How to find inverse Fourier transform I have the function
$$ \delta(f-2) $$
How can we inverse Fourier transform it? It's easy if $f$ is replaced with $w$. But based on my knowledge, $w = 2\pi f$.
The correct answer is 
$$ e^{4\pi i t} $$
Can somebody explain to me what happened? Thanks.
 A: The inverse Fourier transform of $\delta(f-2)$ is
$$\mathcal F^{-1}[\delta](t) = \int \delta(f-2) e^{i2\pi ft} \, df = e^{i2\pi2t} = e^{i4\pi t}$$
The 2nd equality holds by definition of the delta function.
A: Two commonly used definitions of the Fourier transform of $x(t)$ are
$$\begin{align*}
X(\omega) &= \int_{-\infty}^{\infty} x(t) e^{-i\omega t} \mathrm dt, 
&x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} X(\omega) e^{i \omega t}\mathrm d\omega\\
\hat{X}(f) &= \int_{-\infty}^{\infty} x(t) e^{-i 2\pi ft} \mathrm dt,
&x(t) = \int_{-\infty}^{\infty} X(f) e^{i 2\pi f t}\mathrm df.
\end{align*}$$
The two functions are related as $\hat{X}(f) = X(2\pi f)$ and $X(\omega) = \hat{X}(f/2\pi)$.
I think your question essentially is: if you have a table that tells you the inverse Fourier transform of $X(\omega) = \delta(\omega-\omega_0)$ is $\frac{1}{2\pi}e^{i\omega_0t}$ from which it is easy to deduce that the inverse Fourier transform 
of $\delta(\omega-2)$
is $\frac{1}{2\pi}e^{i2t}$, how do you deduce from this that the inverse Fourier 
transform of $\hat{X}(f-f_0)=\delta(f-f_0)$ is $e^{i 2\pi f_0 t}$ in general, and that the 
inverse Fourier transform of $\delta(f-2)$ is $e^{i 4\pi t}$? As williamdemeo showed
you, and Willie Wong emphasized to you, just computing the Fourier integral
$$x(t) = \int_{-\infty}^{\infty} \hat{X}(f) e^{i2\pi f t} \mathrm df
= \int_{-\infty}^{\infty} \delta(f-2) e^{i2\pi f t} \mathrm df 
= e^{i 2\pi 2 t} = e^{i4\pi t}$$ is far easier than fussing with tables.
But if you are dead set on using tables only, then note that if
$$x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} X(\omega) e^{i \omega t}\mathrm d\omega$$
is known to you, then since it does not matter what we call the variable
of integration
$$
x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} X(\omega) e^{i \omega t}\mathrm d\omega
= \frac{1}{2\pi}\int_{-\infty}^{\infty} X(f) e^{i f t}\mathrm df
= \frac{1}{2\pi}\int_{-\infty}^{\infty} X(f) e^{i 2 \pi f (t/2\pi)}\mathrm df
$$
Let $y(t)$ denote the rightmost integral.  Then we have that $y(t)$ is
the inverse Fourier transform of $X(f)$ evaluated at $t/2\pi$, and it
happens to equal $2\pi x(t)$. So, 

given that 
  $$x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} X(\omega) e^{i \omega t}\mathrm d\omega$$ is the inverse Fourier transform of $X(\omega)$, the inverse
  Fourier transform of $X(f)$ is
  $$\int_{-\infty}^{\infty} X(f) e^{i 2 \pi f t}\mathrm df
= 2\pi \cdot x(2\pi t).$$

In particular, given that the inverse the inverse Fourier transform 
of $\delta(\omega-2)$ is $\frac{1}{2\pi}e^{i2t}$, the inverse Fourier
transform of $\delta(f-2)$ is 
$2\pi \frac{1}{2\pi}e^{i2\cdot 2\pi t} = e^{i4\pi t}$.
