Question

kainuiiiiiiiiiii, I need you to explain to me what exactly you did with this integral haha.

Answer 1

@Kainui

Answer 2

i 4get sry

Answer 3

Haha, no seriously, what's up?

Answer 4

\[\huge \int\limits_{-\infty}^{\infty}e ^{-x ^{2}}dx\]

Answer 5

Ahh, so now just multiply that same integral by itself and take the square root of it. Since the integral doesn't depend on x, make the other integral you multiply it by just with y's instead.

Answer 6

That's just to make it more obviously different from it if that makes any sense at all lol

Answer 7

I see, so would this method work even without the limits?

Answer 8

I dunno, I tried and couldn't really figure it out.

Answer 9

Ah, yeah this integral has been haunting me for quite some time..

Answer 10

You could try the more general version though: \[\int\limits _\infty ^\infty ae^{-bx^2}dx\] Then you could differentiate the answer you get with respect to a or b depending on what comes out. Actually I'll try that out now. =P

Answer 11

well obviously that integral up there =0 but w/e

Answer 12

You know what I meant lol

Answer 13

Mhm wonder if you would still get...\[\frac{ 1 }{ a^2 }(au-1)e ^{au}\]

Answer 14

And yeah lol

Answer 15

+C

Answer 16

idk where did you get that 1/a^2 thing?

Answer 17

Back of my calc textbook lol, I'm looking at tables of integrals :P

Answer 18

\[\sqrt{\pi}*a^{-1/2}=\int\limits_{-\infty}^{\infty} e^{-ax^2}dx\] This is what I'm getting so check this out

Answer 19

\[\frac{1}{2}\sqrt{\pi}*a^{-3/2}=\int\limits_{-\infty}^{\infty} x^2e^{-ax^2}dx\]