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I really appreciate it if someone help me solving this integral:

$ \int \frac 1x \cdot \operatorname{Erfc}^n x\, dx,$

where $\operatorname{Erfc}$ is the complementary error function, defined as $\operatorname{Erfc}=\frac 2{\sqrt \pi}\int_x^{+\infty}e^{-t^2}dt$.

thank you

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    Do you need the indefinite integral (as stated), or the definite integral, e.g., from 0 to $\infty$?2012-08-11

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A Taylor series at $x=0$ may be found here: $ \int \frac{\text{Erfc}^n(x)}{x}dx=\log(x)-\frac{2nx}{\sqrt{\pi}}+\frac{(n-1)nx^2}{\pi}+\cdots $ There is also a result for $n=1$ given: $\log(x)-\frac{2x}{\sqrt{\pi}}{ _2F_2}\left(1/2,1/2;3/2,3/2;-x^2\right)$.

EDIT: You get a series expansion for $\text{Erfc}^n(x)$ at $x=\infty$ here: $ \text{Erfc}(x)^n=\left(1-2 \sum _{k=0}^{\infty } \frac{(-1)^k x^{2 k+1}}{\sqrt{\pi }(2 k+1) k!}\right){}^n $

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    @davood, you're welcome.2012-03-29