Simply Beautiful Art's realm of integrals

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dydxx

00:00

Could you do this step by step lol...

isn't the first integral a standard tan integral

arctan*

Simply Beautiful Art

Pretty much

$$\int_{-\infty}^{+\infty}\frac1{x^2+a^2}\ dx=\sqrt{\frac\pi a}$$

From arctan

Take the derivative with respect to a and you get

$$\int_{-\infty}^{+\infty}\frac{-2a}{(x^2+a^2)^2}\ dx=-\frac12\sqrt{\frac\pi{a^3}}$$

Let $a=1$ and divide both sides by $-2$ to get

$$\int_{-\infty}^{+\infty}\frac1{(x^2+1)^2}\ dx=\frac{\sqrt\pi}4$$

Done, and we never directly evaluated that integral

dydxx

ooo

interesting stuff

Simply Beautiful Art

(also use this technique for series problems of course)

My favorite example!

$$\int_0^\infty\frac{\sin(x)}x\ dx$$

Consider the harder integral:

$$\int_0^\infty\frac{\sin(x)e^{-xa}}x\ dx$$

dydxx

how did u know to put a random

e^-xa

Simply Beautiful Art

Differentiate with respect to $a$, solve, then integrate with respect to $a$.

@dydxx you know, practice, recognition, all that

dydxx

00:10

lol

could u go through that one

Simply Beautiful Art

Well, after differentiating, we get

$$-\int_0^\infty\sin(x)e^{-xa}\ dx$$

$$=\frac{-1}{1+a^2}$$

(integration by parts)

Integrate: $$\int\frac{-1}{1+a^2}\ da=C-\arctan(a)$$

As $a\to\infty$, our original integral goes to zero, hence $C=\frac\pi4$.

At $a=0$, we finally get $$\int_0^\infty\frac{\sin(x)e^0}x\ dx=\frac\pi4-\arctan(0)=\frac\pi4$$

Or maybe it was $\pi/2$. I don't remember

dydxx

:35589905

Simply Beautiful Art

dydxx

where did the 1/x goes

wait nvm

chain rule

got u

okay

woah pre COOL

Simply Beautiful Art

:P Tricky integration techniques. Never clear when and how to apply XD

Oh, wait, $C=\frac\pi2$, so the integral is $\frac\pi2$.

:P Close enough

dydxx

00:21

hmm

any other cool integral techniques

Simply Beautiful Art

152

Q: Really advanced techniques of integration (definite or indefinite)

Okay, so everyone knows the usual methods of solving integrals, namely u-substitution, integration by parts, partial fractions, trig substitutions, and reduction formulas. But what else is there? Every time I search for "Advanced Techniques of Symbolic Integration" or "Super Advanced Integration ...

dydxx

alright

thanks

Simply Beautiful Art

:P 'advanced techniques'

dydxx

I wanted cool ones :((

Simply Beautiful Art

@dydxx Cauchy's integral formula?

(it's a special case of the residue theorem)

Or Glasser's master theorem?

dydxx

00:26

never heard of that thing...

Simply Beautiful Art

$$P.V.\int_{-\infty}^{+\infty}f(x)\ dx=P.V.\int_{-\infty}^{+\infty}f\left(x-\frac1x\right)\ dx$$

dydxx

what integral

could I use this on

Simply Beautiful Art

I have no idea

dydxx

alright

how do you find school work there

Simply Beautiful Art

I've personally never applied Glasser's master theorem yet

school work?

dydxx

00:29

I assume you're top of your class or something

Simply Beautiful Art

Math, physics, sure

not really history

dydxx

You have friends with similar math knowledge

to yours

Simply Beautiful Art

dydxx

feels

Simply Beautiful Art

lonely?

dydxx

00:32

I don't know as much as you however it math class was quite boring in last year highschool

Since I knew the whole syllabus inside out

and I tried talking to my calculus teacher about taylor series and complex analysis stuff

but he didn't know that much...

let alone anyone in my class haha

Simply Beautiful Art

Lol, same

heh. I'm in Calculus II!

XD And I'm kinda chilling in there

dydxx

They offer Calc 2 in highschool?

Simply Beautiful Art

yeah?

dydxx

sorry not sure about usa curriculum

since i live in NZ

Simply Beautiful Art

At least, AP/advanced placement

I heard

dydxx

00:33

thats what AP stands for

i see

Simply Beautiful Art

I could've skipped the class I'm in two years ago

but its the last math class they offer, so they won't let me

dydxx

oh rip

i imagine the integrals in calc 2 are a piece of cake :P

Simply Beautiful Art

:P of course

Occassionally I do some magic

dydxx

Such as?

Simply Beautiful Art

Differentiation under the integral

Perhaps instead of apply integration by parts

(I actually did the entire integration by parts chapter without integration by parts XD)

dydxx

00:39

Explain further?

Simply Beautiful Art

$$\int x\cos(x)\ dx$$

Instead, take $\int\sin(xt)\ dx$ and take the derivative with respect to $t$.

dydxx

xcos(xt)

Simply Beautiful Art

And set $t=1$ to get our integral

This is trivial since $\int\sin(xt)\ dx=\frac{-\cos(xt)}t$

$\frac d{dt}\frac{-\cos(xt)}t=\frac{xt\sin(xt)+\cos(xt)}{t^2}$

No integration by parts

dydxx

have you used this in a test

lol

Simply Beautiful Art

>.>

Not on the tests

dydxx

00:46

thought so

LOL

Simply Beautiful Art

but none of my homework ever used integration by parts

Not to mention, for higher derivatives, I just pull stuff in like the general leibniz rule for nth derivatives of products and stuff

So my answer to $\int x^5 e^x\ dx$ is given in summation form XD

dydxx

how so?

Simply Beautiful Art

in This is the Realm of Simply Beautiful Art, 14 secs ago, by Simply Beautiful Art

I actually got to go, I'll be back later

$\int e^{xt}\ dx$

The solution is a product, and differentiating a product 5 times...

just look up general leibniz rule and you will see

dydxx

yeah i get that but just wondering how you would x^5e^x in summation form

Simply Beautiful Art

01:09

@dydxx Note that:

$$\int e^{xt}\ dx=\frac1te^{xt}$$

Treat it as a product and apply general leibniz rule:

$$f=\frac1t,g=e^{xt}$$

dydxx

yup

Simply Beautiful Art

$$\frac{d^5}{dt^5}(fg)=\sum_{k=0}^5\binom5kf^{(k)}g^{(5-k)}$$

Easy enough to show that $f^{(k)}=\frac{(-1)^kk!}{t^{1+k}}$ and $g^{(k)}=x^ke^{xt}$

See how crazy magical I am? XD

I'm not sure what my Calculus teacher thinks of me

@dydxx Feel like you see integration in a whole new light?

dydxx

I don't see f^{k} = ((-1)^k * k!))/(t^{1+k}) and g^{k) = ...

yeee integration is quite beautiful

Simply Beautiful Art

Prove each by induction

They hold true for $k=0$.

Simply Beautiful Art

01:32

Yeah, kinda just scroll up and stuff

2 hours ago, by Simply Beautiful Art

Have you tried differentiation under an integral?

@S.C.B. Start there

S.C.B.

Simply Beautiful Art

Click the "2 hours ago"

1 hour later…

dydxx

02:39

Im back

S.C.B.

03:00

@dydxx So you are.

13 hours later…

Simply Beautiful Art

16:27

@dydxx hey!

@S.C.B. :P

S.C.B.

$:\mathscr{P}$.

Now it looks uglier, @SimplyBeautifulArt

$:\mathcal{P}$.

That looks better.

$:\mathcal{PPPP}$.

Simply Beautiful Art

Ok lol

4 hours later…

dydxx

20:32

good mornig

Simply Beautiful Art

20:55

@dydxx good day

dydxx

21:17

what up

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

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