Integral $\int_{-\infty}^0 e^{-t^2/2}dt$

Question

can someone help with this integral $\int_{-\infty}^0 e^{-t^2/2}dt$? I know that $\int_{-\infty}^{\infty} e^{-t^2/2}dt = \sqrt{2 \pi}$ and $\int_{-\infty}^{\infty} e^{-t^2}dt = \sqrt{\pi}$. I tried to calculate it myself and my result is $\sqrt{\frac{\pi}{2}}$. Is that correct?

Answer 1

Yes, your answer is correct.

$e^{\frac{-x^2}{2}}$ is even function, therefore,

$\int_{-\infty}^{\infty} e^{\frac{-x^2}{2}}dx = 2 \int_{-\infty}^{0} e^{\frac{-x^2}{2}}dx $