# Changing place of limit and integral - problem

1. May 23, 2010

### Petar Mali

$$\delta(x)=\frac{d}{dx}\Theta(x)$$ - delta function like a first derivative of Heaviside step function

$$\Theta(x)=\int dx\delta(x)$$

We use integral representation of delta function

$$\delta(x)=\frac{1}{2\pi}\int^{\infty}_{-\infty}dke^{-ikx}$$

from that we get

$$\Theta(x)=\int dx\frac{1}{2\pi}\int^{\infty}_{-\infty}dke^{-ikx}$$

And now we use one trick!

$$\Theta(x)=\int dx\frac{1}{2\pi}\int^{\infty}_{-\infty}dke^{-ikx}lim_{\epsilon \rightarrow 0^+}e^{\epsilon k}$$

Consider now the expression without the limit

$$\frac{1}{2\pi}\int^{\infty}_{-\infty}dk\int dxe^{-i(k+i\epsilon)x}=\frac{i}{2\pi}\int^{\infty}_{-\infty}dk\frac{e^{-ikx}}{k+i\epsilon}$$

we get that

$$\Theta(x)=\frac{i}{2\pi} lim_{\epsilon \rightarrow 0+}\int^{\infty}_{-\infty}dk\frac{e^{-ikx}}{k+i\epsilon}$$

Using complex integration in lower half plane for $$x>0$$ and upper half plane for $$x<0$$ we get that the upper expression is correct. But why we can change the place of limit and integral? I'm not sure.

2. May 23, 2010

### HallsofIvy

The question of when we can interchange to limits (and the integral, of course, is a limit) is a main topic in Mathmatical Analysis. Here we can do it because that limit converges uniformly.

3. May 25, 2010

### Petar Mali

Can you tell me the answer with little more details? Thanks

4. May 25, 2010

### jostpuur

Nooo!

If you want to justify

$$\lim_{n\to\infty} \int\limits_X f_n(x) d\mu(x) = \int\limits_X \lim_{n\to\infty} f_n(x) d\mu(x)$$

with uniform convergence, you must assume $$\mu(X) < \infty$$

If $$\mu(X)=\infty$$, then uniform convergence is not enough. Counter examples exist. The Lebesgue Dominated Convergence theorem is the standard tool for this situation.

Mali, details of your calculations didn't make sense to me, but I believe that you are in a situation where an expression

$$\lim_{\epsilon\to 0} \int dx\; F_{\epsilon}(x)$$

makes sense, and an expression

$$\int dx\; \lim_{\epsilon\to 0} F_{\epsilon}(x)$$

does not make sense. That means that not everything in your formulas is convergent, and you are not really supposed to justify change of orders of limit and integration. You are merely doing tricks.

5. May 25, 2010

### some_dude

Show that the sequence of functions converges uniformly on every compact (closed and bounded) set in its domain. Then you can exchange limits.

6. May 26, 2010

### jostpuur

Rubbish! The standard counter example to the change of order of limit and integration, which is

$$f_n(x) = \left\{\begin{array}{ll} \frac{1}{n}, & 0\leq x\leq n\\ 0, & n < x\\ \end{array}\right.$$

is also a counter example to this claim that uniform convergence on compact subsets of domain would be enough.

Last edited: May 26, 2010
7. May 26, 2010

### some_dude

I never said my claim held for just any sequence of functions. It holds for the one in question because each integral can be rewritten as the line integral of an analytic function. This is a consequence of Morera's theorem.