- #1

Castilla

- 241

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[tex] \int_{0}^{+\infty}e^{-x^2}dx = \frac{\sqrt{\pi}}{2} [/tex] but the first impasse I find is that, "with the change of variable" [tex]\sqrt{n}x = y [/tex] they justify this equality:

[tex] {\frac{1}{\sqrt{n}}\int_{0}^{+\infty}e^{-y^2}dy = \int_{0}^{+\infty}e^{-nx^2}dx [/tex].

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