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Is there any closed form expression for the following integral?

$$ \int\limits_t^\infty \left(1- \operatorname{erf}(\log x) \right )dx $$

or equivalently:

$$ \int\limits_t^\infty \operatorname{erfc}(\log x ) dx $$

I just wish to know if there is any way I can do better than sampling the values at particular points and calculating the area under the curve numerically. Any help is appreciated. Thanks.

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2  
Please, don't yell!. –  Arturo Magidin Apr 10 '12 at 3:08
    
sorry for that, my bad! did not notice!! –  Sandeep Apr 10 '12 at 3:09
    
@Sand I edited. To introduce a new operator or function name use \operatorname{blah} –  Pedro Tamaroff Apr 10 '12 at 3:10
4  
Mathematica can evaluate this explicitly as $$e^{1/4} \operatorname{erfc}\left(\log t-\frac{1}{2}\right) - t \operatorname{erfc}(\log t)$$ for $t \geq 0$. –  Antonio Vargas Apr 10 '12 at 3:21
1  
@Michael, the first integrand involves "erf" and the second involves "erfc". –  Antonio Vargas Apr 10 '12 at 3:39

1 Answer 1

up vote 4 down vote accepted

Not terribly hard; one only needs simple substitutions and an application of integration by parts here. To wit,

$$\begin{align*} \int_t^\infty \mathrm{erfc}(\log u)\,\mathrm du&=\int_{\log\,t}^\infty \exp\,v\;\mathrm{erfc}\,v\,\mathrm dv\\ &=\left.\exp\,v\;\mathrm{erfc}\,v\right|_{\log\,t}^\infty+\frac2{\sqrt\pi}\int_{\log\,t}^\infty \exp\,(v-v^2)\,\mathrm dv\\ &=-t\;\mathrm{erfc}(\log\,t)+\frac{2\sqrt[4]{e}}{\sqrt\pi}\int_{\log\,t-\frac12}^\infty \exp\,(-w^2)\,\mathrm dw\\ &=\sqrt[4]{e}\,\mathrm{erfc}\left(\log\,t-\frac12\right)-t\;\mathrm{erfc}(\log\,t)\\ \end{align*}$$

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thanks for the step by step derivation. –  Sandeep Apr 10 '12 at 21:13

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