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Question on erf(x) function

  1. Apr 28, 2010 #1
    If we consider the error function [tex]\mathrm{erf}(x)=\int_{0}^{x}e^{-t^2}dt[/tex]

    How can I re-express the following in terms of the erf function?

    [tex]f(x)=\int_{ib}^{x+ib}e^{-t^2}dt = \\ ?[/tex]

    I have troubles with this kind of integrals. How should I treat an integral with complex bounds?
  2. jcsd
  3. Apr 28, 2010 #2


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    Let s = t-ib, then the complex function will appear in the exponent, wile the s integral is from 0 to x.
  4. Apr 28, 2010 #3
    How about if you consider a closed path from zero straight up to the point ib, straight across to the point x+ib, and then diagonally down back to the origin. The integral over that path is zero since the integrand is entire. We can then write [itex]\int_0^{ib} f(z)dz+\int_{ib}^{x+ib} f(z)dz+\int_{x+ib}^0 f(z)dz=0[/itex] or [itex]\int_{ib}^{x+ib}f(z)dz=\int_0^{x+ib} f(z)dz-\int_0^{ib}f(z)dz=erf(x+ib)-erf(ib)[/itex] given your definition of erf and [itex]f(z)=e^{-z^2}[/itex].
    Last edited: Apr 28, 2010
  5. Apr 28, 2010 #4
    ...then you get


    I cannot relate that integral to the erf function.
  6. Apr 28, 2010 #5
    thanks jackmell!
    your solution is interesting!
    Unfortunately I have some troubles interpreting the erf function with a complex argument: in particular, could you explain how the limit of [itex]2(erf(x+ib)-erf(ib))[/itex] for [itex]x\rightarrow \infty[/itex] gives the result of the gaussian integral, which is [itex]\sqrt{\pi}[/itex] ?
  7. Apr 28, 2010 #6
    Hi. I'm not an expert at this but I believe this is correct:

    [itex]\lim_{x\to\infty} \int_{ib}^{x+ib} e^{-t^2}dt=\lim_{x\to\infty}\left\{\int_0^{x+ib} e^{-t^2}dt-\int_{0}^{ib} e^{-t^2}dt\right\}[/itex]

    [itex]=\lim_{x\to\infty}\left\{\int_0^x +\int_x^{x+ib}-\int_0^{ib}\right\}[/itex]

    and as [itex]x\to\infty [/itex], the center integral goes to zero. Then:

    [itex]\lim_{x\to\infty} \int_{ib}^{x+ib} e^{-t^2}dt=\frac{\sqrt{\pi}}{2}-erf(ib)[/itex]

    If you like you can parameterize the path for the second integral from the origin, straight up to the point ib by letting t=iy and dt=idy then:

    [itex]erf(ib)=\int_0^{ib}e^{-t^2}dt=i\int_0^b e^{y^2}dy[/itex]
  8. Apr 29, 2010 #7
    thanks a lot!
    now it is clear, and your solutions looks correct.
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