Integration notation - integrating over $k'$ instead of $k$

Question

My physics professor has defined a Fourier transform as: $$\tag{1} D_{i}(\boldsymbol{r}, t)=\frac{1}{(2 \pi)^{2}} \iiiint D_{i}\left(\boldsymbol{k}^{\prime}, \omega^{\prime}\right) e^{i\left(\boldsymbol{k}^{\prime} \cdot \boldsymbol{r}-\omega^{\prime} t\right)}\, d \omega^{\prime}\, d^{3} k^{\prime} $$ with the inverse transform of eq. $(1)$ being: $$\tag{2} D_{i}(\boldsymbol{k}, \omega)=\frac{1}{(2 \pi)^{2}} \iiiint D_{i}(\boldsymbol{r}, t) e^{-i(\boldsymbol{k} \cdot \boldsymbol{r}-\omega t)}\, d t\, d^{3} r $$ What is the purpose of using the notation $\omega'$ and $k'$ in eq. $(1)$ while simply integrating over $t$ and $r$ in eq. $(2)$?

Answer 1

It's all about not conflating a free variable with a bound dummy integration variable. If (1) holds, (2)'s RHS is$$\begin{align}&\int_{(\Bbb R^4)^2}D_i(k^\prime,\,\omega^\prime)\frac{e^{i((k^\prime-k)\cdot r-(\omega^\prime-\omega )t)}}{(2\pi)^4}dtd^3rd^3k^\prime d\omega^\prime\\&=\int_{\Bbb R^4}D_i(k^\prime,\,\omega^\prime)\delta^{(3)}(k^\prime-k)\delta(t^\prime-t)d^3k^\prime d\omega^\prime\\&=D_i(k,\,\omega).\end{align}$$