Fourier Transform of $f(t+a)$ if $f(t)$ has tranform $F(k)$?

Question

I know the formula $$f(t) = \int^{+\infty}_{-\infty} F(k)e^{ikt} \, dk$$ and I've seen that for computing $f'(t)$ it's a case of differentiating $e^{ikt}$ inside the integral, so $f'(t)=ikF(k)$

Can someone explain how I'd quickly work out $f(t+a)$ given $f(t)=F(k)$?
What if it was $f(at)$?
I presume it's going to involve some kind of substitution in the integral but I'm unsure how to go about it in either case.

Answer 1

The Fourier transform of $f(t+a) $ is $F(k) e^{ika} $

$$f(t) = \int^{+\infty}_{-\infty} F(k)e^{ikt} \, dk$$ $$ f(t+a) = \int^{+\infty}_{-\infty} F(k)e^{ik(t+a)} \, dk= \int^{+\infty}_{-\infty} F(k)e^{ika}e^{ikt} \, dk$$

And the Fourier transform of $f(at) $ is $\frac{1}{a}F(\frac{k}{a})$ $$f(at) = \int^{+\infty}_{-\infty} F(k)e^{ikat} \, dk= \int^{+\infty}_{-\infty} \frac{1}{a}F(\frac{h}{a})e^{iht} \, dh$$ with $h=ka$ so $dh=adk$

Answer 2

Let $g(t) = f(t+a), h(t) = f(at)$, then:

$G(k) = \displaystyle\int_{-\infty}^{\infty} g(t)e^{-ikt} dt = e^{ika}\displaystyle\int_{-\infty}^{\infty} f(t+a)e^{-ik(t+a)} dt = e^{ika}\displaystyle\int_{-\infty}^{\infty} f(t)e^{-ikt}dt = e^{ika}F(k)$

$H(k) = \displaystyle\int_{-\infty}^{\infty} h(t)e^{-ikt} dt = \displaystyle\int_{-\infty}^{\infty} f(at)e^{-i(k/a)(at)} dt = \dfrac{1}{a}\displaystyle\int_{-\infty}^{\infty} f(t)e^{-i(k/a)t} dt = \dfrac{1}{a}F\left(\dfrac{k}{a}\right)$