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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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What $n$? What makes you think a primitive exists, involving only usual functions? –  Did Mar 28 '12 at 9:06
    
Is $n$ an integer? Real? Complex?! You really should be specific... –  J. M. Mar 28 '12 at 15:18
    
n is a real number –  davood Mar 28 '12 at 17:22
    
Do you need the indefinite integral (as stated), or the definite integral, e.g., from 0 to $\infty$? –  Fabian Aug 11 '12 at 4:43
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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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thank u draks.but i need a simpler answer if it exists –  davood Mar 28 '12 at 17:25
    
@davood So let me quote Didier: What makes you think a primitive exists, involving only usual functions? –  draks ... Mar 28 '12 at 17:32
    
yes you are right.but of course i have to integrate it from a known value to infinity. in this way a series expansion at x=0 is not allowed to use. do i have another choice? –  davood Mar 29 '12 at 6:57
    
@davood you can try this... –  draks ... Mar 29 '12 at 7:05
    
@davood: There are series expansions at $x=\infty$. –  draks ... Mar 29 '12 at 7:38
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