Contour Integration Example

[spare shard]

Solve this using Contour Integration:-

$\int_{0}^{\infty}\frac{\sin{x}}{x}dx$

[slate canyon]

AstroVortex

[high kiln]

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[polar ivy]

@spare shard er this is a standard integral
heard of the Borwein integrals?

$\int^{\infty}_{-\infty}\frac{\sin(x)}x dx = \pi$

[slate canyon]

No Lifer#gtfoanemia#ViataGod

[polar ivy]

now,

$\frac{\sin(x)}x$ is symmetrical along the y-axis

[slate canyon]

No Lifer#gtfoanemia#ViataGod

[polar ivy]

$\int^{\infty}0\frac{\sin(x)}x dx = \int^0{-\infty}\frac{\sin(x)}x dx$

[slate canyon]

No Lifer#gtfoanemia#ViataGod

[polar ivy]

$\int^{\infty}_{-\infty}\frac{\sin(x)}x dx = \pi$

$\int^{\infty}0\frac{\sin(x)}x dx + \int^0{-\infty}\frac{\sin(x)}x dx = \pi$

$2\int^{\infty}_0\frac{\sin(x)}x dx = \pi$

yee enough

[slate canyon]

No Lifer#gtfoanemia#ViataGod

[tawdry marsh]

This is a well known dirichlet integral, but he asked for a solution using contour integration

@spare shard do you know the cauchy residue theorem?

[spare shard]

Yeah....

But idk what a residue precisely means

[tawdry marsh]

It just means a value that x or z cannot be

so for example if you have z-i in the denominator z cannot equal i so that's a residue

and you only care about the residues that are a part of your contour

for this actually a residue is at 0, so you can just make a singularity

one sec

but first you have to define your f(z)

[spare shard]

So singularity and residue are the same??

[tawdry marsh]

well basically, but for some integrals using singularitys are better for others using residues

it depends on your contour and residues

[spare shard]

Okay..

[tawdry marsh]

one sec, Imma write something out so you can make sense of it

[safe gulch]

Ah the residue theorem it’s really cool one of my fav theorems imo

[spare shard]

Yeah maybe....

[safe gulch]

Do you know it ?you can use it here to evaluate your integral

[spare shard]

2πi•Res sin x /x but idk how to find the residue

[tawdry marsh]

you can do like this with a singularity in stead of a residue

the the contour integral is =0

and also since sinx is the imaginary part of e^ix you need the imaginary part of the sum of those integrals

[spare shard]

Kay makes sense

[tawdry marsh]

notice that when R->infinity and epsilon->0 those two integrals at the back add up to become an expanded version of sinx/x integral

and you just need to find the imaginary part

[spare shard]

But why did we choose f(z) = e^iz/z?

[tawdry marsh]

because if you take the imaginary part of f(z) you get sinz/z

basically sinx/x because those two integrals are along the real axis

[spare shard]

Okay...

And I had a question

Why do we choose our contour like that? Like we could make it like a semi circle and give it an outward bulge such that it contains the singularity

[spare shard]

Ok I'm getting it

[tawdry marsh]

so the key right now is to evaluate integrals over big and little gamma, then put them to the LHS and find the imaginary part

and you get your answer

[spare shard]

Bro it's midnight Imma sleep now type the solution I'll check it up in the morning

[tawdry marsh]

okay

I'll put it as a spoiler

[spare shard]

Thanks bro

[tawdry marsh]

It's so good right

[spare shard]

Gn

[tawdry marsh]

gn

[indigo apex]

Where do they teach contour integration in high school

[tawdry marsh]

They only teach easy integration here

[tawdry marsh]

nowhere probably

[tawdry marsh]

Do you have any questions from the solution?

[spare shard]

I didnt understand it at all 😭

[tawdry marsh]

Ok, so the integral over big gamma is the integral of f(z) over big gamma right?

[spare shard]

yes..

[tawdry marsh]

Now in the complex realm the contour is going from zero to pi

that's why the integral is from o to pi

and any complex number z that's on the big gamma can be represented as Re^iphi

[spare shard]

ok...

[tawdry marsh]

Because the radius of big gamma contour ir R

then we just make a substitution z=Re^iphi and dz=iRe^iphi dphi

[spare shard]

bro do one thing please, instead of typing the solution give me a book link which teaches complex analysis from scratch

[tawdry marsh]

Oh, so you should just start from there xd

not from integrating using complex analysis xd

ok, serge langs complex analysis ir pretty good

but also it is kinda advanced, but you can definitely learn from that, i think there might be a pdf

I have the book myself

[spare shard]

Oh nice Ill ask for that as my christmas present

thanks bro

and sorry for wasting your time

😭

[tawdry marsh]

It's fine it was fun integrating

[spare shard]

+close

Hey why isnt it closing

[tawdry marsh]

Also youtube videos and watching how people apply different contours to different integrals is very useful