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

1 files changed, 9 insertions, 3 deletions

diff --git a/content/know/concept/laplace-transform/index.pdc b/content/know/concept/laplace-transform/index.pdc
index bd7673b..5e91a04 100644
--- a/content/know/concept/laplace-transform/index.pdc
+++ b/content/know/concept/laplace-transform/index.pdc
@@ -33,6 +33,12 @@ This is solved by restricting the domain of $\tilde{f}(s)$
 to $s$ where $\mathrm{Re}\{s\} > s_0$,
 for an $s_0$ large enough to compensate for the growth of $f(t)$.
 
+The **inverse Laplace transform** $\hat{\mathcal{L}}{}^{-1}$ involves complex integration,
+and is therefore a lot more difficult to calculate.
+Fortunately, it is usually avoidable by rewriting a given $s$-space expression
+using [partial fraction decomposition](/know/concept/partial-fraction-decomposition/),
+and then looking up the individual terms.
+
 
 ## Derivatives
 
@@ -42,7 +48,7 @@ This is especially useful for transforming ODEs with variable coefficients:
 
 $$\begin{aligned}
     \boxed{
-        \tilde{f}'(s) = - \hat{\mathcal{L}}\{t f(t)\}
+        \tilde{f}{}'(s) = - \hat{\mathcal{L}}\{t f(t)\}
     }
 \end{aligned}$$
 
@@ -107,9 +113,9 @@ $$\begin{aligned}
     \hat{\mathcal{L}} \big\{ f^{(n)}(t) \big\}
     &= \int_0^\infty f^{(n)}(t) \exp\!(- s t) \dd{t}
     \\
-    &= \big[ f^{(n - 1)}(t) \exp\!(- s t) \big]_0^\infty + s \int_0^\infty f^{(n-1)}(t) \exp\!(- s t) \dd{t}
+    &= \Big[ f^{(n - 1)}(t) \exp\!(- s t) \Big]_0^\infty + s \int_0^\infty f^{(n-1)}(t) \exp\!(- s t) \dd{t}
     \\
-    &= - f^{(n - 1)}(0) + s \big[ f^{(n - 2)}(t) \exp\!(- s t) \big]_0^\infty + s^2 \int_0^\infty f^{(n-2)}(t) \exp\!(- s t) \dd{t}
+    &= - f^{(n - 1)}(0) + s \Big[ f^{(n - 2)}(t) \exp\!(- s t) \Big]_0^\infty + s^2 \int_0^\infty f^{(n-2)}(t) \exp\!(- s t) \dd{t}
 \end{aligned}$$
 
 And so on.