Your H function behaves exactly like the Heaviside Step Function.
i just started learning about PDE in complex variables and im very confused by it.
why isit that for f(x) = integrate from 2pi to 0 ( H(x-t)) dt
that if t<x, H(x-t) = 1 and for t>x, H(x-t) = 0?
what does that mean?
also, is there any books on notes that anyone can recommend which explains this concept becos i cant seem to find any materials on this.
Ok, the real integral being claimed is
The reason you can do this is that the Heaviside step function crunches everything down to zero as soon as reaches . Therefore, integrating beyond shouldn't pick up any more area. Think of it this way:
Make sense now?
Because the upper limit of the original integral is . If the variable of integration is , and the lower limit is and the upper limit is , then it follows that for that integral,how do you know that in this integration, that if t>x, the answer of this integration will be 0
I wouldn't say that "the answer of this integration will be 0". Because I'm not saying that, in your case, the integral from 0 to x of your integrand is zero. I'm saying that if you extend the upper limit beyond the original upper limit, you'd better have an integrand of zero, or you'll change the value of the integral.