From 3f6a54c456016a616273a73b52cd92ec74d7d4f9 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Gi=C3=B9=20Marcer?= <giuliamarcer@yahoo.it>
Date: Wed, 3 Jun 2020 14:49:07 +0200
Subject: [PATCH] notes: correct integrals spacing

---
 notes/sections/1.md | 3 ++-
 notes/sections/4.md | 6 +++---
 notes/sections/6.md | 2 +-
 3 files changed, 6 insertions(+), 5 deletions(-)

diff --git a/notes/sections/1.md b/notes/sections/1.md
index 9bc0878..ae54584 100644
--- a/notes/sections/1.md
+++ b/notes/sections/1.md
@@ -189,8 +189,9 @@ The cumulative probability was computed by quadrature-based numerical
 integration of the PDF (`gsl_integration_qagiu()` function in GSL). The function
 calculate an approximation of the integral:
 $$
-  I(x) = \int_x^{+\infty} f(t)dt
+  I(x) = \int\limits_x^{+\infty} f(t)dt
 $$
+
 [^1]: This is neither necessary nor the easiest way: it was chosen simply
 because the quantile had been already implemented and was initially
 used for reverse sampling.
diff --git a/notes/sections/4.md b/notes/sections/4.md
index 5c39187..e02b8f1 100644
--- a/notes/sections/4.md
+++ b/notes/sections/4.md
@@ -114,10 +114,10 @@ $$
 The integral $I$ can now be computed. Note that the domain is implicit in the
 characteristic functions:
 $$
-  I(x) = \int_{-\infty}^{+\infty} dP_v \, f_{P_h , P_v} (x, P_v)
-       = \int \limits_{- \sqrt{P_{\text{max}}^2 - P_h}}
+  I(x) = \int\limits_{-\infty}^{+\infty} dP_v \, f_{P_h , P_v} (x, P_v)
+       = \hspace{-20pt} \int \limits_{- \sqrt{P_{\text{max}}^2 - P_h}}
          ^{\sqrt{P_{\text{max}}^2 - P_h}}
-         dP_v \, f_{P_h , P_v} (x, P_v)
+         \hspace{-20pt} dP_v \, f_{P_h , P_v} (x, P_v)
 $$
 
 With some basic calculus and the identity
diff --git a/notes/sections/6.md b/notes/sections/6.md
index d124686..e59d7db 100644
--- a/notes/sections/6.md
+++ b/notes/sections/6.md
@@ -55,7 +55,7 @@ though, $\theta$ must be uniformly distributed on the half sphere, hence:
   \frac{d^2 P}{d\omega^2} = \frac{1}{2 \pi}
   &\thus d^2 P = \frac{1}{2 \pi} d\omega^2 =
   \frac{1}{2 \pi} d\phi \sin{\theta} d\theta \\
-  &\thus \frac{dP}{d\theta} = \int_0^{2 \pi} d\phi \frac{1}{2 \pi} \sin{\theta}
+  &\thus \frac{dP}{d\theta} = \int_0^{2 \pi} \!\!\! d\phi \frac{1}{2 \pi} \sin{\theta}
   = \frac{1}{2 \pi} \sin{\theta} \, 2 \pi = \sin{\theta}
 \end{align*}