1
$\begingroup$

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

  • 1
    What $n$? What makes you think a primitive exists, involving only usual functions?2012-03-28
  • 0
    Is $n$ an integer? Real? Complex?! You really should be specific...2012-03-28
  • 0
    n is a real number2012-03-28
  • 0
    Do you need the indefinite integral (as stated), or the definite integral, e.g., from 0 to $\infty$?2012-08-11

1 Answers 1

1

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: $$ \text{Erfc}(x)^n=\left(1-2 \sum _{k=0}^{\infty } \frac{(-1)^k x^{2 k+1}}{\sqrt{\pi }(2 k+1) k!}\right){}^n $$

  • 0
    thank u draks.but i need a simpler answer if it exists2012-03-28
  • 0
    @davood So let me quote Didier: *What makes you think a primitive exists, involving only usual functions?*2012-03-28
  • 0
    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?2012-03-29
  • 0
    @davood you can try [this](http://tinyurl.com/d8owgjq)...2012-03-29
  • 0
    @davood: [There](http://tinyurl.com/d9e7jsg) are series expansions at $x=\infty$.2012-03-29
  • 0
    @davood, you're welcome.2012-03-29