- Empty

Quick little thing explaining how to get the gamma function integral @astrophysics

- chestercat

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- Empty

Ok solve this integral, a is a constant with respect to x. $$\int_0^\infty e^{-ax}dx$$

- Astrophysics

Wait wait do this

- Astrophysics

|dw:1437992640744:dw| i saw this and wanted to do it

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## More answers

- Empty

Yeah I know, trust me

- Empty

lol

- Astrophysics

by parts

- Astrophysics

ok

- Astrophysics

ima do it then

- Empty

By parts is one way to do it, but this way is much cooler

- Empty

Hahaha trust me you will be doing the same thing but cooler

- Empty

don't you wanna be cool

- Empty

all the kids are doing it

- Empty

it's just one integral

- Empty

one can't hurt

- Empty

Woah this is OPEN STUDY not PEER PRESSURE ANSWER

- Empty

c'mon what's the integral of e^-ax already hurrrrry

- Empty

ty

- Astrophysics

|dw:1437993094157:dw| I forgot to take the limit the first time ok, relax it's been a while since I've done an improper integral

- Empty

haha yeah no problem this looks perfect!

- Astrophysics

Jk I did one earlier...|dw:1437993189968:dw|

- Astrophysics

|dw:1437993463484:dw|

- Empty

Boom ok perfect now we have this nice looking thing:
$$a^{-1} = \int_0^\infty e^{-ax}dx$$
This is where the magic happens, this is the cool way. We differentiate both sides WITH RESPECT TO A. This is weird if you're not used to it, so it'll be good practice. If you don't make a mistake doing this, then you will be officially better than me cause I messed this up the first time I saw it lol.

- Astrophysics

I am better

- Empty

lolol woahhhh champ

- Empty

I was gonna give you a medal + fan but now I'm considering giving it to Jhan

- ParthKohli

JUST DO IT.

- Jhannybean

|dw:1437993589194:dw|

- Astrophysics

I take that back, please I need my medals and fans ok leggo

- Empty

lolol

- Astrophysics

I tried putting the a over the t JB

- Empty

Parth gets the medal

- Empty

sry

- Empty

I haven't got all night

- ParthKohli

It's because I'm better than him.

- Empty

I'm gonna pull a Michele_Laino on you if you don't hurry up

- Astrophysics

Go do your discriminants and quit discriminating against me so I can do this

- Astrophysics

|dw:1437993707615:dw|

- Astrophysics

Differentiate both sides you say?

- Empty

yeah with respect to a

- Empty

and with respect to me, your sensei

- Astrophysics

|dw:1437993761110:dw| I'm better

- ParthKohli

FLAWLESS VICTORY

- Jhannybean

Why take the limit when you can differentiate - his logic.

- Empty

|dw:1437994060905:dw| You evaluated the left side, but not the right side and now Parth is enjoying his birthday wish pssssh

- ParthKohli

Yes, my birthday wish was a free medal. <3

- Astrophysics

That was the right side

- Astrophysics

|dw:1437994295728:dw|

- Empty

$$\frac{d}{da} \int_0^\infty e^{-ax}dx = \int_0^\infty \frac{\partial}{\partial a} (e^{-ax})dx$$

- Astrophysics

Yeah ok I see what you're doing, you want me to do use leibniz rule pretty much and not just plug and chug haha

- Jhannybean

defining a is indeed a constant and not just another variable

- Empty

a is not a constant, it is just an independent variable from x

- Astrophysics

|dw:1437994507765:dw| oOoOoOo

- Empty

Constant with respect to x though, you could say that, also https://www.youtube.com/watch?v=Lcw6xBaCaXM

- Empty

Boom good job, ok now we have something good

- Jhannybean

Ah, I see.

- Empty

\[a^{-1} = \int_0^\infty e^{-ax}dx\]
\[a^{-2} = \int_0^\infty xe^{-ax}dx\]
...?
Keep going do this 2 or 3 more times until you get the picture.

- Empty

differentiate with respect to a I mean haha

- Astrophysics

Integrate it, or keep taking derivatives?

- Empty

keep taking derivatives :D

- Astrophysics

K good lol

- Astrophysics

|dw:1437994844847:dw|

- Empty

noice

- Astrophysics

I think we can write a series for this...

- Empty

no don't!

- Empty

just keep taking derivatives

- Jhannybean

how do partials work again? \(a^{-2} = \int_0^\infty \color{red}{x}e^{-ax}dx\) work again? The highlighted red portion I mean...

- Astrophysics

Ok I'll take one more lol

- Empty

x is a constant when we're differentiating with respect to a @Jhannybean since a and x are independent of each other nothing happens

- Jhannybean

Oh that's right,

- Jhannybean

I kept reading \(\frac{ \partial}{\partial x}\) instead of \(\frac{\partial}{\partial a}\) and I was confusing myself. -_-

- Astrophysics

Taking derivatives in integrals is amazing, I saw this technique once before, did not know you would treat it like a constant in the integral, awesome.
|dw:1437995191725:dw|

- Astrophysics

Oki, what's next

- Empty

Do one more!

- Astrophysics

|dw:1437995573741:dw|

- Astrophysics

|dw:1437995632368:dw|

- Empty

\[a^{-1} = \int_0^\infty e^{-ax}dx\]
\[a^{-2} = \int_0^\infty xe^{-ax}dx\]
\[6a^{-3} = \int_0^\infty x^2e^{-ax}dx\]
\[24a^{-4} = \int_0^\infty x^3e^{-ax}dx\]
...?

- Empty

What are we doing again

- Empty

something something gamma is the continuous version of the factorial something something whatever I already forgot

- Astrophysics

Ok I see, there's a factorial relationship, so there's a reoccurrence relationship.

- Empty

Reoccurrence relationship?

- Empty

\[a^{-1} = \int_0^\infty e^{-ax}dx\]
\[a^{-2} = \int_0^\infty xe^{-ax}dx\]
\[6a^{-3} = \int_0^\infty x^2e^{-ax}dx\]
\[24a^{-4} = \int_0^\infty x^3e^{-ax}dx\]
...?
This is the pattern, what's the general form?

- Astrophysics

Haha recurrence*

- Empty

No recurrence relation here either

- Empty

Whoops I had been writing it wrong lol oh well don't look at what I write you wrote it yourself correctly anyways
\[a^{-1} = \int_0^\infty e^{-ax}dx\]
\[a^{-2} = \int_0^\infty xe^{-ax}dx\]
\[2a^{-3} = \int_0^\infty x^2e^{-ax}dx\]
\[6a^{-4} = \int_0^\infty x^3e^{-ax}dx\]
\[24a^{-5} = \int_0^\infty x^4e^{-ax}dx\]
...
\[\frac{n!}{a^{-???} }= \int_0^\infty ???dx\]
Here's my hint, what the hell is this last thing!

- Astrophysics

(n-1)!

- Empty

Stop looking at what it should be and look at what it actually is, forget the gamma function for a minute

- Empty

Just find the truth don't worry about being right right now

- Empty

idk check it: Here's what you KNOW
\[2a^{-3} = \int_0^\infty x^2e^{-ax}dx\]
try out your false formula to find out how to fix it

- Astrophysics

Yeah I tried, that didn't work..one sec let me go over it again..6 am xD

- Empty

Pythagoras didn't have anyone to check his work I think on the whole a^2+b^2 thing lol so you're at least fortunate :P

- Astrophysics

\[\frac{ n! }{ a^{-(n+1)} } = \int\limits_{0}^{\infty} x^{n+1} e^{-ax} dx\]

- Astrophysics

I tried by plugging in integers haha

- Empty

Known: \[\frac{2!}{a^3}= \int_0^\infty x^2e^{-ax}dx\]
Yours:
\[\frac{ n! }{ a^{-(n+1)} } = \int\limits_{0}^{\infty} x^{n+1} e^{-ax} dx\]
Plugged in n=2
\[\frac{ 2! }{ a^{-3} } = \int\limits_{0}^{\infty} x^{3} e^{-ax} dx\]
Same or not?

- Astrophysics

x^n but the known should be a^-(...) ye?

- Empty

???

- Empty

Sorry I don't know what you're saying X_X

- Astrophysics

Oh no I was just looking at when you said \[\frac{n!}{a^{-???} }= \int\limits_0^\infty ???dx\]

- Astrophysics

\[\frac{ n! }{ a^{-(n+1)} } = \int\limits_{0}^{\infty} x^n e^{-ax} dx\]

- Empty

better but why are you just going with my mistake?

- Empty

like negative in the denominator is like "double dividing"

- Astrophysics

Haha, I get scared so I thought it was right

- Astrophysics

That works out

- Astrophysics

Tried n = 2

- anonymous

My gawd this is a long question

- Empty

Hahaha ok that's what I like to hear, everything leading to this point doesn't really matter now, it's like the ladder we can kick out from under us and if you like you can pretend we were never wrong on the path to this.
@isaac4321 yeah we should probably have flushed halfway through for courtesy :(

- anonymous

It's all good at least your gettin edumacated

- Empty

tru

- Astrophysics

|dw:1437998982910:dw|

- anonymous

Um ok

- Astrophysics

Wow, empty, this happens every time spend so much time on one thing..then...yeah it doesn't matter xD

- Astrophysics

You're like a real prof

- anonymous

Who is that

- Astrophysics

google your name sir

- anonymous

Ok I thought it was him

- Empty

ok so now you can derive the gamma function, you're on your own, you can think more about why this works

- anonymous

👍

- Empty

It should be really easy

- Empty

and if it isn't, wellllll think about it until you do

- Astrophysics

Ok so I don't really know what exactly the gamma function is, I mean is it actually the thing I posted above, I thought that was just a problem that was a gamma function haha..well that's cool, learned so much

- anonymous

Obviously it is very easy 😉

- Astrophysics

|dw:1437999236907:dw|

- Empty

Yeah, the gamma function is basically defined as \[\Gamma(n)=(n-1)!\] for positive integers but extends further.

- Empty

do you see where the n-1 is coming into play now?

- Astrophysics

Yeah I think so

- Empty

ok

- Empty

|dw:1437999792138:dw|

- Astrophysics

lmao

- Astrophysics

Howd you get so many medals

- Empty

it's a thumbs up not a pile of poop for the record

- Astrophysics

Why do people always observe you teaching me

- Empty

they're creepers

- Empty

creepin'

- Empty

idk because this site is boring normally lo

- Astrophysics

So I don't really see where exactly to use this function, like would you use it in QM, looks like it may have to do with statistics/ probability.

- Empty

Yeah I have used it in QM which is a statistical theory, so yes to both. You can use it anywhere you would use a factorial.

- Astrophysics

There's also the beta functions...

- Empty

Yeah beta functions are just like the continuous binomial coefficients

- Astrophysics

Well, did a lot of math today, I didn't think this would take this long, was about to go to bed before the post came up xD

- Empty

Yeah if you want some exercises go find the pdf online of Mary Boas book Mathematical Methods in the Sciences, there's a whole section on gamma functions and beta functions, fun exercises really nothing too hard that you could work through to get some basic exposure to finding stuff that fits the form for integrating.
Consider them new closed form solutions to some integrals that you can get to from doing u-substitution and stuff

- Astrophysics

Sounds like fun, thanks, I think we should also start working on some relativity/ QM that would be fun, I'm not so good at finding general forms and such, unless I plug integers in xD, and understanding the theory without it being of use really...

- Astrophysics

as far as this math goes

- Astrophysics

I should say I'm not used to it*

- Empty

yeah actually the QM book we can start going through, the guy reccomends Boas' book in the beginning. It's called Griffith's Intro to QM.

- Astrophysics

The one with the cat

- Astrophysics

Schrodingers cat :(

- Empty

Yeah that one

- Empty

lol

- Astrophysics

I like how it starts with schro eq

- Empty

Yeah it's basically like starting a classical physics class with F=ma

- Astrophysics

Ah so this means avg < > you know what I mean xD

- Empty

Yeah, expected value means average, $$\langle x \rangle$$

- Empty

Standard deviation of some value is \(\sigma_x\) and satisfies this nice equation (which is very nicely shown and explained I thought in his book)
\[\sigma_x^2 = \langle x^2 \rangle - \langle x \rangle^2\]

- Astrophysics

Ok I'll go through this book, it seems pretty straight forward...but lets see how long that lasts haha.

- Empty

well when it stops that's why we're reading together lol

- Astrophysics

Sounds good xD

- Astrophysics

Ok I'm going to sleep, later and thanks!!

- Empty

laters

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