Question

Alright fancy integral for @nincompoop

Answer 1

\[\int\limits_0^\infty x^n e^{-x}dx\] Before I solve it, I guess I'll give it a second to let people try to figure out what the answer is and derive it for fun. Then I'll show the answer.

Answer 2

Let's start with something simple. \[\int\limits_0^\infty e^{-y}dy\]And let's introduce a new parameter, a, just because we can. A is a constant with respect to x and y. After Id o this, I'll only worry about x, y is just a dummy variable here. You'll see why in a moment.

y=ax
dy=adx

\[\int\limits_0^\infty a e^{-ax}dx\]So this should be pretty straight forward to integrate, remember, a is just a constant.

Answer 3

\[a \int\limits_0^\infty e^{-ax}dx=a[\frac{-1}{a}e^{-ax}]_0^\infty\]\[[\frac{-1}{e^{ax}}]_0^\infty=\frac{-1}{e^\infty}- \frac{-1}{e^0}=1\]So the whole integral becomes: \[1=a \int\limits_0^\infty e^{-ax}dx\]divide both sides by a, really just a^(-1) right? \[a^{-1}= \int\limits_0^\infty e^{-ax}dx\]Now this is the trick. It involves partial derivatives, I'll give you a second to take a guess.

Answer 4

\[a^{-1}= \int\limits_0^\infty e^{-ax}dx\] Take the derivative with respect to a. We can do that, since a and x are independent variables of each other. \[(-1)a^{-2}= \int\limits_0^\infty (-x) e^{-ax}dx\] notice that the negative signs will cancel each other out just leaving: \[a^{-2}= \int\limits_0^\infty x e^{-ax}dx\]Let's do a few more derivatives in a row, I'll automatically skip the part where I get rid of the negative signs though.\[2a^{-3}= \int\limits_0^\infty x^2 e^{-ax}dx\]\[2*3a^{-4}= \int\limits_0^\infty x^3 e^{-ax}dx\]\[2*3*4a^{-5}= \int\limits_0^\infty x^4 e^{-ax}dx\] Seems like a pattern is coming up? \[\frac{n!}{a^{n+1}}= \int\limits_0^\infty x^n e^{-ax}dx\]

Cool. Set a=1 \[n!= \int\limits_0^\infty x^n e^{-x}dx\]

Yikes.

Answer 5

I dare you to check it with integration by parts lol.

Answer 6

For a second I doubted you're going to derive laplace transform stuff out of this :P

Answer 7

clearly

Answer 8

Nah I'm sharing the love! This is such a hard looking integral but really it turns out to be really easy and kind of like magical haha. I wish I came up with it, but sadly I'm just a messenger lol.

Answer 9

"easy"

Answer 10

ya... this was about half a board in my class and it wasn't like this

Answer 11

Well you are just solving an e^x integral and then doing partial derivatives after that. I mean it's not as uncomfortable as doing integration by parts or trig sub, it's just new to you right now that's all lol.

Answer 12

apparently
I am still "validating" the steps

Answer 13

Ask questions when/if you have them, or show me a cool integral you know about so I can learn more. =P

Answer 14

adx = dy
somehow, I saw the 1/a but in a different manner...

Answer 15

Oh what do you mean?

Answer 16

this part
\[\frac{ 1 }{ a } \int\limits ae^{-ax}dx\]

Answer 17

you skipped this part or am I hallucinating?

Answer 18

\[1=\int\limits_0^\infty ae^{-ax}dx\] So we're at this point right? Since a is just a constant, we can just pull it out of the integral\[1=a \int\limits_0^\infty e^{-ax}dx\]divide both sides by a:\[\frac{1}{a}=\frac{a}{a}\int\limits_0^\infty e^{-ax}dx\]\[a^{-1}=\int\limits_0^\infty e^{-ax}dx\]

Did that answer your question?

Answer 19

yeah

Answer 20

thank you for sharing

Answer 21

I just sorta noticed something interesting if you replace the x with an r it looks kind of like a polar integral. \[\frac{r!}{a^{n+1}}=\int\limits_0^\infty r^n e^{-ar}dr\]

Noticing that \[2\pi = \int\limits _0^{2\pi} d \theta \] we can multiply this by that integral to get: \[\frac{2\pi \ r^n}{a^{n+1}}=\int\limits_0^{2\pi} \int\limits_0^\infty r^{n-1} e^{-ar} \ r \ dr \ d \theta\] convert to rectangular coordinates:\[\frac{2\pi \ (x^2+y^2)^{n/2}}{a^{n+1}}=\int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty (x^2+y^2)^{(n-1)/2} \ e^{-a \sqrt{x^2+y^2}} dx dy\] Yeah that's just a dirty integral but we have the general formula for this useless thing now lol.

Any ideas on how else to play with this thing?

Answer 22

you're crazy

Answer 23

it's 6:55AM

Answer 24

5:55 AM here.

Answer 25

happens

Answer 26

actually what I just typed makes no sense.