# How to evaluate this integral $\int_{-\infty}^{\infty}\exp(-ay^2)dy$ [duplicate]

I want to evaluate this integral

$$\int_{-\infty}^{\infty}\exp(-ay^2)dy$$

using the error function definition. The problem I am facing is with the coefficient of $y^2$.

Any suggestions?

Fact

$$\int_{-\infty}^{\infty}\exp(-y^2)dy=\sqrt{\pi}$$

using the polar substitution.

• Let $u=\sqrt{a} \,y$. Jul 26 '16 at 14:41

substituting $$\sqrt{a}y=x \qquad \rightarrow \qquad \sqrt{a}dy=dx$$ the integral becomes: $$\frac{1}{\sqrt{a}}\int_{-\infty}^{+\infty}e^{-x^2} dx$$ so from the definition
$$\mbox{erf}(t)=\frac{1}{\sqrt{\pi}} \int_{-t}^{+t}e^{-x^2} dx$$ the integral becomes: $$\frac{1}{\sqrt{a}}\int_{-\infty}^{+\infty}e^{-x^2} dx=\lim_{t \to \infty}\frac{\sqrt{\pi}}{\sqrt{a}} \int_{-t}^{+t}e^{-x^2} dx=\sqrt{\frac{\pi}{a}} \left(\lim_{t \to \infty} \mbox{erf}(t)\right)=\sqrt{\frac{\pi}{a}}$$

• I do not understand the last step. How come you end up from $$\lim_{t \to \infty}\frac{\sqrt{\pi}}{\sqrt{a}} \int_{-t}^{+t}e^{-x^2} dx=\sqrt{\frac{\pi}{a}} \left(\lim_{t \to \infty} \mbox{erf}(t)\right)$$ The limit is the whole real line. How you we end up only on the half real line?
– zhk
Jul 31 '16 at 3:02
• Look at the definition of erf function, it contains the two opposite values $-t$ and $+t$ so for $t\to \infty$ the integral goes from $-\infty$ to $+\infty$. Jul 31 '16 at 9:27

Set y=$Y/\sqrt{a}$ and see what happens ;-)

• This should be a comment Jul 26 '16 at 14:44
• This is a perfectly good answer that completely addresses the question asked. Upvoted. Jul 26 '16 at 15:32
• I agree with Daniel and I have upvoted the answer as well. Jul 26 '16 at 15:47