening and lost in 7 moves
also he spent 7 minutes before playing the losing move
@Enjoys I will upvote it if I think it should be upvoted :)
Maybe check out the content
good question :)
00:50
Thx
I'm trying to relate it more to twin primes
@BalarkaSen also he looked tired and emotional
It's difficult now because we moved to the group ring
what's the main difficulty at the moment?
@geocalc33 look at
In mathematics, specifically abstract algebra, the isomorphism theorems (also known as Noether's isomorphism theorems) are three theorems that describe the relationship between quotients, homomorphisms, and subobjects. Versions of the theorems exist for groups, rings, vector spaces, modules, Lie algebras, and various other algebraic structures. In universal algebra, the isomorphism theorems can be generalized to the context of algebras and congruences. == History == The isomorphism theorems were formulated in some generality for homomorphisms of modules by Emmy Noether in her paper Abstrakter...
So I have that $\ker \Omega \subset \ker \epsilon$
That's one ideal contained in another ideal
So let $R = \Bbb{Z}[G]$
where $G = \{ x^2 : x \in \Bbb{Q}^{\times}\}$ are the squares
okay just keep going and maybe I'll somehow understand all of it at the end
00:53
Just understand now
$G$ is a subgroup of multiplicative rationals right?
Well for any group there's the group ring construction $\Bbb{Z}[G]$.
That's formal sums with basis $G$ and coefficients in $\Bbb{Z}$.
shows up in group cohomology
yeah I guess G is a subgroup of the multiplicative rationals okay
So $R$ is a ring
known as a group ring the way it's constructed
okay fair enough
So take $I = \ker \Omega$. $\Omega : R \to \Bbb{Z}[\Bbb{Z}]$ btw is a ring homomorphism. I've proven that already
00:56
Now there's a 1-1 correspondence between subrings of $R/I$ and subrings of $R$ containing $I$. In addition
$A \geqslant I$ is an ideal iff $A/I$ is an ideal of $R/I$.
Thus we have $\ker \epsilon = A$
as an ideal, so that $A/I$ is an ideal of $R/I$.
$A/I = \{ x + I : x \in A \}$ btw
cosets
So the twin prime conjecture has to do with intersecting in $G$ itself, how can we bring this up to the group ring?
*intersection of certain sets
$\Omega : \Bbb{N} \to \Bbb{N} \cup \{0\}$ counts the number of primes (counting multiples)
so prove that $\Omega(xy) = \Omega(x) + \Omega(y)$ for all $x,y \in \Bbb{N}$. I.e. that it's completely multiplicative
Well the twin prime conjecture is then just htat $\Omega(x - 1) + \Omega(x+1) = 2$ for infinitely many integers $x$.
okay cool
I've taken $\Omega$ extended it to all of $\Bbb{Z}\setminus 0$, then to all of $\Bbb{Q}^{\times}$ where it forms a group hom $\Bbb{Q}^{\times} \to \Bbb{Z}$. See link in post
to other post
So $\Omega(x -1 ) + \Omega(x+1) = \Omega(x^2-1)$
by the multiplicative property of $\Omega$ or that it's now a group hom from multiplicatives to additive integers
okay so what's the problem
01:03
Thus the statement of TPC is $\Omega^{-1}(2) \cap \{x^2- 1 : x \in \Bbb{Z}\}$ is an infinite set
the problem is, translate that statement to the group ring setting $\Bbb{Z}[G]$ where $G = S$ the squares
how are you gonna do that
I know that there is an ideal $I_G$ generated by elements of the form $g - 1 : g \in G$
I'm asking you :|
^_^
Thus if $G = \{ $ squares in $\Bbb{Q}^{\times}\}$. Then $g - 1 $ looks a lot like $x^2 - 1$
So that brought is to the homs of which $I_G$ is the kernel, known as the augmentation maps of the group ring
they take all coefficients of an element $x = \sum_{g \in G} x_g g \in \Bbb{Z}[G]$ and sum them: $\sum_{g \in G} x_g \in \Bbb{Z}$. That is a ring hom
explain that G={squares in Q line....
it's kernel is generated by all $g - 1$ where $g \in G$.
$G = \{ x^2 : x \in \Bbb{Q}^{\times}\}$ do you know what $\Bbb{Q}^{\times}$ is? it's just $\Bbb{Q} \setminus 0$ with the apparent multiplicative structure from the field $\Bbb{Q}$.
So $G$ is clearly a group, same can be done with $x^k$ for any fixed power $k$
positive or negative
or $0$
so tell me this
01:09
?
how am I supposed to solve this. I know barely any math
I'm teaching you, just ask questions
You know about formal power series right?
okay so you want to somehow "bring it up to the group ring?"
what does that mean
Yes
Currently the intersection takes place in $\Bbb{Q}^{\times}$ our original mother space
01:12
But how can we have an equivalent TPC statement but done in the group rings?
See post, there's two group rings: $\Bbb{Z}[S], \Bbb{Z}[\Bbb{Z}]$. In another post someone said $\Bbb{Z}[\Bbb{Z}]$ was isomorphic to the Laurent polynomial ring
so you want to translate from $\Bbb Q^{\times}$ to like the integers or something
Let me teach you about group rings
Take any group $G$ now
finite or infinite
Now take formal sums:
$\sum_{g \in G} x_g \cdot g$ where $x_g \in R$ a ring (in my case $R = \Bbb{Z})$.
They add component wise
formally
and multiply like:
$(\sum_{g \in G} x_g g) (\sum_{g \in G} y_g g) = \sum_{g \in G} (\sum_{ab = g} x_a y_b) g$
Or some such thing
similar to the polynomial mult of power series
alright
Prove on paper that a ring is formed from that $+$ and $\cdot$ but look them up to be sure I've got them right
The sums are formal, so we make $+$ be commutative formally
So for instance if $G = \Bbb{Z}_3$
we have $a\cdot 1 + b\cdot 2 + c \cdot 0$ with $a,b,c \in R$
That lists all elements of the group ring $R[G]$.
If $R$ is finite as well, then so is this group ring
The $a, b,c $ are like coefficients and the $g$'s are like a basis
Alternatively, understand the first few pages of:
Let me know when you have a good idea of group rings, then we can proceed
okay I will
01:21
After understanding group ring basics, then you have to know how $\Omega$ extends first to $\Bbb{Q}^{\times}$ and then to the group ring
You simply say "extend $\Omega$ linearly" and apply it to each basis element while taking the coefficients outside of $\Omega$
There is a unique such linear map that extends $\Omega$ from $G$ to $\Bbb{Z}[G]$. Note that $G \cdot 1 \leqslant \Bbb{Z}[G]$ is definitely a subring, so extending from one to the other makes sense
so are you looking for this map?
No, found $\Omega$
Wanting to make the intersection statement in the group rings
We know given the two augmentation maps $\epsilon: \Bbb{Z}[G] \to \Bbb{Z}$, and $\epsilon' : \Bbb{Z}[\Bbb{Z}] \to \Bbb{Z}$ that $\epsilon$ factors through $\epsilon': \epsilon = \epsilon' \circ \Omega$.
that's what the last post was about
okay I'll try to read up
 
1 hour later…
03:02
@BalarkaSen yeah im stuck in princeton, tried to go home early but decided against it because have immunocompromised family
that sucks, i hope you're safe
you too
im good just bored
lots of math though
haha, yeah at least we have something to do during apocalypse
truuuue that
03:04
im back home which was a good idea because back in bangalore where i study things are turning out badly
oof you hate to see it
500 cases until now and 200 of them can't be traced back to anything; top it all off they happened in the span of last week
community spreading clearly
man shits crazy
sounds like it's gonna spiral
03:06
that sucks
we're officially in a resident evil live action spin off
places are reopening across the us which is such a bad idea
it's just so bad
@ÉricoMeloSilva us is such a something
im like the wack leon
03:08
Very Normal Country^Tm
@ÉricoMeloSilva yah i just think they're (incl my govt) taking every worst possible decisions. im convinced locking down early instead of med facilities/testing and segregation was a political move; it's just to gain more control and surveillance. and now lifting it in the middle of the pandemic is just pure idiocy
its like they want the curve to gain more kurtosis lmao
considering this thing seems to be able to incubate for like multiple weeks it sseems to me that trying to lockdown was just a DOA strategy
i mean locking early and not ramping up testing fast
did you see that when obama was president that he help fund the lab that accidentally released chronoa
did it? i thought they were worried it escaped from that lab but fortunately didn't
03:14
i heard that someone at the lab got bit and went to a food market
like 1 week after
bit by what man
bizarre story
lmao
the bat
this is probably not true
diseases happen to transfer between mammals like pretty often, that's probably all that happened
idt anyone understands the trajectory of the virus; it def has something to do with the bats but mine workers used to get corona all the time and humans never acted as vectors
bizarre mutations prolly
earth is angry
this news place said it started from a lab newsweek.com/… but they could be wrong
03:19
@ÉricoMeloSilva my govt declared situation is under control and stopped fast tests lmao
the world is just maximally dumb all the time
this is after their own team of doctors walked out of modi's conference because he wasn't listening to them
yeah lol idiots
this will go on for a year at this rate
Brazil is like totally fucked too
oh i dont know the stats in brazil
Oh shit it looks bad
the bolsonaristas were like NEVER SHUT DOWN for a while
and it was running rampant through political elites
03:21
Oh yeah Bolsonaro was denying the existence of the virus or something right
also there's maybe ruptures and maybe we'll see the second president being impeached in one decade lol
but then mourão is just as much of a total ghoul if he succeeds bolso
i dont see a foreseeable future of democracy in my country either tbf
its one crackpot after another
on amazon i ordered a trump mask lol it says trump 2020 and trump can't do anything bc everyone is trying to impeach him
haha imagine MAGA masks
brb
idk why but in my name it says wii in it even though i meant to put will
04:04
Why are they protesting? What they would do with the money if they will be dead? It’s very simple; go outside and work and earn so much of money then go in the hospital and die with COVID 19 infection
its more complicated than that. wage workers are stuck in the middle of this shit because of the lockdown
at least in my country
people have to eat
the amount of people who went jobless in the past 4 weeks is insane
Well, it is possible to live without complete food for $1 ~1/2$ months but with COVID19 there are very very less chances of coming out alive
that statement is baloney lmao
covid is not at all fatal to most people
and people cant survive without food. most wage workers in india live on way less food than necessary for an average human being every day
@BalarkaSen dem my bullet rating is now 1681
wait till I go back below 1600 in 3 games
04:13
People can’t survive without food, but they can survive with incomplete food. If Government is doing something then I don’t think it is nice to protest
government isnt doing anything lmao
they started this shit
protest will happen sooner or later, there's no stopping it. its easy to whine about this sitting cozily in front of a computer
it's as if all the governments are incompetent and only came to office because they are rich / have powerful relations
definitely not true though, definitely democratic
Well, I think people will curse the government even more when thousands of people start dying each day.
And that’s what the consequence of taking the lockdown away
People will die
Well, that will happen, precisely because government didn't take any initiative. Lockdown is as effective as hygiene and test+segregation if community spread doesn't start, that's not my opinion that's a fact. They never started testing until 2 weeks in the mess
The first move was to lockdown, which is just tactically a wrong move
Of course they can't lift the lockdown now that'd be suicide. But they will, because that's what will earn them popularity in a few weeks
04:18
I don’t always tend to blame the government, because I don’t consider them more clever than me.
i dont blame them i just believe they dont act on good faith
if you believe that then i think that's a terrible belief lmao
its stupid to think the indian government acts on good faith; their take on the whole covid thing was superbly political
I don’t know much about the internal things of Indian Governemnt, but here in Chicago people went into lockdown by themselves and I do accept low wage workers are fatally affected but they must know a worse situation would have hit them
i dont know how bad this hits US, i have very little understanding of the economic situation in US.
i know its quite, quite, quite bad in India
if people from the lower class stratum do end up protesting they have a very good reason to
i dont think i deserve to have an opinion against it since the effect of lockdown due to covid will be epsilonic on my economic situation.
and also because i generally think the upper middle class and the pmc should stop talking out of their ass so much
I'm trying to make sure I understand this notation: $(g \circ f)^{-1}(H)$ is $g(f(H))^{-1}$, right?
 
1 hour later…
05:35
@CaptainAmerica16 I don't understand what your wrote, but $(g \circ f)^{-1}(H) = f^{-1}(g^{-1}(H))$.
Let $F$ be a homogenous polynomial in 3 variables such that $\nabla F$ is non-zero on the zero set of $F$. I am trying to show that the zero-set of $F$ gives a submanifold of $\mathbb{RP}^2$.
Let $\{U_0, U_1, U_2\}$ be the standard atlas of $\mathbb{RP}^2$.
My idea is to define a function $f : \mathbb{RP}^2 \to \Bbb R$. And show that the zero-set of $F$ is a regular level-set of $f$ (at 0).
05:53
@feynhat That's what I was looking for. I had a feeling l was misunderstanding the notation, so what I typed out doesnt make sense anyway. Thanks!
Suppose $[x_0, x_1, x_2] \in U_0$, then, $f([x_0, x_1, x_2]) = F(1, x_1/x_0, x_2/x_0)$.
In the coordinates, $f(x, y) = F(1, x, y)$.
Then, $\nabla f = \partial F/\partial x_1 + \partial F / \partial x_2$.
But, what if in $U_0$, $\partial F/\partial x_0 \ne 0$ and $\partial F / \partial x_1 = \partial F/\partial x_2 = 0$?
Euler's homogeneous function theorem says $\sum x_i dF/dx_i = \deg(F) \cdot F$. If $F = 0$ and two of your partials vanish on $F^{-1}(0) \cap U_0$, where $x_0 = 1$, the third partial also vanishes
This is standard calculus trick
Oooh.
Thanks.
06:18
why do you check these things anyway? i dont think i have ever checked half the details you check in my life
(to be taken very nonseriously)
im trying to stay awake and rambling just for the hell of it
It didn't seem very obvious to me.
Fair, yeah. I am having similar issues in Riemannian geometry, have to keep checking trash because nothing is really clear to me until I calculate
Plus it's good to be able to know how to put it down in paper anyway; I have definitely faced repercussions for handwaving past details
I wasn't looking for a serious answer :P
But when you do skip checking things like that, do you usually go 'I have a rough sketch of a proof, and will be able to write down, if needed' or is it more hand-wavy like 'the things are nice-enough... and the book says so. So, it must be true'.
Neither to be honest, I am never confident about writing proofs down on the spot. If I see a clear picture I trust it and don't write things down analytically
I don't trust words of books, no
to do math i need to have a clear picture. this is also true if i can in fact write down a proof; if i dont understand what the hell is happening i cant proceed
Fair.
06:35
the good thing about proofs is theres no way its wrong once you have it. pictures can be deceiving on the other hand and i have personally found often that my mental model of things miss finer details/subtleties
over time it improved though so i got more confident about visual intuition
i have grave respect for people who can actually start with a very complicated picture which has no unnecessary details and covers all possible subtleties that can arise in the object
Thurston definitely was one of them
06:59
@BalarkaSen Mate
Is Eos on that album?
07:29
@BalarkaSen K. Buzzard has entered the chat
@EdwardEvans Ya
@LeakyNun if you cant give me an automatic proof of the geometrization conjecture then automatic math is a waste of time
I had a dream about maths and would like help figuring out what (if anything it was about)
good luck coding that in Lean
It's not coherent enough for a mainsite question
At the same time it could just be garbage
im curious, tell
07:37
I think it was some time of optimization problem, but I'm not sure what kind
There were a series of 5 numbers, which got arranged into a segmented triangle that had 3 tiers
The top tier had 1 of the numbers, the middle had like ~5 numbers then the bas had 3 or 4 numbers in it?
Then each of the 3 tiers was assigned a number which then helped figure out a sum?
It may have been that the 3 tiers were hours minutes and seconds from top to bottom
I don't know what sum was done on them either
That's all I remember
@Balarka brilliant song lol
@AncientSwordRage Awesome. Maybe it was something like a magic triangle?
Weird that the middle tier would have more numbers than the base
How is it equivalent saying an integer $n$ is not divisible by 3 and $n \equiv 1$ or $5 \mod{6}$ ?
@EdwardEvans It's the first track, I love it
It caught me off guard when the Sanskrit chant started suddenly
@BalarkaSen no it looked more like voronoi partitioning of each tier
07:46
Oh I see
@BalarkaSen ¯\_(ツ)_/¯
@BalarkaSen like that?
But different numbers
@Balarka saaaaaaaaaaame
Any ideas?
I don't think I have seen anything quite like this before, nope
@EdwardEvans The whole album is great. The second track is also a bamboozle
They also have a cover of Solitude by Black Sabbath
07:56
I've only listened to Eos, for some reason I've avoided listening to anything other than Bergtatt and Kveldssanger
@BalarkaSen thanks anyway, I appreciate the chat room listening to my babbling
@EdwardEvans I tried to listen to Nattens Madrigal but 20 minutes in my head started paining
It's unlistenable
loool
it's like
"Shit our first album was too nice, where's the hello kitty microphone and amp"
3
Loooool
08:16
@BalarkaSen south winds by ihsahn is a tune
08:38
ergh I already don't want to do any analysis and it's only week 2
08:58
Does anyone know any good mathematical software to plot numerical solutions to differential equations?(no closed form solution)
Preferably free, and also some notes on how to use them.
09:19
@EdwardEvans There's some Leprous live shows with ihsahn which are amazing
Yeah they are related in some way
Like ihsahn's wife is the sister of somebody in leprous or something like that
that's cool af
09:46
by the way @Edward or any other person living in Germany, any idea what "salary level E13 TV-L, 75%" means?
No idea what the TV-L part means but I think the E13 is just a rung on a ladder
75% will be (I assume) three quarters of full time under the Kollektivvertrag or smth, which I guess you might have assumed
Quora confirms this
the dream
@Alessandro have you heard anything from Lukas btw? :P or @TedShifrin ?
09:52
Not since February or something like that
I'm trying to understand this statement in Wikipedia "If k = 3: if n ≡ 0 or 3 (mod 4), then..."
In mathematics, the Lucas–Lehmer–Riesel test is a primality test for numbers of the form N = k ⋅ 2n − 1, with k < 2n. The test was developed by Hans Riesel and it is based on the Lucas–Lehmer primality test. It is the fastest deterministic algorithm known for numbers of that form. For numbers of the form N = k ⋅ 2n + 1 (Proth numbers), either application of Proth's theorem (a Las Vegas algorithm) or one of the deterministic proofs described in Brillhart-Lehmer-Selfridge 1975 are used. == The algorithm == The algorithm is very similar to the Lucas–Lehmer test, but with a variable starting point...
What does $:$ mean? Does it mean "and" or "or" or that the statements are interchangeable...?
Like why is there two if's in a row?
I think it's a disjunction of cases
First you look at the case $k=3$
In this case you have $n \equiv 0$ or $3$ mod $4$
@Alessandro just that I haven't heard from him in ages and nobody can get in contact with him haha, he's got a seminar in a couple of weeks and the organiser was asking about him but that's less important lol
Is a manifold with constant positive curvature always a sphere ?
@EdwardEvans Ah I see
10:00
I'm sure he'll pop up again sometime soon, I think he has bigger fish to fry than a seminar on completions of $\Bbb Q$ lol
ok thnx
@EdwardEvans Dunno, but he has this habit of disappearing once in a while
yeah exactly
 
2 hours later…
11:45
Astyx you there?
12:01
Don't mind me. I am testing.
Does the ChatJax thing render MathJax automatically? Or I have to press it every time? $\big[\text{ad}(x),\text{ad}(y)\big]=\text{ad}\big([x,y]\big)$.
Oh, nice.
Can I draw a commutative diagram here? $\require{AMScd}$
\begin{CD}
\mathbb{Z}/p^n\mathbb{Z} @>{\theta_{n+1,\alpha}}>> U_1/U_{n+1}\\
@VVV @VVV\\
\mathbb{Z}/p^{n-1}\mathbb{Z} @>{\theta_{n,\alpha}}>> U_1/U_n
\end{CD}
Oh, great...
Pictures are ok... That's all, guys..
Sorry for the nuisance.
Oh, what about links? Click here to learn about Galois groups.
In mathematics, more specifically in the area of abstract algebra known as Galois theory, the Galois group of a certain type of field extension is a specific group associated with the field extension. The study of field extensions and their relationship to the polynomials that give rise to them via Galois groups is called Galois theory, so named in honor of Évariste Galois who first discovered them. For a more elementary discussion of Galois groups in terms of permutation groups, see the article on Galois theory. == Definition == Suppose that E {\displaystyle...
Ah ha... OK
That was the last try. Bye.
user434058
12:21
@Batominovski BTW, you can do all the testing in the Sandbox: a dedicated chatroom, just for testing :)
12:41
2
Q: Example of an equation that generates solutions?

geocalc33 Q: Is this an example of an equation that generates solutions? And is everything below correct? Part $1:$ Suppose that $x$ and $z$ are independent of each other. $$\phi(x,z)= e^{\frac{z^2}{\ln(x)}}=x,$$ where $x,z\in\Bbb C.$ $x=e^{z},e^{-z}.$ With this information I rewrote the equatio...

bounty
13:23
good Q!
13:57
I'll throw this up here for good measure
@AdamL what's happening?
14:35
I’m waiting for @Semiclassical
me too!
@Knight yes
@JoeShmo! Welcome back
Hello hello!
14:48
Hi @Balarka
@JoeShmo what math have you been thinking about lately?
algebraic topology predominantly, can't get myself to begin studying for the analysis quals
what's stopping you?
algberaic topology, and work
oh I see. well good luck :)
14:55
I wish I could go to school full time, but it's hella expensive
I feel that. In the US it's way too expensive
like it's like US is punishing people that say they want to get education
it's a recipe for disaster.
it's one side of it.. but then you get into the job market and you get a larger paycheck than you do in Europe, and you see the other side of it
I'm talking about the country as a whole. US has rampant inequality. 90% of the people in the US hold about 10% of the wealth.
theyre more generous with the money in Europe though, you could get a stipend as a masters student, which is I think unheard of on this side of the pond
You know what this does. It basically says: "only wealthy people have access to education." And if you have enough wealth to get a degree, then you will get slammed with major debt
on average
trust me if we keep importing talent and not cultivating our own
we will crumble
15:05
7
A: Gaussian-like integral : $\int_0^{\infty} e^{-x^2} \cos( a x) \ \mathrm{d}x$

Random VariableOne approach is to integrate $e^{-z^2}$ around a rectangle in the complex plane with vertices at $R,R+ia,-R+ia$ and $-R$. Another approach is to expand $\cos ax$ in a Maclaurin series and then switch the order of integration and summation. You'll end up with a constant times the Maclaurin series...

Kind of, I'm actually not sure where I stand on this issue. There's certainly a major education inflation in the U.S. You are slammed with nonsense gen eds (that you have to pay for out of pocket), and people are getting B.S degrees (baccalaureate, yes?) in underwater basket weaving and sea turtle psychology
Does anyone mind explaining how does one arrive at that approach and how does it work?
and they are demanding the state subsidize their adventures
I completely agree with cultivating our own talent
15:31
@Archer which approach? There is also a similar question.
15:50
@robjohn Hi :-)
@Knight hey there... how are things?
@robjohn Everything is going on well
How are you?
@Knight feeling like an isolated singularity
@Batominovski As mentioned by robjohn, most people use here some kind of bookmarklet. You can check this post on meta: Should chat have TeX support? I am using robjohn's bookmarklet: math.ucla.edu/~robjohn/math/mathjax.html
@robjohn The rectangle one
I dont see how the integral reduces to 0 on the vertical side of the imaginary axis
2
A: Gaussian-like integral : $\int_0^{\infty} e^{-x^2} \cos( a x) \ \mathrm{d}x$

MoliYou can integrate $$f(z)=e^{-z^2},z\in\mathbb{C}$$ On the boundary of the rectangle $$\{z\in\mathbb{C}:\Re(z)\in[-R,R]\wedge \Im(z)\in[0,a/2]\}$$ Given $a>0,R>a/2.$ Being $f$ entire, this integral is equal to zero. Using the fact that $$\int_0^\infty{e^{-x^2}dx}=\frac{\sqrt\pi}{2}$$ And observin...

15:56
Robbie I got a very bad problem
I get that integral reduces to 0 in case of R to R+ic using ML inequality
but it doesn't in case of ic to 0.
^ are these types of problems meant to be solved or what?
@Knight JEE question (mains)
@abhas_RewCie Someone sent me this and asked me to try it
00:00 - 16:0016:00 - 00:00

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