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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

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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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>>6713376

Doesn't work for k=6

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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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>>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

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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

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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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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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>>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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>>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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>>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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>>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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>>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

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>>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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>>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?

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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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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

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>>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.

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>>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

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>>6718307

In his case, what happens to the -u^2 in the numerator?

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>>6718309

Magic. Or algebraic manipulation. You choose.

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>>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.

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>>6718318

/sci/'s latex is lame.

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>>6718318

Thanks.

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