1
$\begingroup$

If I want to calculate the following integral in terms of the Error function, is this correct?

$$\frac{1}{\sqrt{2\pi}}\int_{f(x)}^{-\infty}e^{-p^2}\mathrm{d}p = \mathrm{Erf}(-\infty) - \mathrm{Erf}(f(x))$$

$\endgroup$
  • 1
    $\begingroup$ Yes, other than the fact that $$\frac1{\sqrt{2\pi}}\int_{f(x)}^{-\infty}e^{-p^2}dp=-\operatorname{Erf}(f(x))+\lim_{L\to-\infty}\operatorname{Erf}(L)$$ $\endgroup$ – clathratus Oct 26 '18 at 16:42
0
$\begingroup$

You can check that $$\frac{1}{\sqrt{2\pi}}\int_{f(x)}^{\infty}e^{-t^2}dt = \frac{1}{2\sqrt{2}}\big(1-\mathrm{erf}(f(x))\big).$$

Hint: Use the definition of $\mathrm{erf}(x)$:

$$\mathrm{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-t^2}dt$$ and the identity: $$\int_{-\infty}^{\infty}e^{-t^2}dt = \sqrt{\pi}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.