Improper integral x^(2) * e^(-x^2)

TsAmE
Messages
129
Reaction score
0

Homework Statement



Show that \int_{0}^{\infty}x^{2}e^{-x^{2}}dx = \frac{1}{2}\int_{0}^{\infty}e^{-x^{2}}dx.

Homework Equations



None.

The Attempt at a Solution



I used substitution:

t = x^{2}

dx = \frac{dt}{2x}

\frac{1}{2}\int_{0}^{\infty}\sqrt{t}e^{-t}dx

Then tried using integration by parts but then I didnt get an answer and got stuck.
 
Physics news on Phys.org
have you tried straight integartion by parts?
 
Try:

x^2*e^(-x^2) = [-0.5*x]*[-2*x*e^(-x^2)dx]

[-0.5*x] = u
[-2*x*e^(-x^2)dx] = dv

Tell us if you got it !

Good Luck :)
 
I tried integration by parts but it didnt work out. I only learn how to integrate e^(-x^2) next year.
 
TsAmE said:
I tried integration by parts but it didnt work out. I only learn how to integrate e^(-x^2) next year.
How did you try? Did you try letting u= x and dv= xe^{-x^2}dx.
 
Try integration by parts with t=x, not t=x^2 !

I remind you:

integral(u*dv) = u*v - integral(v*du)

P.S.: If you learn in the next year how to get the anti-derivative of e^(-x^2), please tell us ;)
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top