notes: correct integrals spacing

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.

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

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*}