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6:32 AM
@Alessandro you were right about Neukirch, he's hard
 
7:18 AM
Hi! I came across an instance where the combination of a variable and a constant (e.g. 5+$x$) is also called a variable. I've never heard of the combination being called a variable before. While 5+$x$ will vary depending on the value of $x$, is it correct to call it a variable?
 
@Paradox101 ill-defined question
 
@Daminark where are you in the book?
 
Starting 1.3
(I didn't even really finish 1.2, kinda skipped the last bit on bases and discriminants but I felt I probably wasn't gonna need it and I'm trying to make a decent amount of progress in... about a week or two)
 
@LeakyNun Okay, how so?
 
@Paradox101 define a "variable"
 
7:33 AM
@LeakyNun Oh ok. In this instance, a variable is defined as a 'symbol which can be given different values, and it represented by letters.' This is for grade 8 so they're just learning about variables and constants for the first time.
 
and a constant?
 
@LeakyNun A constant is simply 'a symbol having a fixed value'
 
7:50 AM
@Alessandro rip me, right after that he proves that rings of integers are Noetherian by referencing the fact that the integral closure of a PID in finite extensions of the field of fractions are finitely generated modules
Guess I gotta go back to that shit
 
@Daminark I guess AM 5.17 is relevant?
 
I'm not quite referencing AM but yeah it's that
 
Determinant trick incoming?
 
actually AM used trace
 
8:06 AM
Yeah same for Neukirch
 
oh so this is like more general than 5.17
 
Neukirch does a lot of stuff in very general settings
He works with extensions of Dedekind rings rather than field extensions and their rings of integers
 
:o
 
@Alessandro Lang does the same in his ANT
and like 90% of his proofs are him proving stuff for maximal ideals and then claiming you can easily localise and talk about primes instead
 
8:21 AM
@ÍgjøgnumMeg sounds reasonable
 
yeah I liked his book (the parts that I read anyway)
but he does randomly make up notation that doesn't occur in the statement of theorems
or anyway else in the book
anywhere*
 
Wait I'm not sure how to parse the statement "they are given as the quotient of an element of $A$ by the determinant $\det(Tr_{L\mid K} (\alpha_i\alpha_j)) = d$"
Does that mean those elements of $A$ are given as $\frac{x}{d}$ for $x\in A$?
Because if so I have no idea why that's the case
Oh oh wait hmm
I misread
I read $Tr_{L\mid K}(\alpha_i\alpha_j)$ and thought that the "they" was referring to those guys
And I'm just like
Tf kinda drugs are you on?
 
But the "they" is referring all the way back to the $a_i$
Okay and this is true because of some kinda Cramer rule thing which seems to hold in rings
Err no this just is a thing actually
Yeah okay we coo Neukirch, we coo
 
 
2 hours later…
10:45 AM
When problem-solving, I keep making this same mistake: I think of a possible solution, I discard it thinking it won't work, then, after an hour of struggle, I find out the solution I thought of to begin with worked. Does anybody have the same experience? Does anybody have any advice regarding this issue?
 
10:55 AM
@TuringMachine holy mother that's me
 
 
1 hour later…
12:11 PM
@LeakyNun $$\int \dfrac{x^2- 4}{x^4 + 24x^2 +16} dx$$
@LeakyNun Do you have any interest in this^?
 
$x^4 + 24x^2 + 16 = (x^2 + 12)^2 - 128$
partial fraction?
 
@LeakyNun no
 
why not
 
@LeakyNun itd be complicated because of $\sqrt {128}$
 
well is that the full question?
 
12:20 PM
@LeakyNun yes
 
indefinite integral?
 
@LeakyNun yes
 
hmm, WA does give a simpler answer.
$x^4 + 24x^2 + 16 = (x^2+4)^2 + (4x)^2$
hi @Mr.Xcoder
 
hi, Leaky
how is it going?
 
@Mr.Xcoder alright
@Abcd If I didn't know the answer, I would have no idea
 
12:27 PM
@LeakyNun that is what Indefinite integration is ...
 
lmao you have a point
 
Integral calculator gives something like $$\frac{\tan^{-1}\left(\frac{x}{2(\sqrt{2}+1)}\right)-\tan^{-1}\left(\frac{x}{2(‌​\sqrt{2}-1)}\right)}{4}+c$$ :||
 
o..o
 
In everyday work who'd ever encounter that in everyday work? why would anyone want to integrate that by hand rather than just plugging it into WA or some other such service :P
 
why would anyone want to integrate that instead of looking at the geometry
 
12:35 PM
@mercio whats the geometry here
I dont see any geometry
 
I forgot yo uwere in high school
 
ok
 
things are prettier than expected because the roots of the denominator are in $\Bbb Q(i, \sqrt 2)$
probably
 
My own language, Physica, uses Numpy SymPy for integration and gives atan(x/4)/4 - atan(x**3/16 + 5*x/4)/4 (+c)
 
$x^4+24x^2+16 = (x-2i-2i\sqrt{-2})(x-2i+2i\sqrt{-2})(x+2i-2i\sqrt{-2})(x+2i+2i\sqrt{-2})$
something like that
so that's why the arguments in the arctangents are $x/(2+2\sqrt 2)$ and his friend and not something ugly
 
12:50 PM
Anyone know the transformation law for a section of a vector bundle
 
did i just write $i\sqrt{-2}$ instead of $i\sqrt 2$ or $\sqrt{-2}$
 
lol
 
Mr. Xcoder's image tricked me into thinking there was another hats event lol
 
1:10 PM
Well, summer bash? :P
 
@Abcd ?
 
@LeakyNun I was asking then what? after writing the denominator that way
is it immediate from there? I dont think so?
 
47 mins ago, by Leaky Nun
@Abcd If I didn't know the answer, I would have no idea
i.e. just look at the answer and you'll know the solution lol
 
52 mins ago, by Leaky Nun
$x^4 + 24x^2 + 16 = (x^2+4)^2 + (4x)^2$
@LeakyNun What to do after this step?
 
did you look at WA?
 
1:15 PM
no
 
look at WA lol
 
@LeakyNun seen, then?
 
then just reverse engineer it lol
 
hmm
@LeakyNun so basically you actually did the differenitation of the antiderivative and checked.
 
I didn't bother to check
I'm way too lazy for that
 
1:22 PM
@LeakyNun Oh okay ...
@LeakyNun But writing that way(referring to your 52 min ago hint) doesn't help (help = 1 second answer), does it?
 
well then you would divide the numerator and the denominator by $(x^2+4)^2$
no it doesn't
 
@LeakyNun oh okay, I could do it that way surely. I was just wonderng if there's some trick or something which I dont know
 
2:17 PM
If you guys have time , could you please look at my question ?
Here is the link : https://bit.ly/2Nxp2Uu
It has been on for 12 days and till now has no answers .
Help would be appreciated . Thank you .
 
It seems to me that $\prod_{i=1}^\infty F_q$ has the cardinality of the continuum for any fixed $q$, where $F_q$ is a field of $q$ elements, but I have two questions I'm less certain about. What about $\prod_{i=1}^\infty F_{p^i}$? Then, I have the same questions for products of countable fields. Can any of these have cardinality different than that of the continuum?
 
This is just about sets, really. Pick a surjection from each set $X_n$ to {0,1}; using choice, $\prod X_n \to \{0,1\}^{\Bbb N} = \mathcal P(\Bbb N)$ is surjective.
On the other hand, $\Bbb N^{\Bbb N}$ has the same cardinality. (You can inject it into the reals using continued fractions.)
 
So even a countable product of $\mathbb N$'s is still the continuum in size?
the continued fractions idea is interesting. I only felt confident about the one I described because I was thinkingof it as decimal expansions
 
2:41 PM
ah, wiki says that $\mathbb Z^\mathbb N$ has continuum cardinality. I suspected as much but was wanted to be careful
 
That's what my proof shows :) it surjects onto P(N) and injects into R, which have the same cardinality via the decimal expansion argument
For continued fractions it is easier to work with positive integers everywhere
 
@MikeMiller Huh, I'd never seen the continued fractions thing, but that makes a lot of sense
 
I'm sure I learned it somewhere on MSE.
 
Everything I know I either learned on MSE or from Number Munchers, there's no in between
 
@rschwieb $2 < \Bbb N < \Bbb R \implies \Bbb R = 2^\Bbb N \le \Bbb N^\Bbb N \le \Bbb R^\Bbb N = \Bbb R \implies \Bbb N^\Bbb N = \Bbb R$
all steps are valid in ZF
and the last step is Schroeder-Bernstein
 
3:01 PM
Oh man Number Munchers
where 1 was prime
@fargle that takes me back.
 
@rschwieb Luckily I didn't learn too much from Number Munchers.
 
If you were real careful, you could exit a square directly into an approaching monster so that you passed through each other
I suppose there is a FPS version of number muchers now
Number snipers?
Number munchers was in color?!!? I had no idea... the screen on the library computer was just black and green.
 
3:29 PM
The actual Number Munchers was a bit too old for my time. We had the slightly more graphically advanced Math Munchers, as well as Math Blaster.
Of course, the name "Math Munchers" set against the name "Number Munchers" just makes me think about a game like, "Alright kids, get ready to blast the aliens using the fundamental theorem of finitely generated abelian groups!"
 
4:04 PM
Hello!
I was learning differential equation and thinking that if there are any shortcuts to find the a differential equation of the given equation.
Or is it possible to eliminate options from several given options to find the differential equation fast
It would be great if anyone help me with this!
 
Where is the question you are talking about ?
 
Please use mathjax. (See the link on the upper right.)
 
@Alphanerd I want to eliminate options here.
@Semiclassical alright!
I thought to eliminate option by $dy/dx$ at $x=0$ from the given equation and from the options.
 
4:27 PM
What I notice is that the solutions tend to contain the ratio y/x. That to me suggests the substitution y=tx
Also, are the options (a)-(c) exclusive?
 
howdy @Semiclassic
 
Hi @ted
 
@Semiclassical well, the answer is given as A which has no $y/x$ ratio
@Semiclassical yes.
 
that’s not really easy to tell given how hard it is to decipher the above
I frankly can’t read what options a-c are supposed to be. Use text, please
 
4:34 PM
(C)$x e^{arcsin (y/x)}=k$
(B)$x+e^{arcsin (y/x)}=k$
@Semiclassical I hope it's clear
(A)$kx=e$
$xdy-ydx=({x^2-y^2})^{1/2}dx$
 
Well, the thing is that option A would mean that x isn’t a function of y. That’s not a problem necessarily, but in that case dx/dy = 0
 
hi, demonic @Alessandro
 
4:49 PM
By compariaon, if you divide your differential equation by dy, you get $x-y(dx/dy)=\sqrt{x^2-y^2}(dx/dy)$
if dx/dy=0, then that becomes x=0. But that’s not in general true for A.
So A is not in general a solution.
 
-3
Q: Limit of a two variable function using the definition

S.ChauhanHow do I find $$\lim_{(x,y) \rightarrow (0,0)} \frac{x^2y}{x^2 + y^2}$$

@TedShifrin How is this problem done? (not my question, just curious)
 
@Semiclassic was right when he said this is a homogeneous differential equation. Let $y/x = z$, or $y = xz$. Then $y' = xz'+z = z + \sqrt{1-z^2}$, so this is separable. You get $dz/\sqrt{1-z^2} = dx/x$. @Jasmine
 
@Semiclassical I haven't found the solution yet . But I was thinking of converting x and y into polar coordinates . x=rcosθ and y=rsinθ. Then we can write xdy-ydx=r^2 dθ and then proceed from there .
 
(I didn't actually say that, but I should have)
@Alphanerd the problem is that what's on the right-hand side is sqrt(x^2-y^2), not sqrt(x^2+y^2)
 
@Abcd: Note that $0\le x^2/(x^2+y^2)\le 1$.
 
4:56 PM
@TedShifrin yes, then?
 
which means that going to polar coordinates isn't necessarily helpful
 
I have answered probably a dozen of such questions on multivariable limits. People are very fond of polar coordinates, but I disparage that unless it's absolutely necessary.
So $|f(x,y)|\le |y|$, @Abcd.
 
@TedShifrin not understanding...Okay leave it ... we dont have multivariable calculus in our syllabus...i just saw this problem and found it interesting...
 
My problem with multivariable limits is that I'm better at proving limits don't exist than proving they do
 
@TedShifrin My question was: How can we show that $0<-\dfrac{x^3}{3}+ \dfrac{x^5}{5}- \dfrac{x^7}{7}...<1$
 
4:58 PM
The former is usually just a matter of finding two ways of approaching (0,0) that don't give the same result
 
If $|f(x,y)|\le |y|$, what happens as $(x,y)\to (0,0)$?
 
The latter is what I don't remember
 
The squeeze principle for limits is universal.
 
that's true
 
@Abcd: What does that have to do with the limit question you asked?
 
4:59 PM
Could you guys look at my question while you are at it ?
bit.ly/2Nxp2Uu
 
I suspect it's his next question
 
LOL
 
@TedShifrin That limits question was not my question. This is my question
 
What do you know about $x$?
 
@Semiclassical no I just found that limits question while scrolling down on Math.SE
$x>0$
 
4:59 PM
When $x$ is near $0$ and positive, you're getting a negative number, so that's wrong.
 
Plus it's not clear what the coefficients are supposed to be
 
Huh?
 
+1/3, +1/5, -1/7, and then...?
 
No, $-1/3$.
 
@Semiclassical -1/3
 
5:00 PM
derp
 
Alternating odd numbers ...
 
this is what i get for leaving chatjax off
 
Anyhow, @Abcd, as stated, it's wrong.
 
@TedShifrin Okay what I am actually trying is: Prove that: $x-\dfrac{x^3}{3}<\arctan x< x$, x>0
I want to prove this using taylor series of arctan x
 
No, don't mess with Taylor series. Just use the first derivative.
 
5:02 PM
@TedShifrin Yes, then its very easy.
But I want to do it using Taylor as well
Because you are literally given first 2 terms of Taylor series of arctan x there
 
on the bottom, yes
but not on the top
 
$$\dfrac{x^5}{5}- \dfrac{x^7}{7}- \dfrac{x^9}{9}... >0 $$
So we have to prove this^
sorry I had wronly added an extra x^3/ 3 before
is it possible to prove this^?
 
So you want $x^5/5 - x^7/7 + x^9/9 -\dots > 0$ and $-x^3/3+x^5/5-x^7/7< 0$.
You had totally the wrong thing.
 
@TedShifrin why do we want the right thing to be less than o?
 
For the second inequality.
 
5:05 PM
because that's what $\arctan x-x<0$ means
 
@TedShifrin Oh yeah
 
Note that the arctan series is only for $|x|<1$. That's crucial.
 
oh
 
Now think about the proof of the alternating series test.
 
What is that?
 
5:06 PM
Leibniz's test?
 
Dont know that
 
Probably worth stressing that arctan(x)<x is true for all x>0, not just 0<x<1
 
If $a_n\ge 0$, then $\sum (-1)^n a_n$ converges if $a_n$ decrease to $0$.
@Semiclassic @Abcd: That's why the first derivative proof, which works for all $x$, is superior.
 
Okay
 
5:07 PM
The proof of that convergence test is very pretty, @Abcd. Best to draw a picture of what the partial sums are doing.
 
@TedShifrin Okay ...
 
I guess another approach would be to use $\tan^{-1}x=\int_0^x \frac{dt}{1+t^2}$ for $x>0$
and then do some shenanigans
 
yes
 
for instance, 1/(1+t^2) < 1 for all t>0
 
Well, that's where the series comes from (for $|x|<1$) :P
 
5:09 PM
so that gives the upper bound immediately
 
Yes, right.
This is equivalent to the derivative proof, of course.
 
Yeah, I buy that
I just like integral representations
 
5
A: How to prove for $x>0$, $\dfrac{x}{1+x^2}<\arctan x<x$?

robjohnWe can use the Mean Value Theorem: $$ \frac{\tan^{-1}(x)-\tan^{-1}(0)}{x-0}=\frac1{1+\xi^2} $$ for some $\xi\in(0,x)$. That is $$ \frac1{1+x^2}\lt\frac{\tan^{-1}(x)}{x}\lt1 $$

 
Obviously, one way to prove that $f(x)<g(x)$ when $x>0$ given $f(0)=g(0)$ and $f'(x)<g'(x)$ for $x>0$ is to integrate :P
 
5:11 PM
@TedShifrin Please explain how he has used Mean Value Theorem here
I have never seen MVT being used that way
 
What do you need explained?
The RHS is $\arctan'(\xi)$.
 
Mostly I like integral representations because they bring contour integrals to mind and those are fun
 
@TedShifrin yes, so?
 
(I'm being a bit silly, but not entirely)
 
This is more fun, @Semiclassic. What's the right picture? :)
So where are you stuck, @Abcd?
 
5:13 PM
...
 
@TedShifrin In the step after that. How has he concluded that inequality?
 
right...fun
 
@TedShifrin yeah, as if derivative must be integrable
 
Because $1/(1+x^2)$ is a monotone decreasing function, @Abcd.
 
@TedShifrin But why should it be less than (arctan x)/x
 
5:14 PM
I guess the picture that comes to mind first is to consider some particular $zw=a$ with $a\neq 0$
 
@Abcd: The original LHS is $\dfrac{\arctan x}x$ and it equals $\dfrac1{1+\xi^2}$.
 
which if I projectivize is just two touching riemann spheres? I feel like I'm saying nonsense
i dunno
 
@Semiclassic: Well, projectivizing isn't going to be helpful. Better to think about removing $z=0$ and what's left deformation retracts to $|z|=\epsilon$, which is a $3$-dimensional cylinder in $\Bbb C^2 = \Bbb R^4$.
 
okaly
 
@TedShifrin derivative of arctan x is 1/(1+x^2) for some $E$ in $(0,x)$
That's what he has proved using MVT
How does he get his inequality from this minor data?
 
5:18 PM
a 3D cylinder as in $S^1\times \mathbb{R}^2$?
(visualizing 3D objects in 4D space is not my forte)
 
What are the minimum and maximum values of $1/(1+\xi^2)$ if $0<\xi<x$, @Abcd?
Yes, @Semiclassic.
 
If I follow that, though, then i guess I'd at least say that the first homology group is just Z.
 
@TedShifrin maximum is 1, minimum is 1/(1+x^2)
 
OK, so you're done, @Abcd.
See my comment 10 lines up.
 
And it's a connected space, so that means H^0 is known by definition (I forget how that works if I'm honest)
 
5:20 PM
Well, but we need to take the union of two such tubes, drilling out stuff near the origin.
 
Back to your research work, Semiclassic :P
 
in truth, my research work right now is supposed to be "research how to get a job"
 
Well, yeah, that's relevant to your life.
 
(I'm not very good at it yet. That's what twenty years of being a student does to you I guess.)
 
5:22 PM
I'm still no good at it. I couldn't even get a volunteer job!! :D
 
I'm finding myself defaulting to educational stuff simply because that's what I know
community college etc
(And there actually is a posting for math adjunct teaching at my old undergrad.)
 
Make sure you play up your computing skills. You should look into consulting stuff. Of course, a lot of this is in the world of investment/finance, which may come crashing down soon.
 
yeah, finance...bleh
 
Well, OK ... what about biotech?
 
@TedShifrin I have certain qualms there, both in terms of the stability of such a job and the social impact of such
 
5:25 PM
tons of math/physics stuff used there.
 
@TedShifrin that's a thought, I guess
 
Sure, I understand.
I've suggested it before. You just ignore me.
 
eh, or I just forget it
 
No, at my age one forgets. At your age one ignores. :P
 
5:27 PM
Well, go look into it :P
 
fair enough
 
@Semiclassical Learn python.
Since abandoning academia, I've worked for a math education software company, a 360-degree video camera company, and now a robotics company. It's been great
 
@rschwieb yeah, that's one thing I keep meaning to do
I downloaded anaconda a while back
 
and I still have time to do math.
That's a good choice if you're on a windows machine
 
but I'm the sort who really needs a project to work on in order to learn a language
 
5:29 PM
me too
build something
 
there is at least one project I could work on
 
After learning the basics, I learned quite a bit by building django websites
 
Oh, a robotics company would be awesomely cool, Semiclassic.
 
which I've tried doing in mathematica but its Fourier transform is kinda annoying
yeah
 
My ring theory database is one project that drove me to learn a lot
 
5:30 PM
QC stuff is another thing I'd find cool
 
Or one of the self-driving car companies ...
Well, get off your duff and start web-searching and résumé-slanting!
 
self-driving car company is a job to aspire to, but I don't know how long they're going to last, or at least how long the current ones are going to last
 
one thing people like to talk up is data science
which...eh. seems fad-ish
 
data science is going to be around to stay for a whiel
2
 
5:31 PM
Yup.
That's a ton of statistics.
 
Yeah, it's a fad, but after all the wannabes drop out, you'll be rolling in it
 
data science is like the new actuarial science
 
No one says this has to be a permanent job for Semiclassic.
 
ironically, my other research-for-fun task has had me learning some stats stuff
nothing deep
 
5:31 PM
Yeah, the beauty of being a mathematician-in-industry is you're usually flexible enough to do tons of different jobs
 
but the Gram matrix stuff connects to correlation matrices and such
 
basic statistics gets you a long way. The rest is probably mostly numerical techinques
 
You can do a ton of stuff with algorithms based on the SVD
I listened a guy at PyCon talk about search algorithms based on SVD, and I was convinced he didn't understand the first thing about what the SVD does, but he understood what it accomplished in searches
 
for context, in quantum information theory one likes to talk about quantum correlation matrices
which basically amounts to being able to factorize such a matrix as $C=A^\top B$
i.e. a matrix whose entries are inner products
(in this case the inner product is just the Euclidean dot product between real vectors. You get to QM by using a generic Hilbert space)
Anyways. You can imagine that the SVD is helpful for the above factorization problem
for better or worse, though, the above has mostly pushed me to learning about more linear algebra/stats than it has python. so not the best project for that. (good thing I have a different one)
@TedShifrin the resume is a big stumbling block for me. I drafted one a while back, but didn't format it
 
vzn
5:57 PM
lol Big Data/ ML/ Deep learning etc are far more than fads, but they do have overhyped elements... just have to separate baby/ bathwater... or signal vs noise :) :P ps recently got repl.it to do scikit-learn code, it wasnt super hard, thought it was awesome :)
 
@Semiclassical yes that's why I thought that how can A be the answer but in my book A is given as the answer
@TedShifrin alright but I thought if A is not that A is not correct
 
well, either the book wrote the wrong answer, or A hasn't been transcribed properly
I have to wonder about the latter, since something like x=ke^(y/x) would fit the pattern of the other answers
as it stands, that "e" in answer A is just sorta sitting there
 
6:14 PM
I got the answer as $e^{arcsin (y/x)}=xk$ and it doesn't match
 
that's the same set I get as well
my guess is that A is supposed to be $x=k e^{\sin^{-1}(y/x)}$
and that for whatever reason the exponent wasn't transcribed
it'd also be consistent with the other two answers, which are similar but not identical to the above
 
6:40 PM
@Semiclassical yes you are correct thank you!
 
7:09 PM
Hey everyone!
 
7:20 PM
1
Q: Multilinear transformations being determined by their values on basis elements

PerturbativeThe following is stated in the book Analysis on Manifolds by James Munkres Just as is the case with linear transformations, a multilinear transformation is entirely determined once one knows its values on basis elements. That we now prove. And then he gives the following lemma. Lemm...

 
 
1 hour later…
8:46 PM
Is power law y = kx^a or y = k(1/x^a) ?
 
9:09 PM
@user963241 both
 
Hey @LeakyNun
 
hi
 
I am reading a paper and they have the following:
$\sum_{m=0}^\infty \left( 3^n b_n + (4 b_n + 2 n c_n) m^{n-1} \right) e^{-\pi m^2}$
n is a natural number and b_n and c_n are just some constants
m is the radius of an n-ball
now they go from this line to the following line
$\le \alpha \sum_{m=0}^\infty \left( 3^n b_n + (4 b_n + 2 n c_n) \right) e^{-\pi m^2/2}$
I understand the exponential /2 part but somehow they pull that m out as a constant, but the sum is dependent on m. How is that even possible?
 
9:25 PM
Hi
Is it true that a bounded Lebesgue null set is also a Jordan null set?
I know that a Jordan null set is also a Lebesgue null set. The opposite is in general not true since there are counterexamples.
 
anyone know graph theory
 
9:44 PM
10
A: Evaluating the integral $I (k)= \int_\infty^0\frac{1}{x^2 +k^2}\ln x ~ dx$

ComplexYetTrivialYou can let $x= k t$ to find $$ I(k) \equiv\int \limits_0^\infty \frac{-k \ln(x)}{k^2+x^2} \, \mathrm{d} x = \int \limits_0^\infty \frac{-\ln(t)}{1+t^2} \, \mathrm{d} t + \int \limits_0^\infty \frac{-\ln(k)}{1+t^2} \, \mathrm{d} t \, .$$ Your substitution $t \to \frac{1}{t}$ shows that the first ...

Wrt this answer, If I try to solve it using Leibniz Rule , I am getting $dI/dk$ to be divergent (after putting x = kt)
So is it not possible to use that method in this question. Any comment?
@LeakyNun Please give it a look if you don't mind
$$\dfrac{dI}{dk}= \int_{\infty}^0 \dfrac{1}{t(1+t^2)}dt $$
never mind then. I will ask on main...
I have done wrong partial differentiation. Got my mistake.
 
 
2 hours later…
11:33 PM
The field of $p$-adic numbers has the same cardinality as the continuum, right?
 

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