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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?

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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
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Looks right to me. –  J. M. Feb 11 '12 at 11:53

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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 $$ \int_x^\infty {\rm erfc}(t)\,dt= {1\over{\sqrt{\pi}}}e^{-x^2} - x{\rm erfc}(x). $$

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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
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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
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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

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