Simply Beautiful Art's realm of integrals

S.C.B.

00:41

Well Hello.

Simply Beautiful Art

hello

So... I must confess, it is easier to prove for $0<a<1$ for division by $0$ reasons

So, I suppose we are going to do this problem with a contour just for fun and your learning?

S.C.B.

Yes. I suppose it's going to be related to this, but I cannot exactly understand that answer.

Simply Beautiful Art

Well, I'm going to choose a different shape for our problem. Still remember what a contour integral sorta means?

S.C.B.

Well, you said that it had to do with something with residues, and integrating over the complex plane, and the arrows became zero or something like that so we had to consider certain points?

Or something along those lines.

Simply Beautiful Art

Yeah, pretty much

S.C.B.

00:46

Pacman.

Simply Beautiful Art

I'm going to use what I'm going to call the Pacman contour

Hahaha

Have your Mathjax set?

S.C.B.

Yes, I managed to add the bookmark.

Simply Beautiful Art

So, $$f(x)=\frac{x^a}{1+x^2}$$

Let $R$ be the radius of the big circle and $r$ be the radius of the small circle along our contour

We also have

$$I_n=\int_{C_n}f(x)\ dx$$

And $C_5$ is the entire path, including all pieces

S.C.B.

What is $I_{n}$ and $C_{n}$? Is $C_{n}$ the shape?

Simply Beautiful Art

$C_n$ is the shape and $I_n$ is the integral along the contour $C_n$.

S.C.B.

00:49

OK.

Simply Beautiful Art

Arrows designate the directions of the integral as well

So, given $r,R$, then $I_1$ is given by... what?

(trust me, it should be your gut feeling. What does $C_1$ represent?

S.C.B.

$(R^2-r^2) \pi$? (Could you clarify what $n$ is, please)?

Simply Beautiful Art

$n$ up there? it's just a place holder for me to put $n=1,2,3,4,5$ in.

$I_1$ should be given by...

S.C.B.

Oh, OK.

Simply Beautiful Art

$$I_1=\int_{C_1}f(x)\ dx=\int_r^Rf(x)\ dx$$

Integrals you know and love with actual bounds instead of weird $C$ things

S.C.B.

00:53

OK.

Simply Beautiful Art

Now, for $I_2$, we use what is called "Jordan's lemma"

Don't freak out, it's incredibly basic and powerful at that

Basically, we use the squeeze theorem

S.C.B.

Jordan's lemma

In complex analysis, Jordan's lemma is a result frequently used in conjunction with the residue theorem to evaluate contour integrals and improper integrals. It is named after the French mathematician Camille Jordan. == Statement == Consider a complex-valued, continuous function f, defined on a semicircular contour C R = { R e i θ ∣ θ ∈ [ 0 , π ] } ...

I'm trying to follow you.

Simply Beautiful Art

Yes...but I'm just gonna teach you the basics real quick

S.C.B.

OK.

Simply Beautiful Art

desmos.com/calculator/fcuncsdbu7

Look at that graph

Consider the integral along that line as $n\to\infty$.

What is the result after $n\to\infty$?

S.C.B.

00:56

$y=0$.

Simply Beautiful Art

Yes

Basically, if the integrand goes to $0$ fast enough, then the limit of the integral is $0$.

S.C.B.

Yes.

Simply Beautiful Art

desmos.com/calculator/nd1psim9kc

What about this one as $n\to\infty$?

What happens to the area?

Slightly harder, huh?

Especially if I don't let you calculate the integral first, then take the limit

In other words, it's a race between how fast the function goes to 0 versus how fast the "length" of the path goes to infinity

S.C.B.

It seems to be increasing, according to calculations.

OK.

Simply Beautiful Art

Well, yes, it is increasing. But that is the rough idea of Jordan's lemma

Now, let's go back to our function

Approximately how fast does the function go to $0$ as $R\to\infty$?

We are looking at $C_2$.

S.C.B.

01:02

I can't understand. Is $R$ fixed?

Simply Beautiful Art

What is $p$?$$\frac{x^a}{1+x^2}=\mathcal O(x^p)$$

S.C.B.

$p=a-2$.

What does $R$ have to do with our integral though?

Simply Beautiful Art

$R$ is the radius of the outer circle

Now, what is the "length" of the contour $C_2$?

S.C.B.

$2 \pi R$?

Simply Beautiful Art

Yup.

Roughly speaking, there is a constant $k$ such that

$$|f(x)|<kx^{a-2}$$

And the integral is bounded:

S.C.B.

01:05

Yes.

Simply Beautiful Art

$$|\int_{C_2}f(x)\ dx|\le|2\pi Rkx^{a-2}|$$

Length times "height" I suppose

Oh, whoops

There we go

Now, what is $|x|$ along $C_2$?

(distance from 0)

S.C.B.

$R$

Simply Beautiful Art

And so it all simplifies nicely:

$$|I_2|\le2\pi kR^{a-1}$$

For every $0<a<1$, as $R\to\infty$, we see then that $I_2\to0$.

S.C.B.

Don't you mean $$Rx^{a-2}$$

Simply Beautiful Art

No, $|x|=R$.

S.C.B.

01:09

Ah.

Simply Beautiful Art

Anyways, we now know that $I_2$ is negligible.

S.C.B.

$|a|<1$?

Yes, OK.

Simply Beautiful Art

Well, yeah. $|a|<1$ is fine.

But for $0<a$, it's easier to see that as the smaller $r$ goes to $0$, $I_4\to0$

I leave it to you to show in the same manner that $I_4\to0$.

Now $I_3$ is a weird little thing IMO

$$I_3=\int_{C_3}f(x)\ dx$$

S.C.B.

@SimplyBeautifulArt Could you clarify the relationship between the circles and $$\int_{0}^{\infty} \frac{x^{a}}{x^2+1}$$?

Simply Beautiful Art

It is almost like $C_1$. Can you guess what the integral comes out to be?

As the big circle tends to infinity, and the small circle tends to $0$, what does $I_1$ tend to become?

(It is an integral from point $r$ to point $R$)

:-|

S.C.B.

01:13

I'm not sure.

Simply Beautiful Art

$$I_1=\int_r^Rf(x)\ dx$$

As $r\to0$ and $R\to\infty$.......

S.C.B.

It is the integral we wish to evaluate?

Simply Beautiful Art

It is.

But not in a manner of your substitutions and differentiations

Now, do you see why the above is true?

I mean, with the $I_1$ and all.

S.C.B.

OK.

Simply Beautiful Art

What is $I_3$ then?

S.C.B.

01:16

Does $I_{3}=I_{1}$?

Simply Beautiful Art

Nope

Arrows for example

S.C.B.

$I_3=-I_1$?

Simply Beautiful Art

Aha!!!!!....nope

I tricked you, it's a bit harder than you might've thought

$$I_3=\int_R^rf(xe^{2\pi i})\ dx$$

The trick here is that it is rotated $2\pi$ radians around, and for non-integer $a$, this makes things interesting

(a great example of why $0\ne2\pi$ in the purest of senses.)

Now, if you plug it all in, we get

$$I_3=-\int_r^R\frac{(xe^{2\pi i})^a}{1+(xe^{2\pi i})^2}\ dx$$

$$=-e^{2\pi ia}\int_r^R\frac{x^a}{1+x^2}\ dx$$

Where I factored that all out and simplified.

$$I_3=-e^{2\pi ia}I_1$$

S.C.B.

OK.

Simply Beautiful Art

At this point, your probably wondering where the trig function in the solution comes in, but don't worry, it's quite a nice surprise at the end if you haven't figured it out

The last challenge is to figure out $I_5$, which is the entire integral from $I_1$ to $I_4$.

We do this with Cauchy's residue theorem.

Cauchy's residue theorem is almost always applicable for closed contours. That is, it has to loop allll the way back to where it started

First, it asks us "for what values of $x$ is $f(x)$ undefined?"

(hint: division by $0$.)

S.C.B.

01:24

$x= \pm i$.

Simply Beautiful Art

As long as $a>0$, we ignore $x=0$.

(we can handle that too, but not now)

The residue here is very simply given:

Residue 1 = $$\lim_{x\to i}(x-i)f(x)$$

Residue 2 = $$\lim_{x\to-i}(x+i)f(x)$$

And $I_5$ is the sum of these times $2\pi i$.

$$I_5=2\pi i(res+res)$$

I have to go now, hope you can figure the rest out

S.C.B.

So $$I_{5}=2 \pi i (\text{Residue 1}+\text{Residue 2})$$ By Cauchy Reisude Theorem?

Simply Beautiful Art

Hint: Complex definition of cosine/Euler's formula for where the trig function comes in

Yup

And...

$$I_5=I_1+I_2+I_3+I_4$$

You may now solve for $I_1$, given all of the above, which is out integral in question

Good night!

Simply Beautiful Art

02:15

@S.C.B. And as a last note, I usually find this more "elementary" than real methods.

For example, you didn't tell me why the Weierstrass product equaled the integral form of the gamma function

And it also extends to more general problems without acknowledging special functions

Simply Beautiful Art

13:16

@S.C.B. if you want, I can give you an integral that is nearly impossible to solve without contours, but easily solved

S.C.B.

15:01

@SimplyBeautifulArt Tell me more.

Simply Beautiful Art

@S.C.B. About what?

I mean, I can give you a problem to try out

S.C.B.

Could you tell me the problem?

Simply Beautiful Art

$$\int_{-\infty}^\infty\frac{1-x}{1-x^3}\ dx$$

Solve this using contours integrals

(first, decide what shape. The $-\infty,+\infty$ should tell you at least one part of the shape)

@S.C.B. When you think you have the right shape, tell me

4

Q: Calculate $\int_0^\infty {\frac{x}{{\left( {x + 1} \right)\sqrt {4{x^4} + 8{x^3} + 12{x^2} + 8x + 1} }}dx}$

Prove $$I=\int_0^\infty {\frac{x}{{\left( {x + 1} \right)\sqrt {4{x^4} + 8{x^3} + 12{x^2} + 8x + 1} }}dx} = \frac{{\ln 3}}{2} - \frac{{\ln 2}}{3}.$$ First note that $$4{x^4} + 8{x^3} + 12{x^2} + 8x + 1 = 4{\left( {{x^2} + x + 1} \right)^2} - 3,$$ we let $${x^2} + x + 1 = \frac{{\sqrt 3 }}{{2\c...

S.C.B.

Wait a moment.

Simply Beautiful Art

Also, you can now answer questions like this

S.C.B.

15:08

I'm somewhat preoccupied.

Simply Beautiful Art

ok

S.C.B.

I'm doing a difficult inequality.

Simply Beautiful Art

15:36

Seems to be an extremely hard inequality. @S.C.B.

S.C.B.

See math.stackexchange.com/questions/2127470/…

I don't like using Lagrange Multipliers, so I am trying to manipulate the condition.

Simply Beautiful Art

mhm....

S.C.B.

The condition is vexing.

Simply Beautiful Art

I imagine

S.C.B.

Usually you establish a bound on $xyz$, but no bound can be found.

Simply Beautiful Art

15:45

@S.C.B. well, good luck

Simply Beautiful Art

16:25

@DHMO scroll to top for solutions to the integral

DHMO

ok

S.C.B.

@SimplyBeautifulArt Well, here again.

Simply Beautiful Art

@S.C.B. ok

So, we have...

$$\int_{-\infty}^\infty\frac{1-x}{1-x^3}\ dx$$

@S.C.B. ideas on the shape?

Simply Beautiful Art

16:57

Man, you've been busy @S.C.B.

S.C.B.

A circle? I don't know.I am going to look up more information.

On contour integrals

What makes you think I have been busy?

Simply Beautiful Art

@S.C.B. rep

Think about it. You need a line going from negative infinity to positive infinity, and some other line to finish the shape

@S.C.B. btw, someone (the one currently in this room you don't know) wants to do Discord chat with you.

Transcript for

$Mathematics$ Simply Beautiful Art's realm of integ…