# Question on erf(x) function

1. Apr 28, 2010

### 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!

2. Apr 28, 2010

### mathman

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

3. Apr 28, 2010

### jackmell

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 $f(z)=e^{-z^2}$.

Last edited: Apr 28, 2010
4. Apr 28, 2010

### mnb96

...then you get

$$\int_{0}^{x}e^{-(s+ib)^2}ds$$

I cannot relate that integral to the erf function.

5. Apr 28, 2010

### mnb96

thanks jackmell!
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 $\sqrt{\pi}$ ?

6. Apr 28, 2010

### 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:

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

7. Apr 29, 2010

### mnb96

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