Given a time-dependent pattern in a 2D function $b(x,t)$ that follows a path with a constant velocity $v$, which I define with the use of the delta function as: $$b(x,t)=b_0(x)\delta(x-v t)$$ where $b_0(x)$ is the function at time $t=0$.
I'm confronted to a frequency-domain representation of it with the fourier transform: $$B(k,w) = \iint_{-\infty}^{\infty} b(x,t)e^{-i2\pi(kx + wt)}dx \ dt $$ with $k,w$ the 1D space & time frequencies. I'm seeing the following result that I cannot prove:
$$B(k,w) = B_0(k) \delta(w + k v)$$
I tried starting by simply replacing $b(x,t)$ with its delta representation:
$$B(k,w) = \iint_{-\infty}^{\infty} b_0(x)e^{-i2\pi(kx + wt)} \delta(x-v t) dx \ dt $$
I tried first to have a more explicit separation of space and time with
$$B(k,w) = \int_{-\infty}^{\infty} b_0(x)e^{-i2\pi kx} dx \int_{-\infty}^{\infty} \delta(x-v t) e ^{-i2\pi w t} dt $$
That would get me at least half-way there:
$$B(k,w) = B_0(k) \int_{-\infty}^{\infty} e ^{-i2\pi w t} \delta(x-v t) dt $$
So this problem seems to come down to proving that the Fourier transform of $\delta(x-vt)$ acts as $\delta(w + kv)$, which eludes me.
[Update 1]: I cannot separate the way I did as $\delta(x - vt)$ depends on both $x$ and $t$.
So I shoud let $B(k, w)$ written as: $$B(k,w) = \iint_{-\infty}^{\infty} b_0(x) \delta(x-v t) e ^{-i2\pi(kx + w t)} dx \ dt$$
[Update 2]: In the latter expression, if I apply the fundamental property of the Dirac distribution that for a test function $f(x)$ continuous at $x=a$ it does $\int_{-\infty}^{\infty} f(x)\delta(x-a) dx = f(a)$, could I consider $a = v t$ and get:
$$\begin{eqnarray} B(k,w) &=& \int_{-\infty}^{\infty} b_0(vt) e ^{-i2\pi(kvt + w t)} \ dt \\ &=& \int_{-\infty}^{\infty} b_0(vt) e ^{-i2\pi t(kv + w)} dt \end{eqnarray}$$
So rearranging the argument of the exponential shows an effect of $\delta(kv+w)$ but I don't see the formal way to get the final result of $B_0(k) \delta(kv+w)$.
