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I'm assuming $\frac{d}{dw}$ should be written as $\frac{\partial}{\partial \omega}$

I'm a bit confused by the part highlighted in green. I'm think i'm right in saying that when we integrate wrt one variable ($x$ in this case) we leave the other variable ($\omega$) constant. However the gren underline implies we that we can treat $\omega$ as a variable inside the integral which is integrating wrt $x$?

usainlightning
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1 Answers1

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Think about an easier example. You want to rewrite $\int_0^1 2xy dx$ as $\frac{df}{dy}$ for some function $f$. (I realize this example is artificial, please bear with me.)

You notice that $2xy=\frac{\partial}{\partial y} xy^2$, so you have

$$\int_0^1 2xy dx = \int_0^1 \frac{\partial}{\partial y} xy^2 dx.$$

Now you justify taking the derivative out of the integral to get

$$\int_0^1 2xy dx = \frac{d}{dy} \int_0^1 xy^2 dx.$$

At no point here is $y$ actually changing inside the integration, we're just writing one function of $y$ as the derivative of another function of $y$.

The same happens in your example: you're rewriting $f(x,\omega)=xe^{-i \omega x}$ as $\frac{\partial}{\partial \omega} \left ( i e^{-i \omega x} \right )$. The value of $\omega$ isn't actually changing.

Ian
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  • I don't understand your explanation, however would it make sense to think of $\omega$ only as a constant wrt x and not a constant in the normal sense of the word. Then $\frac{\partial}{\partial \omega}$ does not suggest I am differentiating wrt a constant. – usainlightning Mar 19 '15 at 21:15
  • @usainlightning You can think of it that way if you like. Another way is to think about approaching from the right side. You want to differentiate the Fourier transform, which is a function of $\omega$. When you do that, just like you would have a sum of derivatives if you had a finite sum, you get an integral of derivatives since you have an integral. – Ian Mar 19 '15 at 21:48