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2:46 AM
Unfreeze!
 
@PeterTamaroff Well blow me away! A person! here...in chat!
2
 
@amWhy Heh.
=)
 
:D
 
@amWhy What are you up to?
 
@PeterTamaroff Not too much; may make it to bed on the earlier side. It's so incredibly slow on main!
 
2:54 AM
@amWhy True. Been fiddling around with Bessel functions myself.
 
 
3 hours later…
5:40 AM
@anon are u there ?
 
depends on my interest
 
Orthogonal Trajectories
interesting ?
 
go on
 
I am still on approach 1
where they calculate gradient
$f(x)$ is the family of curves for whom we want to find orthogonal trajectories
what is $g(x)$
@anon
 
f(x) is not technically a family of curves
it's a function that takes in a point and spits out a value
the curves are the level-curves of f(x) (aka level-sets or many other things)
 
5:48 AM
oh ok ..
 
so g(x) is another function we wish to find, with the property that f(x)'s level curves intersect g(x)'s level curves at right angles (wherever they do intersect)
for example, let f(x,y)=x^2+y^2
the level curves are of the form x^2+y^2=A, i.e. circles
 
maybe I will have to do the practical approach section :)
 
and the level curves of g(x,y)=y/x are nonvertical lines through the origin
which intersect these circles (centered at the origin) at right angles
 
So.. the algorithm will be the same in case of complex functions, too ??
I mean, I have to do this for curves in complex plane. I thought learning this on real curves would be easier :)
@anon
 
I mean, I have to do this for curves in complex plane It doesn't say that in the article. Is there more context behind your questions?
 
5:53 AM
none
just complex numbers.
 
why do you believe that you "have to" do it with complex numbers? certainly there is not a gun to your head.
 
Yes, there is. University is holding it. They will make sure I flunk in the exam :D
 
aha! so there is context behind your questions. you have somehow been assured that you need to learn about orthogonal trajectories in the context of complex analysis because that topic may be on a test
 
but so far complex numbers were an 'extension'. What applied to reals applied to complex and reals are easier to learn
Yes.
:-p
 
but you will need to be more specific about what you mean about "the case of complex functions," since they map the plane to other points in the plane, rather than the plane to real values (so you cannot define level curves in the same sense). it is true that the real and imaginary parts will each be real-valued functions of the plane whose level sets form orthogonal trajectories, though
 
5:58 AM
I have no idea what the uni means by orthogonal trajectories. They just put it under the chapter of complex analysis. What do u think I should learn ? Your best judgement ?
:)
@anon
 
really, all you need to know is the "case" with real-valued functions, and how that relates to complex variables (the relation that I just mentioned, real and imaginary parts)
for example, see "4. Orthogonal trajectories and harmonic functions" here. it's like not even a whole page of material.
 
6:38 AM
Hi @BrianM.Scott how are you?
 
7:06 AM
@anon A complex function is analytic if it satisfies the Cauchy-Reimann equations.. right ? :)
 
mmhmm
 
That's all I have to do to prove a function analytic ? If it does not satisfy the equations, it is not analytic
@anon
 
yes, a function f:U->C (where U is an open subset of C) is complex-analytic at z (z in C) iff f's partial derivatives at z satisfy the cauchy-riemann equations
 
woohoo !!! one more topic done !!
Next: harmonic functions
@anon what is a level set
 
If f:A->B is a function, a level set is a subset of A of the form {a from A such that f(a)=b} for some (arbitrary) element b from B
so like I mentioned earlier, the level sets of f(x,y)=x^2+y^2 are of the form x^2+y^2=k, i.e. circles centered at the origin
 
7:16 AM
ohh.. so if I plug in a from A then I should get some element from B
 
"if I plug in a from A then I should get some element from B" describes any function f:A->B before even beginning to talk about level sets (aka level curves, fibers, many other names)
 
 
3 hours later…
10:08 AM
2 hours later...
1 hour later...
:D
 
11:00 AM
at 11:00AM GMT...
 
@skullpatrol D
 
11:26 AM
@N3buchadnezzar wazzup?
 
MAthz
 
just chillin' killin'
 
11:42 AM
@skullpatrol trippin, sippin, slippin
 
Yo @amWhy wazzup?
 
12:17 PM
Wazzahhhhhhhhap!?
just chillin......
 
cool
 
12:53 PM
Does anyone have a hint for why for all $(m+2\le p \le 2m+1)$, $p| \binom{2m+1}{m}$?
'hint for'? Perhaps 'hint about why' would have been better.
 
@Alyosha
you there?
 
Hello
 
Alyosha, @Ethan may be able to answer your question...
Hello, @Ethan
 
@amWhy hey
@Alyosha basically your asking why $\binom{2n}{n}$ is divisible by every prime between $n$ and $2n$
 
It's probably something to do with Pascal's tringle mod $p$
 
1:02 PM
no
well i wouldn't say no
maybe if you tryed hard enough you could find a relation
nvm anyway
write out $\binom{2n}{n}$ in terms of factorials as,
You get $\frac{(2n)!}{(n)!^2}$
do you follow?
 
Yes
 
ok (2n)! is divisible by every prime between 1 and 2n yes
 
Yes.
 
we also have (2n)!/(n!)^2 is clearly an integer sense its a binomial coeiffient
 
Thanks, I understand now.
 
1:06 PM
which means all the prime factors in n!^2 must cancle with factors in (2n)!
 
I was confused by the later division by $n+1$, but then realised $p\ge n+2$
 
but n!^2 doesn't have any prime factors between $n$ and $2n$, its only divisible by the primes between 1 and n
 
Thanks for the poke in the correct direction
 
so the remaining integer must be divisible by them all
ye
This fact can be used to show the asymptotic order of the prime counting functions
Basically you can show $$\pi(x)/(\frac{x}{\ln(x)})$$ is bounded between constants
only using the previous fact and some partial summation
erdos also used it in his elementry proof of bertands postulate
@amWhy what kind of math do you study?
 
That's where I found this
 
1:09 PM
@Alyosha I have answered a question related to this in which I derived the asymptotic order of $\pi(x)$ only using this, if your interested
 
@Ethan All "kinds". I particularly like logic and abstract algebra.
 
Sure
 
@amWhy lol I waste so much printer paper, I usually just erase my work and write over it instead of getting new paper, there are eraser crumbs all over my desk
do you write on slate?
@Alyosha here alyosha, sec
 
hahahaha. I understand. I go through erasers as quickly as I go through pencils!! No, I don't use slate, though I really should think about a portable black-board (or whiteboard)...but I like my "fine point" pencils.
 
lol my mom got me a whiteboard for christmass, she had this guy hang it on the wall, I never use it
Its to big of a pain in the ass to stand up and write
I am usually reading and studying while I write
ye I like my pencils to
I have a pretty big callous on my right hand lol
@Alyosha actually mind if I ask what your studying right now
 
1:19 PM
@Ethan Basic complex analysis, some number theory, physics olympiad problems and x-ray crystallography (which means Fourier transforms later). Right now I'm doing number theory. You? A
 
lol from number theory to x-ray crystallography? are they related? or are you just doing a bunch of different stuff
 
@Ethan since
 
A bunch of different things
 
why?
 
Wait, will return in 10 minutes
 
1:23 PM
@Ethan (Nice explanation by the way) =)
 
lol
how about you peter what are you studying
analysis?
 
@Ethan Mostly, yes. Reading Polya and Szego, learning from the pros =).
Gotta run!
Sawry.
 
alright later
 
@Ethan The Crystallography's for a school project, and there's (from my still fairly ignorant viewpoint) a lot of overlap between (analytic) number theory and complex analysis
Although really, I'm just sampling different areas of maths and physics and sticking with what seems interesting. What are you studying?
 
1:39 PM
guys in p->q , how can we determine if p is true or false?
 
Anyway, thanks for the help
 
@PeterTamaroff what is polya
 
@Alyosha sorry, was busy im pretty much just studying number theory and bits of analytic number theory, hopefully I will have time to study some enumerative combinatorics before my break is over, I have only been studying mathematics really on my own time for about 2 or 3 years now.
 
Hello There, could someone please explain to me what the Cesàro summation means
 
is a special summation technique used to 'assign' values to divergent series
 
1:46 PM
@Ethan: I don't understand how one "assigns" a value to a divergent series
 
i am not an expert on the topic, but I think sometimes there is ambiguity when one has a function represented by some expression valid for one interval, and then sees it being expressed in this matter for things not in that interval
 
..that is very ambiguous
 
like it would make sense to say $$\frac{1}{1-2}=-1$$
but not $$-1=\frac{1}{1-2}=1+2+2^2+2^3+2^4...$$
The function $$f(x)=\frac{1}{1-x}$$ Has a maclaurin series representation that converges to f(x) for all $|x|<1$
 
@Ethan are u math undergradiate?
 
no, im still in high school
@Nick and people could clearly write $$f(\frac{1}{2})=1+\frac{1}{2}+\frac{1}{2^2}...$$
But for short hand if $f(x)$ didn't have a simple representation as $$\frac{1}{1-x}$$, one might write
Say, $-1=1+2+2^2+2^3..$
this is a bad example actually lol, you wont see someone writing this, because it has a simple representation
take the zeta function
$$\zeta(s)=\sum_{n=1}^\infty\frac{1}{n^s}$$
$\zeta$ can be defined this way for all $s$ in a specific interval
but for certain $s$, this expression doesn't converge to the value of $\zeta$ as we have defined it
I can't really talk freely on this, I don't really know enough about it
A keyword is analytic continuation
 
1:53 PM
@Ethan What are you using to learn (analytic) number theory?
 
@Alyosha Tom Apostol - Introduction to analytic number theory
And 'a freindly introduction to number theory' by silverman
though I wouldn't really reccomend the latter
I am only using it, because I have it in book
The first one by apostol is the good one, though I only have it on pdf
 
@Ethan: Thank you so so very much.
@EthanYou've really helped me
 
i doubt it lol, there are a variety of summation techniques though i know of
'ramanujan summation'
'cesario summation' as you said
I know this comes up somewhere in physics lol
and I think I have seen it in asymptotic analysis
 
I've yet to see it in physics but I think I can face it when it comes
 
and there are probably a variety of other uses, I don't know to much complex analysis ( or analysis in general ) so I don't typically deal with this sort of thing
@nick en.wikipedia.org/wiki/Zeta_function_regularization, "An example of zeta-function regularization is the calculation of the vacuum expectation value of the energy of a particle field in quantum field theory. More generally, the zeta-function approach can be used to regularize the whole energy-momentum tensor in curved spacetime."
has some more stuff about physics, summation methods, and the zeta function in particular
Though I barely know some basic mechanics, so I can't really say anything about the physics stuff lol
@Alyosha have you learned about euler products?
 
2:00 PM
You mean the prime-number expression of $\zeta(n)$?
 
yes, thats one euler product
 
Yes
 
$$\zeta(s)=\prod_{p}\frac{1}{1-p^{-s}}=\frac{1}{1-2^{-s}}\frac{1}{1-3^{-s}}\frac{1}{1-5^{-s}}\frac{1}{1-7^{-s}}...$$
 
@ethan: It is is intriguing to read. Thank you.
 
in general if $f(xy)=f(x)f(y)$ when x and y are coprime, then
$$\sum_{n=1}^\infty f(n)=\prod_{p}(1+f(p)+f(p^2)+f(p^3)+f(p^4)...)$$
2
ignoring convergence lol
 
2:03 PM
@Ethan is this directed at me or are you continuing the answer to Nick?
 
it was directed at you, lol
have you learned about arithmetic functions?
 
For any arbitrary function?
No
 
that gets interesting
lol
 
How are they useful?
 
do you know the divisor function is?
 
2:05 PM
Yes
 
the divisor sigma function
$$\sigma_x(n)=\sum_{d\mid n} d^x$$
the sum of the xth powers of the divisors of n?
 
Indeed
 
and you know its representations, in terms of the factorization of n, $$\sigma_x(n)=\prod_{p\mid n} \frac{p^{x(v_p(n)+1)}-1}{p^x-1}$$
 
How does that equation take into account factors that aren't prime?
 
what do you mean?
$v_p(n)$ is the exponent of the prime p in the factorization of $n$
traditionally called the p adic order of $n$
you can expand the inside expression
so that
 
2:10 PM
Sorry, I thought it was a $\Sigma$.
Yes, the formula makes sense
Seems very similar to the proof of the Euler product
 
ye there is a direct link between the euler product representation of dirichlet series' for multiplictive functions and sums over there divisors,
in general if $f(xy)=f(x)f(y)$ when x and y are coprime, ie f is multiplictive
$$\sum_{d\mid n} f(d)=\prod_{p\mid n}(1+f(p)+f(p^2)+f(p^3)...f(p^{v_p(n)}))$$
 
Is $f$ an arithmetic function?
 
ho ho ho santa is here
 
ye any multiplictive function so that the domain of $f$ is the integers and its range is the complex numbers
anyway well it turns out there are some very nice identities that turn out from the theory of modular forms, like $$\sum_{k=1}^{n-1}\sigma_3(k)\sigma_3(n-k)=\frac{\sigma_7(n)-\sigma_3(n)}{120}$$
 
anyone into trig today ?
 
2:15 PM
I quite like spherical trigonometry
 
@Alyosha I know nothing about that.
I asked a question about sin
on main that is.
 
@Alyosha this also ventures into the theory of elliptic functions which is very beautiful, and which I know nothing about, things like $$\sum_{n=1}^\infty\frac{1}{e^{\pi n^2}}=\frac{\pi^{1/4}}{2\Gamma(3/4)}-\frac{1}{2}$$ amaze me
 
I yearn for the day when that equation seems a trivial case.
 
Which is a jacobi theta function identity $$\theta_3(q)=\sum_{n=-\infty}^\infty q^{n^2}$$
 
@Ethan You're doing very well if this is a plug for Apostol. Do you read it in class?
 
2:20 PM
@Alyosha no I read on my own time lol, most of this is from wolframalpha/wikipedia/ varrious blogs
@Alyosha The fastest known algorithms for calculating varrious constants like $\pi$ use the theory of elliptic functions
Ramanujans formula for $\frac{1}{\pi}$ is an example of one
if youve seen it
 
I haven't
 
$$\frac{1}{\pi}=2\frac{sqrt{2}}{9801}\sum_{n=1}^\infty \frac{ (4n)!(1103+26390n)}{n!^4396^{4n}}$$
 
Anything formula with Ramanujan in it fills me with dread
 
If you calculate the first term, its already correct to 6 decimal places
 
Have elliptic functions anything to do with elliptic integrals?
 
2:24 PM
yes
@Alyosha the series is related to heegner numbers, which are related to several mathematical 'coincidences' like the almost integer, $$e^{\pi\sqrt{163}}=262537412640768743.999999999999..$$
though I don't know much about this either lol
 
What's the name of the multiplicative-$f$ theorem?
 
@Alyosha this sort of thing eisenstein series/ elliptic functions etc are definitely topics I want to study, though I am lacking in complex analysis and a variety of other fields commonly used in there study
@Alyosha which theorem?
 
Seems abstract algebra is also heavily interlinked
The starred one
 
@Alyosha have you studied any combinatorics or possibly a little basic set theory?
 
$\Sigma f(n)= \Prod \Sigma f(p^x)$
 
2:32 PM
yes
i know
its not really considered a theorem, more so just common knowledge
i can give you a short proof
 
I'm happy with the proof
 
do you know any combinatorics or a little set theory?
 
Yes
 
do you know what it means to take the Cartesian product of two sets
 
Yes
 
2:36 PM
alright you know that, $$(\sum_{a\in A}a)(\sum_{b\in B}b)(\sum_{c\in C}c)(\sum_{d\in D}d)...=\sum_{(a,b,c,d,..)\in (A\times B\times C\times D\times...)}abcd...$$
where the rhs sum runs over all the tuples (a,b,c,d...) that are formed by the cartesian product of all the previous sets A,B,C,D,..
 
It seems true by inspection, yes
 
Well the fundamental theorem of arithmetic says any integer $n$, can be written in a unique way as a product of prime powers $$n=p_1^{a_1}p_2^{a_2}p_3^{a_3}...=\prod_{k=1}^\infty p_k^{a_k}$$, where only a finite number of the $a_i$ are non zero, and for $n=1$, we can take them all to be zero
If two integers $n$ have the same factorization they are the same, and every integer has a factorization like this
 
I've been aquainted with that fact
 
If $f$ is multiplictive we can write
$$f(n)=\prod_{k=1}^\infty f(p_k^{a_k})$$
each exponent is a positive integer
Now lets expand the product $$\prod_{p}(1+f(p)+f(p^2)+f(p^3)...)=\prod_{p}(\sum_{j\in \mathbb{Z}_+}f(p^j))$$
$$=(\sum_{j\in \mathbb{Z}_+}f(2^j))(\sum_{j\in \mathbb{Z}_+}f(3^j))(\sum_{j\in \mathbb{Z}_+}f(5^j))(\sum_{j\in \mathbb{Z}_+}f(7^j))...$$
$$=\sum_{(h,i,j,k,...)\in \mathbb{Z}_+^\infty}f(2^h)f(3^i)f(5^j)f(7^k)...$$
$$=\sum_{(h,i,j,k,...)\in \mathbb{Z}_+^\infty}f(2^h3^i5^j7^k...)$$
And this sum runs over all the distinct tuples of exponents in the product of prime powers where each is a positive integer
but every integer corresponds to one of these tuples, because every integer can be written as a product of primes, with exponents being positive integers
 
Wait, you did know I was happy with the proof of that theorem? I just asked so I could find more examples of special cases.
 
2:48 PM
lol
ye substitute say $$\sigma_x(n)$$
@aloysha $$\sum_{n=1}^\infty\frac{\sigma_x(n)}{n^s}=\prod_{p}(1+\frac{\sigma_x(p)}{p^s}+.‌​..+\frac{\sigma_x(p^3)}{p^{3s}}...)=\zeta(s)\zeta(s-x)$$
 
Wow
 
For example $$\sum_{n=1}^\infty \frac{d(n)}{n^s}=\zeta(s)^2$$
@Alyosha theres a certain type of inverse integral transform, that involves complex analysis, and can be used to gives estimates on $$\sum_{n\leq x} f(n)$$ From $$\sum_{n=1}^\infty \frac{f(n)}{n^s}$$
 
I don't see the final equality $=\zeta(x)\zeta(s-x)$
 
@Alyosha google.com/… is the transform, basicly if $$F(s)=\sum_{n=1}^\infty\frac{f(n)}{n^s}$$ then $$\sum_{n\leq x}^* f(n)=\frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} F(z)\frac{x^z}{z} \ dz$$
The star on the sum indicates that the last term of the sum must be multiplied by 1/2 when x is an integer.
 
Surely it depends on which contour you integrate over?
 
2:55 PM
If x is not an integer it works fine, and gives the normal $$\sum_{n\leq x} f(n)$$
@Alyosha the formula requires that $F(s)$ be absoloutely convergent for $\Re(s)>c$
but other then that you can chose whatever c you want
@Alyosha these techniques were used in the first proof of the prime number theorem, and are still typically used for giving proofs of the prime number theorem
 
Do you know how complex analysis can be used to evaluate real integrals if the contour integrated over affects the integral?
 
sorry I don't really know that much, its not my specialty lol
 
I'm probably being especially dim today, but I still don't see $\approx$ 5 equations ago how you got $\zeta(s) \zeta(s-x)$
 
@Alyosha basicly in the proof of the prime number theorem, using this transform you can give an exact formula for the number of primes less then or equal to x, and a variety of other interesting functions, in terms of the zeros of the zeta function, the good thing is that the main term in most of these is a real well behaved expression like for $\pi(x)$ its $\text{Li}(x)$ the logarithmic integral
@Alyosha In the starred expression on the right substitute $$f(n)=\frac{\sigma_x(n)}{n^s}$$
And notice that for prime powers $p^j$
$$\sigma_x(p^j)=\frac{p^{x(j+1)}-1}{p^x-1}$$
then sum up the factors in the euler product using the geometric series formula, and simplify with a little bit of algebra
note that $$\zeta(s)=\prod_{p}\frac{p^s}{p^s-1}$$
so, $$\zeta(s)\zeta(s-x)=\prod_{p}\frac{p^{2s-x}}{(p^s-1)(p^{s-x}-1)}$$
@Alyosha one of arguably the most famous problems in mathematics is the reimann hypothesis which says the real part of all the non trivial zeros of the reimann zeta function will always have a value of $\frac{1}{2}$, if true this would give much better estimates then the best known now for $\pi(x)$ and other prime counting functions, it would also solve a bunch of other open problems in number theory and I think other fields to
though this is again getting pretty out of my scope of knowledge
When I say non trivial zeros, I am referring to zeros that have an imaginary part not equal to zero.
 
Intuitively, why would they be on $\Re(s)=\frac{1}{2}$?
Or does a Perelman
 
3:12 PM
@Alyosha i am not sure I don't know much about the zeros of complex functions, this would be something I imagine covered in a text on complex analysis, I know however that there are many intuitive arguments for why it should be $\frac{1}{2}$ , and equivilent statements to the rh (short for reimann hypothesis)
@Alyosha here en.wikipedia.org/wiki/Riemann_hypothesis, on the same page if you scroll down it shows the explict formula for a varient of the prime counting function, and gives information on how its derived
Along with Reimann's century and a half old original paper written in german I think
 
I don't know anything about the Mobius function
 
you will learn about it when you study arithmetic functions (if you do lol)
its a useful arithmetic function
Though as I said again practically all of this is way out of my reach, as I have only been studying for a very short while, and haven't ever taken a course on or read anything thorough on analysis.
 
Know anything about random walks?
 
arithmetic functions, some basic stuff on euler products, congruences, estimates certain sums I know more about that sort of thing
 
I've always preferred integrals to sums
 
3:20 PM
@Alyosha not really, but I remember reading somewhere that the summatory function for the mobius function (the mertens function its called), behaves like a random walk or somthing about it behaves that way, whatever that means lol
@Alyosha Reimann steiljes integrals are used in number theory sometimes to calculate sums
 
Good, here's a fun problem that won't seem trivial then http://www.physics.harvard.edu/academics/undergrad/probweek/prob4.pdf
it looks interesting
Or perhaps it will seem trivial, I don't know
 
not sure, lol not really my thing, I think there should be some powers of $2$ and maybe some factorials for the first question though
 
Neither was I
Sure, that is
 
3:44 PM
I'm off now, bye
 
alright later
 
4:49 PM
hi
 
5:08 PM
@DominicMichaelis $\sup$?
 
@PeterTamaroff hi
Did you see my question ?
 
@DominicMichaelis Nope.
 
@PeterTamaroff here
 
@DominicMichaelis Heh, no clue on that.
 
@PeterTamaroff got and answer and I am ashamed of how easy it is
 
6:02 PM
Heya
 
@robjohn
You derped on an answer, sire!
 
robjohn does not "derp"
 
@skullpatrol le proof
 
In university I learned that every complex integral is a path integral
 
@N3buchadnezzar Isn't that the way they are defined?
 
6:15 PM
Does that mean that $\int_\mathbb{R} e^{-ix^2} \mathrm{d}x$ is really a path integral? (As in a simple one, straight line $y = 0$)
 
Complex integrals are defined as "path integrals" over a curve $\gamma$.
 
@peter you can also integrate over a subset of the complex plane
 
It always bugged me when we use treat integrals with complex variables as real integrals.. Must be something I am missing somewhere.
 
Using the Riemann-Stieltjes integral IIRC, one defines $$\int_\gamma f =\int_{\gamma(a)}^{\gamma(b)}f(\gamma)d\gamma$$
 
Yeah, as in a curve from a to b along gamma.
 
6:20 PM
@N3buchadnezzar Right.
@DominicMichaelis Like double integrals?
 
@PeterTamaroff something like this, but a Integral over an subset is not necesarrily a double integral
 
Measuretheory !
<3
@PeterTamaroff What fields do you study?
 
@N3buchadnezzar "Fields"?
Analysis.
Next four months I'll be taking Analisis II and Linear Algebra.
 
heh, cool.
I will be taking Quantum mechanics I and Statistics for the next few months.
 
@N3buchadnezzar But now I am studying from Polya and Szego.
@N3buchadnezzar QM? Dope?
 
6:30 PM
Dope?
I have already taken some analysis, two courses in basic analysis and an intermediate course in functional analysis. Looking forward to learning more.
Thinking about reading rudin next summer.
 
@N3buchadnezzar Baby Rudin? (I guess not?)
 
Maybe, I will try the red(?) one first and if it is too dry I will try the baby version
 
6:45 PM
Guys, is there any software that can solve differential eqns ?? :)
 
@LittleChild Wolfram.
@N3buchadnezzar Ah.
 
@PeterTamaroff cool !
@PeterTamaroff I am unable to get it to work -_- I am unable to integrate:
$\int_{-\pi}^{\pi}t^2.e^{-int}$
help ?
 
@PeterTamaroff I derped?
 
@LittleChild Integrate by parts a few times.
 
I am doing it manually. :)
 
6:57 PM
@robjohn Yes, kinda. First, the integral blows up at $x=1$; not $x=0$.
 
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