Rewrite "Lindhard function", split off "dielectric function"

1 files changed, 15 insertions, 15 deletions

diff --git a/content/know/concept/convolution-theorem/index.pdc b/content/know/concept/convolution-theorem/index.pdc
index 2712c21..a14e960 100644
--- a/content/know/concept/convolution-theorem/index.pdc
+++ b/content/know/concept/convolution-theorem/index.pdc
@@ -26,7 +26,7 @@ and $A$ and $B$ are constants from its definition:
 $$\begin{aligned}
     \boxed{
         \begin{aligned}
-            A \cdot (f * g)(x) &= \hat{\mathcal{F}}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\} \\
+            A \cdot (f * g)(x) &= \hat{\mathcal{F}}{}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\} \\
             B \cdot (\tilde{f} * \tilde{g})(k) &= \hat{\mathcal{F}}\{f(x) \: g(x)\}
         \end{aligned}
     }
@@ -41,12 +41,12 @@ We expand the right-hand side of the theorem and
 rearrange the integrals:
 
 $$\begin{aligned}
-    \hat{\mathcal{F}}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\}
-    &= B \int_{-\infty}^\infty \tilde{f}(k) \Big( A \int_{-\infty}^\infty g(x') \exp(i s k x') \dd{x'} \Big) \exp(-i s k x) \dd{k}
+    \hat{\mathcal{F}}{}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\}
+    &= B \int_{-\infty}^\infty \tilde{f}(k) \Big( A \int_{-\infty}^\infty g(x') \exp\!(i s k x') \dd{x'} \Big) \exp\!(-i s k x) \dd{k}
     \\
-    &= A \int_{-\infty}^\infty g(x') \Big( B \int_{-\infty}^\infty \tilde{f}(k) \exp(- i s k (x - x')) \dd{k} \Big) \dd{x'}
+    &= A \int_{-\infty}^\infty g(x') \Big( B \int_{-\infty}^\infty \tilde{f}(k) \exp\!(- i s k (x - x')) \dd{k} \Big) \dd{x'}
     \\
-    &= A \int_{-\infty}^\infty g(x') f(x - x') \dd{x'}
+    &= A \int_{-\infty}^\infty g(x') \: f(x - x') \dd{x'}
     = A \cdot (f * g)(x)
 \end{aligned}$$
 
@@ -55,11 +55,11 @@ this time starting from a product in the $x$-domain:
 
 $$\begin{aligned}
     \hat{\mathcal{F}}\{f(x) \: g(x)\}
-    &= A \int_{-\infty}^\infty f(x) \Big( B \int_{-\infty}^\infty \tilde{g}(k') \exp(- i s x k') \dd{k'} \Big) \exp(i s k x) \dd{x}
+    &= A \int_{-\infty}^\infty f(x) \Big( B \int_{-\infty}^\infty \tilde{g}(k') \exp\!(- i s x k') \dd{k'} \Big) \exp\!(i s k x) \dd{x}
     \\
-    &= B \int_{-\infty}^\infty \tilde{g}(k') \Big( A \int_{-\infty}^\infty f(x) \exp(i s x (k - k')) \dd{x} \Big) \dd{k'}
+    &= B \int_{-\infty}^\infty \tilde{g}(k') \Big( A \int_{-\infty}^\infty f(x) \exp\!(i s x (k - k')) \dd{x} \Big) \dd{k'}
     \\
-    &= B \int_{-\infty}^\infty \tilde{g}(k') \tilde{f}(k - k') \dd{k'}
+    &= B \int_{-\infty}^\infty \tilde{g}(k') \: \tilde{f}(k - k') \dd{k'}
     = B \cdot (\tilde{f} * \tilde{g})(k)
 \end{aligned}$$
 </div>
@@ -73,10 +73,10 @@ the convolution theorem can also be stated using
 the [Laplace transform](/know/concept/laplace-transform/):
 
 $$\begin{aligned}
-    \boxed{(f * g)(t) = \hat{\mathcal{L}}^{-1}\{\tilde{f}(s) \: \tilde{g}(s)\}}
+    \boxed{(f * g)(t) = \hat{\mathcal{L}}{}^{-1}\{\tilde{f}(s) \: \tilde{g}(s)\}}
 \end{aligned}$$
 
-Because the inverse Laplace transform $\hat{\mathcal{L}}^{-1}$ is
+Because the inverse Laplace transform $\hat{\mathcal{L}}{}^{-1}$ is
 unpleasant, the theorem is often stated using the forward transform
 instead:
 
@@ -95,20 +95,20 @@ because we set both $f(t)$ and $g(t)$ to zero for $t < 0$:
 
 $$\begin{aligned}
     \hat{\mathcal{L}}\{(f * g)(t)\}
-    &= \int_0^\infty \Big( \int_0^\infty g(t') f(t - t') \dd{t'} \Big) \exp(- s t) \dd{t}
+    &= \int_0^\infty \Big( \int_0^\infty g(t') f(t - t') \dd{t'} \Big) \exp\!(- s t) \dd{t}
     \\
-    &= \int_0^\infty \Big( \int_0^\infty f(t - t')  \exp(- s t) \dd{t} \Big) g(t') \dd{t'}
+    &= \int_0^\infty \Big( \int_0^\infty f(t - t')  \exp\!(- s t) \dd{t} \Big) g(t') \dd{t'}
 \end{aligned}$$
 
 Then we define a new integration variable $\tau = t - t'$, yielding:
 
 $$\begin{aligned}
     \hat{\mathcal{L}}\{(f * g)(t)\}
-    &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp(- s (\tau + t')) \dd{\tau} \Big) g(t') \dd{t'}
+    &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp\!(- s (\tau + t')) \dd{\tau} \Big) g(t') \dd{t'}
     \\
-    &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp(- s \tau) \dd{\tau} \Big) g(t') \exp(- s t') \dd{t'}
+    &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp\!(- s \tau) \dd{\tau} \Big) g(t') \exp\!(- s t') \dd{t'}
     \\
-    &= \int_0^\infty \tilde{f}(s) g(t') \exp(- s t') \dd{t'}
+    &= \int_0^\infty \tilde{f}(s) \: g(t') \exp\!(- s t') \dd{t'}
     = \tilde{f}(s) \: \tilde{g}(s)
 \end{aligned}$$
 </div>