Solving Int. e^(-x^2) w/ Change of Variable: Help Needed

[Castilla]

I am trying to follow a proof of
\int_{0}^{+\infty}e^{-x^2}dx = \frac{\sqrt{\pi}}{2} but the first impasse I find is that, "with the change of variable" \sqrt{n}x = y they justify this equality:
{\frac{1}{\sqrt{n}}\int_{0}^{+\infty}e^{-y^2}dy = \int_{0}^{+\infty}e^{-nx^2}dx.

Maybe you can help me to see how they did it? Thanks.

 

[Galileo]

\int_{0}^{+\infty}e^{-x^2}dx
is just the half of
\int_{-\infty}^{+\infty}e^{-x^2}dx

The second is usually done with Fubini's theorem and polar substitution.

 

[Castilla]

Er... but the book from where I take the problem only deals with functions of one real variable... they don't use Fubini...

 

[VietDao29]

\sqrt{n}x = y, so, you have: d(\sqrt{n}x) = \sqrt{n}dx = dy
Then, you have:
\frac{1}{\sqrt{n}} \int_{0} ^ {+ \infty} e ^ {-y^2} dy = \frac{1}{\sqrt{n}} \int_{0} ^ {+\infty} e ^ {-nx ^ 2} (\sqrt{n} dx)
= \sqrt{n} \times \frac{1}{\sqrt{n}} \int_{0} ^ {+\infty} e ^ {-nx ^ 2} dx = \int_{0} ^ {+\infty} e ^ {-nx ^ 2} dx.
Can you get it now?
Viet Dao,

 

[Castilla]

Yes, Viet Dao. Really thanks.