Question

Please help me integrate this

Answer 1

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

Answer 2

Put 2ax^2=z

Answer 3

then x^2=z/2a
2xdx=dz/2a implies x^2 dx=xdz/4a=
\[\sqrt{\frac{ z }{ 2a }}dz/4a\]

Answer 4

so the integral turns
\[\int\limits_{-\infty}^{\infty} e ^{-z} \sqrt{\frac{ z }{ 2a}}\frac{ dz }{ 4a }\]

Answer 5

rest is just formula

Answer 6

take the constant terms out and you have inside the integral e^-z z^(3/2-1)

Answer 7

@ememlove

Answer 8

im unclear, which formula? isit the uv-integ(vdu)?

Answer 9

well at first you just do a thing, since the given function is an even function
\[\int\limits_{-\infty}^{\infty} f(x)dx=2\int\limits_{0}^{\infty } f(x)\]

Answer 10

at last you use the formula
\[\int\limits_{0}^{\infty} e ^{-x} x ^{n-1} dx\]

Answer 11

hello you take out pen and paper and try to solve it, i told you the way @ememlove

Answer 12

im trying but i dont get ur second comment

Answer 13

\[\Gamma (n)=\int\limits_{0}^{\infty} e ^{-x} x ^{n-1} dx\]

Answer 14

@ememlove since you substituted x in terms of z, you have to change x^2dx in terms of z and dz as well. The second step is just that

Answer 15

okay i get that, but how about this e^-z z^(3/2-1)?

Answer 16

its gamma(3/2)=(1/2) gamma (1/2) and you know the value of gamma(1/2)

Answer 17

omgomg i thinkw e havnt study that @@. anw thanks!

Answer 18

If you haven't studied gamma function yet, how can you solve this problem?. :O

Answer 19

its a part of my quantum mechanics question, i thought the integral can solve by simple integration, i dont know abt gamma functions @@

Answer 20

ohkay, you study gamma functions, its very easy. just few formulae to solve integrals involving infinity or a finite number of discontinuities @ememlove

Answer 21

there's no other way to solve this without knowing gamma func?

Answer 22

no sorry... :)

Answer 23

haha okay! thanks anw!(: