Why it's not equal?

**Mengqi** · Oct 9th 2013, 07:49 AM

HI all,

I encountered an integral
$\gamma(x,t)=\pi^{-0.5} \frac{x}{2} \int^t_0 t'^{-\frac{3}{2}} e^{-\frac{x^2}{4t'}} dt' = -\frac{2}{\sqrt{\pi}} \int^t_0 e^{-(\frac{x}{2\sqrt{t'}})^2} d \frac{x}{2\sqrt{t'}} = \frac{2}{\sqrt{\pi}} \int^{\infty}_l e^{-l'^2} d l' = erfc({\frac{x}{2\sqrt{t}}})$
So it's a complementary error function finally.

But it seems that when x=0, the lhs of the equation is equal to 0, while at the rhs, it's equal to 1. So x=0 may not be valid for the error function, since then the integral variable x/2/sqrt(t) is all the time 0. By arguing this, I believe that when x=0, gamma = 0, but then I think that the rhs should be valid for all the finite x, which means that when x is very small but not zero, the rhs and lhs should be the same, which seems not to be the case here.... what went wrong here? Thanks

**Shakarri** · Oct 9th 2013, 10:39 AM

How does $\frac{1}{2}xt'^{\frac{-3}{2}}dt'=d\frac{x}{2\sqrt{t'}}$

What variable is t' being differentiated with respect to?

**Mengqi** · Oct 10th 2013, 05:25 AM

Hi Shakarri,

Thanks. I think one can view x as a constant here, so one can put it after the differentiation symbol, isn't it?
Yesterday I use spectrum method to directly integrate gamma as in the form on the lhs. The results are attached. It seems very stiff at small x, at large x, the numerical results give the same value as the complementary error function. I even used N=10000, which is incredibly high, but it can still not resolve the discontinuity at x=0.

.

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