diff --git a/notes/ewald.lyx b/notes/ewald.lyx
index db9efeb..5b1dfdc 100644
--- a/notes/ewald.lyx
+++ b/notes/ewald.lyx
@@ -1021,6 +1021,103 @@ now the Stirling number of the 2nd kind
 \begin_inset Formula $\kappa>t$
 \end_inset
 
+.
+\end_layout
+
+\begin_layout Plain Layout
+What about the gamma fn on the left? Using DLMF 5.5.5, which says 
+\begin_inset Formula $Γ(2z)=\pi^{-1/2}2^{2z-1}\text{Γ}(z)\text{Γ}(z+\frac{1}{2})$
+\end_inset
+
+ we have 
+\begin_inset Formula 
+\[
+\text{Γ}\left(2-q+n\right)=\frac{2^{1-q+n}}{\sqrt{\pi}}\text{Γ}\left(\frac{2-q+n}{2}\right)\text{Γ}\left(\frac{3-q+n}{2}\right),
+\]
+
+\end_inset
+
+so
+\begin_inset Formula 
+\begin{eqnarray*}
+\pht n{s_{q,k_{0}}^{\textup{L}\kappa,c}}\left(k\right) & = & \frac{\kor{\text{Γ}\left(2-q+n\right)}}{\kor{2^{n}}k_{0}^{q}}\kor{\pi}\sum_{s=0}^{\infty}\left(-1\right)^{s}k^{-2+q-2s}\kappa!\left(-1\right)^{\kappa}\left(\frac{\left(\frac{2-q+n}{2}\right)_{s}\left(\frac{2-q-n}{2}\right)_{s}}{\kor{\text{Γ}\left(\frac{3-q+n}{2}\right)}\text{Γ}\left(\frac{q+n}{2}\right)\text{Γ}\left(\frac{1}{2}+s\right)s!}\sum_{t=0}^{2s}\binom{2s}{t}\begin{Bmatrix}t\\
+\kappa
+\end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s-t}-\frac{\left(\frac{3-q+n}{2}\right)_{s}\left(\frac{3-q-n}{2}\right)_{s}}{\kor{\text{Γ}\left(\frac{2-q+n}{2}\right)}\text{Γ}\left(\frac{q+n-1}{2}\right)\text{Γ}\left(\frac{3}{2}+s\right)s!}\sum_{t=0}^{2s}\binom{2s+1}{t}\begin{Bmatrix}t\\
+\kappa
+\end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s+1-t}k^{-1}\right)\\
+ & = & \frac{2^{1-q}}{k_{0}^{q}}\sqrt{\pi}\sum_{s=0}^{\infty}\left(-1\right)^{s}k^{-2+q-2s}\kappa!\left(-1\right)^{\kappa}\left(\frac{\kor{\text{Γ}\left(\frac{2-q+n}{2}\right)\left(\frac{2-q+n}{2}\right)_{s}}\left(\frac{2-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n}{2}\right)\text{Γ}\left(\frac{1}{2}+s\right)s!}\sum_{t=0}^{2s}\binom{2s}{t}\begin{Bmatrix}t\\
+\kappa
+\end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s-t}-\frac{\kor{\text{Γ}\left(\frac{3-q+n}{2}\right)\left(\frac{3-q+n}{2}\right)_{s}}\left(\frac{3-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n-1}{2}\right)\text{Γ}\left(\frac{3}{2}+s\right)s!}\sum_{t=0}^{2s}\binom{2s+1}{t}\begin{Bmatrix}t\\
+\kappa
+\end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s+1-t}k^{-1}\right)\\
+\mbox{(D5.2.5)} & = & \frac{2^{1-q}}{k_{0}^{q}}\sqrt{\pi}\sum_{s=0}^{\infty}\left(-1\right)^{s}k^{-2+q-2s}\kappa!\left(-1\right)^{\kappa}\left(\frac{\text{Γ}\left(\frac{2-q+n}{2}+s\right)\left(\frac{2-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n}{2}\right)\text{Γ}\left(\frac{1}{2}+s\right)s!}\sum_{t=0}^{2s}\binom{2s}{t}\begin{Bmatrix}t\\
+\kappa
+\end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s-t}-\frac{\text{Γ}\left(\frac{3-q+n}{2}+s\right)\left(\frac{3-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n-1}{2}\right)\text{Γ}\left(\frac{3}{2}+s\right)s!}\sum_{t=0}^{2s}\binom{2s+1}{t}\begin{Bmatrix}t\\
+\kappa
+\end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s+1-t}k^{-1}\right)
+\end{eqnarray*}
+
+\end_inset
+
+The two terms have to be treated fifferently depending on whether q
+\begin_inset Formula $q+n$
+\end_inset
+
+ is even or odd.
+ 
+\end_layout
+
+\begin_layout Plain Layout
+First, assume that 
+\begin_inset Formula $q+n$
+\end_inset
+
+ is even, so the left term has gamma functions and pochhammer symbols with
+ integer arguments, while the right one has half-integer arguments.
+ As 
+\begin_inset Formula $n$
+\end_inset
+
+ is non-negative and 
+\begin_inset Formula $q$
+\end_inset
+
+ is positive, 
+\begin_inset Formula $\frac{q+n}{2}$
+\end_inset
+
+ is positive, and the Pochhammer symbol 
+\begin_inset Formula $\left(\frac{2-q-n}{2}\right)_{s}=0$
+\end_inset
+
+ if 
+\begin_inset Formula $s\ge\frac{q+n}{2}$
+\end_inset
+
+, which transforms the sum over 
+\begin_inset Formula $s$
+\end_inset
+
+ to a finite sum for the left term.
+ However, there still remain divergent terms if 
+\begin_inset Formula $\frac{2-q+n}{2}+s\le0$
+\end_inset
+
+ (let's handle this later; maybe D15.8.6–7 may be then be useful)! Now we
+ need to perform some transformations of variables to make the other sum
+ finite as well
+\end_layout
+
+\begin_layout Plain Layout
+Pár kroků zpět:
+\begin_inset Formula 
+\begin{eqnarray*}
+\pht n{s_{q,k_{0}}^{\textup{L}\kappa,c}}\left(k\right) & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{\kor{\text{Γ}\left(2-q+n\right)}}{\kor{2^{n}}k_{0}^{q}}\kor{\pi}\sum_{s=0}^{\infty}\left(-1\right)^{s}k^{-2+q-2s}\left(\sigma c-ik_{0}\right)^{2s}\times\left(\underbrace{\frac{\left(\frac{2-q+n}{2}\right)_{s}\left(\frac{2-q-n}{2}\right)_{s}}{\kor{\text{Γ}\left(\frac{3-q+n}{2}\right)}\text{Γ}\left(\frac{q+n}{2}\right)\text{Γ}\left(\frac{1}{2}+s\right)s!}}_{\equiv c_{q,n,s}}-\underbrace{\frac{\left(\frac{3-q+n}{2}\right)_{s}\left(\frac{3-q-n}{2}\right)_{s}}{\kor{\text{Γ}\left(\frac{2-q+n}{2}\right)}\text{Γ}\left(\frac{q+n-1}{2}\right)\text{Γ}\left(\frac{3}{2}+s\right)s!}}_{č_{q,n,s}}\frac{\left(\sigma c-ik_{0}\right)}{k}\right)\\
+ & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{2^{1-q}}{k_{0}^{q}}\sqrt{\pi}\sum_{s=0}^{\infty}\left(-1\right)^{s}k^{-2+q-2s}\left(\sigma c-ik_{0}\right)^{2s}\times\left(\frac{\text{Γ}\left(\frac{2-q+n}{2}\right)\left(\frac{2-q+n}{2}\right)_{s}\left(\frac{2-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n}{2}\right)\text{Γ}\left(\frac{1}{2}+s\right)s!}-\frac{\text{Γ}\left(\frac{3-q+n}{2}\right)\left(\frac{3-q+n}{2}\right)_{s}\left(\frac{3-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n-1}{2}\right)\text{Γ}\left(\frac{3}{2}+s\right)s!}\frac{\left(\sigma c-ik_{0}\right)}{k}\right)
+\end{eqnarray*}
+
+\end_inset
+
 
 \end_layout
 
@@ -1313,7 +1410,7 @@ where the spherical Hankel transform
  2)
 \begin_inset Formula 
 \[
-\bsht lg(k)\equiv\int_{0}^{\infty}\ud r\,r^{2}g(r)j_{l}\left(kr\right).
+\bsht lg(k)\equiv\int_{0}^{\infty}\ud r\, r^{2}g(r)j_{l}\left(kr\right).
 \]
 
 \end_inset
@@ -1323,7 +1420,7 @@ Using this convention, the inverse spherical Hankel transform is given by
  3)
 \begin_inset Formula 
 \[
-g(r)=\frac{2}{\pi}\int_{0}^{\infty}\ud k\,k^{2}\bsht lg(k)j_{l}(k),
+g(r)=\frac{2}{\pi}\int_{0}^{\infty}\ud k\, k^{2}\bsht lg(k)j_{l}(k),
 \]
 
 \end_inset
@@ -1336,7 +1433,7 @@ so it is not unitary.
 An unitary convention would look like this:
 \begin_inset Formula 
 \begin{equation}
-\usht lg(k)\equiv\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}\ud r\,r^{2}g(r)j_{l}\left(kr\right).\label{eq:unitary 3d Hankel tf definition}
+\usht lg(k)\equiv\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}\ud r\, r^{2}g(r)j_{l}\left(kr\right).\label{eq:unitary 3d Hankel tf definition}
 \end{equation}
 
 \end_inset
@@ -1390,7 +1487,7 @@ where the Hankel transform of order
  is defined as
 \begin_inset Formula 
 \begin{equation}
-\pht mg\left(k\right)=\int_{0}^{\infty}\ud r\,g(r)J_{m}(kr)r\label{eq:unitary 2d Hankel tf definition}
+\pht mg\left(k\right)=\int_{0}^{\infty}\ud r\, g(r)J_{m}(kr)r\label{eq:unitary 2d Hankel tf definition}
 \end{equation}
 
 \end_inset