Understanding the Complex Bounds of the erf(x) Function

[mnb96]

If we consider the error function \mathrm{erf}(x)=\int_{0}^{x}e^{-t^2}dt

How can I re-express the following in terms of the erf function?f(x)=\int_{ib}^{x+ib}e^{-t^2}dt = \\ ?

I have troubles with this kind of integrals. How should I treat an integral with complex bounds?
Thanks!

 

[mathman]

Let s = t-ib, then the complex function will appear in the exponent, wile the s integral is from 0 to x.

 

[jackmell]

If we consider the error function \mathrm{erf}(x)=\int_{0}^{x}e^{-t^2}dt

How can I re-express the following in terms of the erf function?f(x)=\int_{ib}^{x+ib}e^{-t^2}dt = \\ ?

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 \int_0^{ib} f(z)dz+\int_{ib}^{x+ib} f(z)dz+\int_{x+ib}^0 f(z)dz=0 or \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) given your definition of erf and f(z)=e^{-z^2}.

 

[mnb96]

Let s = t-ib, then the complex function will appear in the exponent, wile the s integral is from 0 to x.

...then you get

<br /> \int_{0}^{x}e^{-(s+ib)^2}ds<br />

I cannot relate that integral to the erf function.

 

[mnb96]

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 2(erf(x+ib)-erf(ib)) for x\rightarrow \infty gives the result of the gaussian integral, which is \sqrt{\pi} ?

 

[jackmell]

Hi. I'm not an expert at this but I believe this is correct:

\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\}

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

and as x\to\infty, the center integral goes to zero. Then:

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

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:

erf(ib)=\int_0^{ib}e^{-t^2}dt=i\int_0^b e^{y^2}dy

 

[mnb96]

thanks a lot!
now it is clear, and your solutions looks correct.