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Interchanging diff. and integr. rigorously

  1. Apr 12, 2005 #1
    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.
  2. jcsd
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