Evaluation on a basis of gaussian integral

Question

Knowing that $$\int_{-\infty}^\infty e^{-x^2} dx= \pi^{\frac{1}{2}}$$ Find: $$\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{\frac{-x^2}{2}} dx$$ And my question is how does this help if have the value of gaussian integral?

Answer 1

Hint: $2=(\sqrt{2}\,)^2$; now make use of a $u$-substitution in the integral.

Answer 2

Notice, we have $$\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt \pi$$

Now, $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-x^2/2}dx$$

$$=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\left(x/\sqrt2\right)^2}dx$$

Let, $\frac{x}{\sqrt 2}=t\implies \frac{x}{\sqrt 2}dx=dt\implies dx=\sqrt 2dt$ $$=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-t^2}\sqrt 2 dt$$ $$=\frac{\sqrt 2}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-t^2}dt$$ Changing variable from $t$ to $x$

$$=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-x^2}dx$$ $$=\frac{1}{\sqrt \pi}(\sqrt \pi)=1$$