I want to evaluate this integral


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

Any suggestions?



using the polar substitution.

  • 3
    $\begingroup$ Let $u=\sqrt{a} \,y$. $\endgroup$ Jul 26, 2016 at 14:41

2 Answers 2


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: $$ \frac{1}{\sqrt{a}}\int_{-\infty}^{+\infty}e^{-x^2} dx=\lim_{t \to \infty}\frac{\sqrt{\pi}}{\sqrt{a}} \int_{-t}^{+t}e^{-x^2} dx=\sqrt{\frac{\pi}{a}} \left(\lim_{t \to \infty} \mbox{erf}(t)\right)=\sqrt{\frac{\pi}{a}} $$

  • $\begingroup$ 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? $\endgroup$
    – zhk
    Jul 31, 2016 at 3:02
  • $\begingroup$ 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 $+\infty$. $\endgroup$ Jul 31, 2016 at 9:27

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

  • 2
    $\begingroup$ This should be a comment $\endgroup$
    – Yuriy S
    Jul 26, 2016 at 14:44
  • $\begingroup$ This is a perfectly good answer that completely addresses the question asked. Upvoted. $\endgroup$ Jul 26, 2016 at 15:32
  • $\begingroup$ I agree with Daniel and I have upvoted the answer as well. $\endgroup$ Jul 26, 2016 at 15:47

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