Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Could you please check the below and show me any errors? $$ \int_ x^ \infty {\rm erfc} ~(t) ~dt ~=\int_ x^ \infty \left[\frac{2}{\sqrt\pi} \int_ t^ \infty e^{-u^2} du \right]\ dt $$ If I let dv=dt and u equal the term inside the bracket, and do integration by parts, $$ \int u ~dv ~=uv - \int v~ du $$ v=t and du becomes $$ -\frac{2}{\sqrt\pi} e^{-t^2} $$ This was obtained from using the Leibniz rule below, $$ \frac {d} {dt} \left[ \int_ a^ b f(u)du \right]\ = \int_ a^ b \frac {d} {dt} f(u) du + f \frac {db} {dt} - f \frac {da} {dt} $$

Then, $$ \frac {d} {dt} \left[\frac{2}{\sqrt\pi} \int_ t^ \infty e^{-u^2} du \right]\ = \frac{2}{\sqrt\pi} \left[ \int_ t^ \infty \frac {d} {dt} \left( e^{-u^2} \right) du + e^{-\infty ^2} * 0 - e^{-t^2}*1 \right]= \frac{2}{\sqrt\pi} \left[0~+~0~- e^{-t^2} \right]$$ Is the first and second term going to zero correct? The upper limit b=infinity, and is db/dt=0 in the second term correct?
The integral becomes $$ \left[~ t~ \frac{2}{\sqrt\pi} \int_ t^ \infty e^{-u^2} du ~\right] _{x}^\infty + \int_ x^ \infty t \left[\frac{2}{\sqrt\pi} e^{-t^2} \right]\ dt =$$ $$ \left[~ t~ \frac{2}{\sqrt\pi} \int_ t^ \infty e^{-u^2} du ~\right] _{x}^\infty - \left[\frac{1}{\sqrt\pi} e^{-t^2} \right] _{x}^\infty =$$ $$ \left[ 0 - ~ x~ \frac{2}{\sqrt\pi} \int_ x^ \infty e^{-u^2} du ~\right] - \left[ 0 - \frac{1}{\sqrt\pi} e^{-x^2} \right] = $$ (Is the first limit going to zero OK? infinity times 0 = 0). The above becomes $$ -x~ {\rm erfc}~(x) + \frac{1}{\sqrt\pi} e^{-x^2} $$ Is everything correct here? Could you please give explanation to the questions I listed?

share|cite|improve this question

migrated from Feb 11 '12 at 11:24

This question came from our discussion, support, and feature requests site for people studying math at any level and professionals in related fields.

This is the second time you are posting a math question here on the meta site. Please post such questions on the main site, which has colours, and not on meta, which is grey! – Rahul Feb 11 '12 at 10:54
Looks right to me. – J. M. Feb 11 '12 at 11:53

1 Answer 1

up vote 2 down vote accepted

You can also achieve this result by an interchange of the integrals as follows. $$ \int_x^\infty {\rm erfc}(t)\,dt = {2\over\sqrt{\pi}}\int_x^\infty\left(\int_t^\infty e^{-u^2}\,du\right)dt = \iint_{x<t<u} e^{-u^2}\,dudt. $$ Now interchange the order of integration to get $$ \int_x^\infty {\rm erfc}(t)\,dt ={2\over\sqrt{\pi}}\int_x^\infty\left( \int_x^udt\right) e^{-u^2}\,du = {2\over\sqrt{\pi}}\int_x^\infty (u-x)e^{-u^2}\,du, $$ hence $$ \int_x^\infty {\rm erfc}(t)\,dt= {1\over{\sqrt{\pi}}}e^{-x^2} - x{\rm erfc}(x). $$

share|cite|improve this answer
Thank you very much. This certainly is a much better way. But, I don't understand how you proceed from the second item to the third item in the last line. Also, how you get the second item either. Could you please explain it more?? I also don't understand the last item of the first equation. I apologize for more questions. – Tony Feb 11 '12 at 14:24
Since you are integrating over $x < t < u < \infty$, to reverse, you write the inequalities $x < t < u$ for the inner $dt$ integral and $x < u < \infty$ for the outer $du$ integral. – ncmathsadist Feb 11 '12 at 14:58
Many thanks for your response. In the 2nd item of your last equations, then is the inner integral from x to u missing dt? Hew..too advanced for me..still unsure about your change of limits.. – Tony Feb 11 '12 at 15:22
t<u<infinity, so u extends from u=t line(at a 45 deg angle) to infiniy. And x<t<infinity. Thus the region of integration is bounded by t=x(some constant?, a vertical line) and u=t line(at a 45 deg angle) to infinity on the graph where u is the ordinate and t is the abscissa, shaped like a V except the left line is vertical. By reversing the order of integration, t is from x to u(u=t line), and u is from x to infinity. I hopt this is correct. I think I understand the answer now. Many thanks to @ncmathsadist!! – Tony Feb 13 '12 at 6:06
Added some missing $dt$ symbols, and got rid of the conflict of notation between $x$ the limit of the integral and $x$ the dummy variable in the same integral. – Did Feb 14 '12 at 14:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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