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You are currently reading a thread in /sci/ - Science & Math

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I noticed that \int_0^1 \frac {dx}{(1+x^2)(1+x^k)} is always \frac{?}{4} \forall k \in \mathbb{R}. Why is that? Also, does it really happen for all k? Is there a proof?
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>>6713376
bump
>>
Can't believe that all the shitposts receive all this attention and a genuine problem like this doesn't.
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>>6713404
bump
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I don't know the answer, but I would like to know
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>>6713376
But for k=1, it is \frac{1}{8}(\pi + log4))
(Hope, LaTeX work here)
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>>6713431
$\frac{1}{8}(\pi + \log(4))$
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>>6713435
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>>6713376
Doesn't work for k=6
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>>6713439
>>6713444
It works for both.
>>
if I'm not mistaken, your claim is pretty far from being true.

You could try to differentiate w.r.t. k under the integral.
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>>6713435
>>6713431
on /sci/ we use [ math ] and [ /math ] tags (without the spaces) for inline latex and [ eqn ] and [ /eqn ] for block latex. Double click on any math to see it's latex. The formatting is all done with jsmath so not all latex stuff is supported.

\frac{1}{8}(\pi + \log(4))
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>>6713449
No, neither works for k=1 to 10.
Seems like it converges to pi/4 with x->infinity
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>>6713452
Ah, thank you very much!
\frac{\pi}{\pi} [\math]
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>>6713455
Ah shit :D
\frac{\pi}{\pi}
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>>6713453
"that's clear"
because if 0<x<1, then x^9000 is close to zero and you're left with integrating over 1/(1+x^2), which is one of the nice definitions of pi.
(In fact, there is a wikipedia page listing integrals which evaluate to pi. can't find it right now, sadly.)
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>>6713457
Yea, that confuses me too. It makes sense though since other boards have stuff like [spoiler] tags and [code] tags.

I'm going to write moot an e-mail asking him to make it so that hitting ctrl+m generates the tags around highlighted text (spoiler tags work the same way on other boards via ctrl+s, try it on /co/).
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OP here, my bad. the top limit of the integral is \infty not 1.
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>>6713469
Here is the correct picture.
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>>6713473
bump. noone is interested now?
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>>6713473
Integration by parts gives a contradiction after one step.

You're really not trying are you?
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>>6713501
I'm enjoying this game of "how many times can I fuck up the LaTeX and delete it before nobody notices".
I converted it to this form thinking it might be easier.
\int_0^\infty \frac{sech(x)}{1+sinh(x)^k} dx

I don't have any other ideas
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>>6713502
is it possible you mean -inf to inf
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>>6713502
According to mathematica, it always gives out \frac{\pi}{4} for any k. Dunno what you mean.
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>>6713531
yet again
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Where are the poles? All on the unit circle... suspect easy to do contour integral along semi-circle and get result. Details left to reader.
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>>6713560
What do I do with k?
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>>6713563
Evaluate the residues at the poles. The poles are at +/- i, and at
z=exp(i*pi*(2*n+1)/k); for n=0,1,...,k-1.
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>>6713579
That is way beyond my knowledge. Isn't this solvable any other way?
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>>6713580
I transformed it into this form, it may be easier to solve. \int_0^{\infty } \frac{2 e^x}{\left(e^{2 x}+1\right) \left(2^{-1} \left(e^x-e^{-x}\right)^1+1\right)} \, dx
>>
fuck. This is the right form;

\int_0^{\infty } \frac{2 e^x}{\left(e^{2 x}+1\right) \left(2^{k} \left(e^x-e^{-x}\right)^k+1\right)} \, dx
>>
For the case where k is divisible by 4, half the roots of z^k+1 are
strictly above the x-axis and half below. The ones above come in
pairs of the form a+ib and -a+ib, with b>0 and a^2+b^2=1. The residues
of the integrand at these pairs are additive inverses and so sum to
zero. This leaves the residue of 1/((z-i)*(z+i)*(1+z^k)) at z=i,
which is 1/((i+i)*(1+1))=1/(4 i) = -i/4. Integration over the curved
part of a semicircle clearly goes to zero for large radius. Also, the
integral over [0,\infty] is half of the integral over the whole line.
Thus the integral we want is 2*pi*i*(-i/4)*1/2 = pi/4, which is what we
wanted.
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Put x=tan(t)
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>>6713865
and then use the a+b-x property for integrals. Literally this simple.
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>>6713865
get
\int 1/(1+tan^k(t)) dt ? Then what?
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>>6713851
This is correct.
The only way to solve these kind of problems (at least that I know of) is using complex integration. So you have to brush up on that first if you aren't familiar with it/
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Google pink monkey definite integral properties and use 12 from there.Then add both of expressions u have for instant ans(pretty common trick)
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>>6713976
\int_0^{\pi/2} (1+\tan^k(t))^{-1} dt
=\int_0^{\pi/2} (1+\tan^k(pi/2-t))^{-1} dt
=\int_0^{\pi/2} (1+\cot^k(t))^{-1} dt
...
then a miracle happens
...
pi/4
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>>6713976
>>6714031
>>
OP here. So now I have to prove that \int_0^{\pi/2} (1+\cot^k(t))^{-1} dt=\frac{\pi}{4} \forall k \in \mathbb{R}. Is this at least simpler?
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>>6714132
Can you show how you worked it into this form?

I feel like I'm missing something here.
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>>6714250
A guy showed me a property I could use earlier in this thread; \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a-b+x) dx.
After reaching \int_0^{\pi/2} (1+\tan^k(t))^{-1} dt with the substitution x=\tan(t), we apply the above property and we get \int_0^{\pi/2} (1+\tan^k(t))^{-1} dt
=\int_0^{\pi/2} (1+\tan^k(pi/2-t))^{-1} dt
=\int_0^{\pi/2} (1+\cot^k(t))^{-1} dt
which is equivalent to the first integral. So we now have to show that the last one is always \frac{\pi}{4} to prove the first one.
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>>6714280
The property is \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx, sorry for typing it wrong.
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>>6714282
>>6714280
This is pretty cool, thanks. I was looking at this problem and I think maybe contour integration could be applied to this.

The only problem is that the function becomes odd with odd values of k, and I would not know what contour to use for that.
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>>6713560
Can't do a semicircular contour here, for odd values of k there is a pole at 0.

You'd have to do a semicircular contour with an indent at z=0.
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>>6714294
Read the thread mate. We want to avoid contour integration, we already solved it with complex analysis.
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>>6714303
Thanks.
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\displaystyle{\int_{0}^{\infty}\frac{dx}{\left ( x^2+1 \right )\left ( 1+x^r \right )}=\int_{0}^{1}\frac{dx}{(x^2+1)(x^r+1)}+\int_{1}^{\infty}\frac{dx}{(x^2+1)(x^r+1)}}, substitute y=\frac{1}{x}.
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>>6714319
\int_{0}^{\infty} \frac{dx}{\left ( x^2+1 \right ) \left ( 1+x^r \right )}= \int_{0}^{1} \frac{dx}{(x^2+1)(x^r+1)}+ \int_{1}^{\infty} \frac{dx}{(x^2+1)(x^r+1)}
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>>6714319
>>6714322
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>>6714301
There is no pole at zero for any k. There is a pole at z=-1 for odd k.

The case where k is a multiple of 4 is here
>>6713851

I'm not seeing how the a+b-x and tan(t) thing is going anywhere.
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>>6714280
Add the two integrals, simplify, and then you get integral of 1. Since this is twice your wanted integral, the answer follows
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>>6714322
This works. Brilliant!

Do the second integral first:
x=u^(-1)
dx = - u^(-2) du
\int_1^0 -u^(-2) / (1+1/u^2) /(1+1/u^k)
=\int_0^1 1/ (u^2+1) /(1+1/u^k) du
=\int_0^1 1/ (x^2+1) /(1+1/x^k) dx

Now add the fist integral in
\int_0^1 1/ (x^2+1) /(1+1/x^k) dx
+ \int_0^1 1/ (x^2+1) /(1+x^k) dx
=\int_0^1 1/(1+x^2) [via a miracle of simplification]
=arctan(1)-arctan(0)
=pi/4
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>>6714322
>>6714280

I posted the one solution, but this tan^k one got my attention. You guys found anything to do with that?
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>>6714358
Oh, missed the cot^k part because jsmath fucked up, but I would reach there anyway. Still, got anywhere with that?
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>>6714358
>>6714360
see
>>6714338
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>>6714361
I posted the solution in >>6714322 lol.
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>>6713851
How do I into residues?
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>>6714411
I'll try to explain it for simple poles and contours. For short, a residue is a value at a pole (spot where f(z) divides by 0.). It can be calculated as follows:

Res(f, \; z_0) = \displaystyle \frac{1}{(n-1)!}\lim_{ z \rightarrow z_0} \displaystyle \frac{d^{n-1}}{dz^{n-1}}((z-z_0)^nf (z))

A contour integral can be expressed as the sum of the residues inside the contour times 2?i:

\displaystyle \oint_C f(z) dz = 2 \pi i \left ( \displaystyle \sum_{k=0}^{n} Res(f, \; z_k) \right )

Where zk is the location of a pole.

For the evaluation of real valued functions we need to use clever contours, I'm not going to go to into detail on that but some can be viewed here: http://en.wikipedia.org/wiki/Methods_of_contour_integration

One of the very important integrals to take note of is the integral on the Methods of contour integration page under Direct Methods, Cz1dz , this is an important result in complex analysis as it is where the 2?i coefficient comes from.

Look up things like Cauchy's integral formula, residue calculus, complex analysis, Laurent series, residue theorem to get a better idea.
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>>6714453
Has anyone ever told you "I love you"? Because I love you. I got some intuition and you spent time to type it clearly and informatively.
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>>6714461
Sarcasm?

I'm terrible at explaining things, but I try.
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>>6714461
I can give you an example problem if you want.
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>>6714467
It wasn't sarcasm, I really liked your attitude and answer. And an example would be great, because I didn't understand what we already know, what we have to find and what's our solution yet.
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>>6714456
What's n in the first formula, and how is someone supposed to sum all these residues when each one would take a lot of time to find?
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>>6714477
>and how is someone supposed to sum all these residues when each one would take a lot of time to find
That's the shitty part.

The n in the first equation is the order of the pole. It's basically the power of the spot where f(z) has a pole.

For example f(z)= \displaystyle \frac{x^2}{(z^2+1)^3} has a pole of order 3 at z=i.

I guess I'll show this problem as an example: \displaystyle \int^{ \infty}_{ - \infty} \displaystyle \frac{x^2}{(x^2+1)^3} dx
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>>6714491
That x in the first equation should be a z.
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>>6714366
Impressive... and worth TeXing...

Break up the integral as

\int_{0}^{\infty} \frac{dx}{\left ( x^2+1 \right ) \left (x^k+1 \right
)}= \int_{0}^{1} \frac{dx}{(x^2+1)(x^k+1)}
+ \int_{1}^{\infty} \frac{dx}{(x^2+1)(x^k+1)}
.

Then using x=u^{-1},\ dx = -u^{-2} du, the second integral
becomes

\int_{1}^{0} \frac{-u^2 du}{(u^{-2}+1)(u^{-k}+1)}
= \int_{0}^{1} \frac{dx}{(x^2+1)(x^{-k}+1)}
.

Adding the two integrals together then gives

\int_{0}^{1} \left( \frac{1}{(x^2+1)(x^{k}+1)} +
\frac{1}{(x^2+1)(x^{-k}+1)} \right)\,dx

=\int_{0}^{1} \frac{1}{x^2+1} \left( \frac{1}{x^{k}+1} +
\frac{x^k}{x^{k}+1} \right)\,dx

=\int_{0}^{1} \frac{1}{x^2+1}\,dx
=\tan^{-1}(1)-\tan^{-1}(0)=\pi/4.
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To evaluate \displaystyle \int^{ \infty}_{ - \infty} \displaystyle \frac{x^2}{(x^2+1)^3} dx we will observe the following equation and contour in the complex plane:

\displaystyle \oint_{C} \displaystyle \frac{x^2}{(x^2+1)^3} dx

The contour C is shown in pic related. In this contour we will let R \rightarrow \infty to enclose the entire real axis. It is important to notice that as R \rightarrow \infty the semicircular part of C goes to 0 only leaving the straight part in the sum, this is why C was chose as our contour.
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>>6714509
So 2 poles at +-i, to evaluate the integral we'll need 3 cuz n=3. Where is the other pole?
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>>6714508
dang it...

\int_{1}^{0} \frac{-u^{-2} du}{(u^{-2}+1)(u^{-k}+1)}
= \int_{0}^{1} \frac{dx}{(x^2+1)(x^{-k}+1)}
.
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>>6714509
That contour integral should be:
\displaystyle \oint_{C} \displaystyle \frac{z^2}{(z^2+1)^3} dz
I think I'm going to do the rest in word and post it as an image because I keep fucking up the LaTeX.

>>6714513
Yes. The only poles are at +-i, I didn't show the other in the image because its not enclosed in the contour, so it's not important.
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>>6714518
latex to gif

http://www.codecogs.com/latex/eqneditor.php
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>>6714523
This will work.
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>>6714518
I got residues: Res(f, -i)= \frac{i}{16}, Res(f, i)= \frac{-i}{16}. But summing these would give 2?i*0 which is counter intuitive. This is my first time evaluating this in my life, I might have screwed up big timr.
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>>6714528
I got -i/16 too, so the integral is
2 pi i (-i/16) = pi/8.
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>>6714538
Don't I have to sum all the residues? they cancel each other...Assuming I'm wrong(which is really highly likely), and the answer is ?/8, what does it mean? Graphically I mean.
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>>6714528
Those are incorrect. I will post a rough outline of the solution using Cauchy's integral formula, then I will post the full solution that I have written out if anyone wants it.

Here's the basic outline:

It's important to note that \displaystyle \frac{z^2}{(z^2+1)^3} = \displaystyle \frac{z^2}{(z-i)^3(z-(-i))^3} .

Using Cauchy's integral formula (http://en.wikipedia.org/wiki/Cauchy's_integral_formula) with our contour we get:

\displaystyle \oint_C \displaystyle \frac{ \left ( \displaystyle \frac{z^2}{(z+i)^3} \right ) }{(z-i)^3} dz = 2 \pi i \displaystyle \frac{f''(i)}{2!}

Where f(z) = \displaystyle \frac{z^2}{(z+i)^3}

Upon evaluation, one finds that f''(i) = \displaystyle \frac{-i}{16}

On second thought you guys not need to see the full solution because you've already got the gist of it.

>>6714528
The residue at -i is not in the sum as it is not enclosed in C.
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>>6714543
Ah fuck, I messed up the last line, but you get the idea.
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>>6714543
>Those are incorrect
I thought that said i/18 and -i/18, fml.

>>6714528
Those are correct.
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>>6714544
I guess you meant f''(i) = \frac{-i}{16}. I do get what you mean and this seems pretty interesting, thought I have simply no idea how can these formulas work, so I'll have to study them.
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>>6714546
>thought I have simply no idea how can these formulas work, so I'll have to study them

Yeah, studying their origins/derivations will help a lot. They are incredibly useful.

The conclusion for this problem is \displaystyle \int^{ \infty}_{ - \infty} \displaystyle \frac{x^2}{(x^2+1)^3} dx = \frac{ \pi}{8} as shown by the residue theorem.

There are a few free complex analysis books on the web that you can download as PDF format, just search 'complex analysis book pdf' and stuff like (http://math.sfsu.edu/beck/papers/complex.pdf, http://www.math.uiuc.edu/~jpda/jpd-complex-geometry-book-5-refs-bip.pdf, http://www.math.wustl.edu/~sk/books/guide.pdf, http://elibrary.bsu.az/azad/new/2556.pdf) comes up. I never took complex analysis, but I used these to teach myself the subject.

Some interesting problems can be found here: http://residuetheorem.com/
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>>6714546
Going back at the original problem, we have +-i poles from x^2+1, how do I find the poles of z^k+1 and how many are all of them together? Also, what's different when I integrate from 0 to infinity?(in example we integrate from -inf to inf)
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>>6714556
I've been trying to tackle this and I have not yet found a general solution with complex integration.
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>>6714557
poles would be \pm i, (-1)^{1/k}, but the residues get fucked up by the k.
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>>6714567
Yeah, that (-1)^{1/k} makes things messy. I'm out of ideas for the night and I'm going to turn in. If I get any ideas later I'll post it here if the thread happens to be up.
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>>6714542
Only one pole (at z=i) is in the semi-circle. So don't add in the other residue.
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>>6714572
The poles are here:
>>6713579
>The poles are at +/- i, and at
>z=exp(i*pi*(2*n+1)/k); for n=0,1,...,k-1.
You only need the ones with non-negative imaginary part. Unfortunately, sometimes exp(i*pi*(2*n+1)/k) can equal i or -1, and that will complicate things.

When k is divisible by 4, this doesn't happen and you can get
>>6713851
>>
>>6714456
>http://en.wikipedia.org/wiki/Methods_of_contour_integration
Can anyone tell me why on example III
\displaystyle \frac{dz}{dt} = iz,\, dt = \frac{dz}{iz}

And what are they doing when they consider the singularities at 3^{-1/2i}, -3^{-1/2} and not the others?
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>>6714625
z = e^(it)
dz/dt = i e^(it) = i z

The other singularities are outside the unit circle, which is the contour in the example.
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>>6714625
In example III they let z = e^(it). dt = dz/(iz) follows from this.
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>>6714633
>>6714635
Thanks.

Is there any particular reason for choosing the unit circle?
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>>6714651
Problem III is an integral where t runs from -pi to pi. If z=exp(i t), then z runs around the unit circle as t runs from -pi to pi. It's not like one of the integrals from 0 to infinity in that sense
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>>6714672
I find that to be very clever.
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>>6714681
I like studyimg maths too. All the examples given on contour integration ITT are trivial and direct applications of the theorens. Smart would be someone to solve OP's problem with contour integration.
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>>6713503
How did you manipulate it into this?
>>
Let w=exp(i*pi*(2*n+1)/k) for some n=0,1,...,k-1, and assume 1+w^2 is not zero. Then 1+w^k=0 and
f(z) = 1/( (1+z^2)(1+z^k) ) has a simple pole at w. The residue is
Res(w)
= lim_{z->w} 1/(1+z^2) * (z-w)/(1+z^k)
= 1/(1+w^2) * lim_{z->w} (z-w)/(1+z^k)
= 1/(1+w^2) * lim_{z->w} 1/(k z^(k-1) [L'Hopital's rule]
= 1/((1+w^2)*k*w^(k-1))
= 1/k * w^(1-k)/(1+w^2)
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>>6715090
How do you apply Cauchy's theorem after that?
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>>6715189
Draw a big semicircle. Add the residues therein.
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>>6715841
what
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>>6715881
Like in this image >>6714509
>>
yeah this is the kind of problems that should be solved with the residue theorem.
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>>6715841
When k is odd the integrand is not an even function and it's not clear you can use a semi circle
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>>6716697
Good point. Semi-circular contours are out of the question in this context.
>>
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>>6714508
Could somone elaborate on how the integral on the right is yielded from the integral on the left?
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>>6717883
Should be u^-2 in the numerator, multiply the numerator by u^2 and you get the expression on the right
>>
>>6713851

When k is even and *not* divisible by 4, residues inside the large semi-circle in the upper half plane still cancel, except at what is now a double pole at z=i. The residue at this double pole is still -i/4 however, and so you still get pi/4.

So that completes the result for all even k. For odd k though, I don't see a contour integral way to get the answer.
>>
>>6717914
I'm still lost as to how the x^-k ended up on the right.
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>>6718263
he just switched back to x instead of using u to bring the integrals together
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>>6718288
Then shouldn't u^-k have turned into x^k?
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>>6718300
\int_{a}^{b} f(x) dx = \int_{a}^{b} f(t) dt
>>
>>6718307
In his case, what happens to the -u^2 in the numerator?
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>>6718309
Magic. Or algebraic manipulation. You choose.
>>
>>6718309
u^{-2}=\frac{1}{u^2} , \frac{1}{u^2(\frac{1}{u^2}+1)(u^{-k}+1)} multiply (\frac{1}{u^2}+1) with u^2 and tada!. Can't believe I latex pre-high school stuff.
>>
>>6718318
/sci/'s latex is lame.
>>
>>6718318
Thanks.