Interchanging diff. and integr. rigorously

[LouisR]

I have some problems with what a prof. asked us in class. This is a class of analysis that is dealing with measure theory and stuff like Fourrier transform:
I am really stuck with the following ( really have problem to explain the steps rigorously ):

(d/dy){g(x,y)} means differentiates(partially) g with respect to y
i is the usual square root of -1

1-) For this one, x is in R^n, y is in R, f(x) is a Schwarz function ( or if you prefer in the Schwarz class, so as consequence, f is integrable over R^n and it is infinitely differentiable). Could anyone gives me the rigorous steps for this please:

integral( (d/dy){e^(-i*x*y)*f(x) dx} ) = (d/dy){integral(e^(-i*x*y)*f(x) dx)}



2-)Here is a simpler case: Let [a,b] be a close interval in the real. Integration is from a to b. Suppose that f(x,y) and (d/dx){f(x,y)} are continuous on R^2. Then again what are the rigorous justifications for this:

(d/dy){integral(f(x,y)dx)} = integral((d/dy){f(x,y)dx})

A really big thanks in advance if anybody could explain me this.

 

[jvicens]

I understand your confusion with the steps provided by your professor. Let me try to explain them to you in a more rigorous manner.

1-) In this case, we have a function g(x,y) = e^(-i*x*y)*f(x), where x is in R^n, y is in R, and f(x) is a Schwarz function. We want to find the partial derivative of g with respect to y, which is denoted by (d/dy){g(x,y)}. The rigorous steps for this are as follows:

1. First, we need to define the partial derivative (d/dy){g(x,y)}. This is defined as the limit of the difference quotient as h approaches 0, where h is the change in y. Mathematically, this can be written as (d/dy){g(x,y)} = lim(h->0) (g(x,y+h) - g(x,y))/h.

2. Next, we can substitute the given function g(x,y) = e^(-i*x*y)*f(x) into the above definition. This gives us (d/dy){g(x,y)} = lim(h->0) (e^(-i*x*(y+h))*f(x) - e^(-i*x*y)*f(x))/h.

3. Now, we can use the properties of limits to simplify this expression. Since f(x) is infinitely differentiable, we can write f(x) as a Taylor series expansion around x. This gives us f(x+h) = f(x) + hf'(x) + (h^2/2!)f''(x) + ... + (h^n/n!)f^(n)(x), where f^(n) denotes the nth derivative of f. Substituting this into our expression, we get (d/dy){g(x,y)} = lim(h->0) (e^(-i*x*y)*[f(x) + hf'(x) + (h^2/2!)f''(x) + ... + (h^n/n!)f^(n)(x)] - e^(-i*x*y)*f(x))/h.

4. Simplifying further, we get (d/dy){g(x,y)} = lim(h->0) (e^(-i*x*y)*hf'(x) + (h^2/2!)*e^(-i*x*y)*f''(x

 

[M98Ranger]

In order to interchange differentiation and integration rigorously, we must use the fundamental theorem of calculus and the Leibniz integral rule. Let's first consider the first problem:

1-) For this one, x is in R^n, y is in R, f(x) is a Schwarz function (or if you prefer in the Schwarz class, so as consequence, f is integrable over R^n and it is infinitely differentiable). Could anyone gives me the rigorous steps for this please:

To start, let's rewrite the integral using the Leibniz integral rule:
integral( (d/dy){e^(-i*x*y)*f(x) dx} ) = integral( (d/dy){e^(-i*x*y)} * f(x) dx) + integral( e^(-i*x*y) * (d/dy){f(x)} dx)

Now, we can use the fundamental theorem of calculus to evaluate the first integral on the right-hand side:
integral( (d/dy){e^(-i*x*y)} * f(x) dx) = e^(-i*x*y) * f(x) + C

Substituting this back into our original integral, we have:
integral( (d/dy){e^(-i*x*y)*f(x) dx} ) = e^(-i*x*y) * f(x) + C + integral( e^(-i*x*y) * (d/dy){f(x)} dx)

Using the Leibniz integral rule again, we can interchange the differentiation and integration:
integral( e^(-i*x*y) * (d/dy){f(x)} dx) = (d/dy){integral(e^(-i*x*y)*f(x) dx)}

Therefore, we have:
integral( (d/dy){e^(-i*x*y)*f(x) dx} ) = e^(-i*x*y) * f(x) + C + (d/dy){integral(e^(-i*x*y)*f(x) dx)}

Now, since we are dealing with a Schwarz function, we know that it is integrable over R^n and infinitely differentiable. This means that the constant C is equal to 0, and we can rewrite our integral as:
integral( (d/dy){e^(-i*x*y)*f(x) dx} ) = e^(-i*x*y) *