Equation formatting

_posts/2019-02-15-ch2-notes.md

@@ -114,11 +114,13 @@ real $$x$$.
 Let's see what the Fourier transform of the conjugate of a function looks like.
 
 $$
+\begin{align*}
 (\mathcal{F} \overline{f})(x)
-= \int_\mathbb{R} \overline{f(x')} e^{-2 \pi i x' x} \mathrm{d} x'
-= \int_\mathbb{R} \overline{f(x') e^{2 \pi i x' x}} \mathrm{d} x'
-= \overline{\int_\mathbb{R} f(x') e^{2 \pi i x' x} \mathrm{d} x'}
-= \overline{(\mathcal{F}^{-1} f)(x)}
+&= \int_\mathbb{R} \overline{f(x')} e^{-2 \pi i x' x} \mathrm{d} x' \\
+&= \int_\mathbb{R} \overline{f(x') e^{2 \pi i x' x}} \mathrm{d} x' \\
+&= \overline{\int_\mathbb{R} f(x') e^{2 \pi i x' x} \mathrm{d} x'} \\
+&= \overline{(\mathcal{F}^{-1} f)(x)}
+\end{align*}
 $$
 
 This implies that $$ \mathcal{F}^{-1} f$$ is the complex conjugate of $$\mathcal{F} \overline{f}$$.