Reflection Rule of a Fourier Transform

[BustedBreaks]

I feel a bit dumb, but could someone help me see this:

$$G(s):= \int_{-\infty}^{\infty}f(-x)e^{-2\pi isx}dx = \int_{-\infty}^{\infty}f(u)e^{-2\pi i(-s)u}du = F(-s)$$

 

[HallsofIvy]

I feel a bit dumb, but could someone help me see this:

$$G(s):= \int_{-\infty}^{\infty}f(-x)e^{-2\pi isx}dx = \int_{-\infty}^{\infty}f(u)e^{-2\pi i(-s)u}du = F(-s)$$

You're about to feel even dumber! (But actually be smarter.)

Let u= -x. Then $-2\pi isx= -2\pi i(-s)u$, f(-x) becomes f(u), of course.

As x goes to $\infty$, u goes to $-\infty$ and vice-versa so the limits of integration are switched. That's the reason the "-" in front of the first integral disappears.

 

[BustedBreaks]

I feel dumber for not realizing this, and even more dumbest for this sentance. A little smarterest though for the learning...

Thanks!

 

[HallsofIvy]

Hey, you've got a long way to go before you are as dumberized as I am!