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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!2012-02-11
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    Looks right to me.2012-02-11

1 Answers 1

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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

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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.2012-02-11
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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.2012-02-11
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    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..2012-02-11
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    t2012-02-13
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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.2012-02-14
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    @did, thank you very much for your help!!2012-08-09