Guest_1013
2026-08-10
\begin{aligned} \int_0^R \frac{r}{\sqrt{r^2+z^2}} \text{d}r &\xlongequal{r=|z|\tan \phi} \int_0^{\arctan \frac{R}{z}} \frac{\tan \phi}{1+\tan^2 \phi} \text{d} \phi \\ &= \int_0^{\arctan \frac{R}{z}} \sin \phi \text{d} \phi \\ &= -\cos \phi \Big|_0^{\arctan \frac{R}{z}} \\ &= 1 - \frac{|z|}{\sqrt{R^2+z^2}} \end{aligned} \tag{1}
\begin{aligned} \int_0^r r^k \exp (-\frac{r}{a}) &= \int_0^r r^k \d (\exp (-\frac{r}{a})) \cdot (-a) \\ &= -a(r^k \exp(-\frac{r}{a}) \Big|_0^r - \int_0^a k r^{k-1} \exp(-\frac{r}{a})\d r) \\ &= ... \end{aligned} \tag{2}
\int (\frac{1}{u^2}+\frac{2}{u})\exp(-2u) \d u = \frac{\exp(-2u)}{u} \tag{3}
---
title: "04 Some Useful or Interesting Integrals"
author: Guest_1013
date: 2026-08-10
---
# Trigonometry-related
$$
\begin{aligned}
\int_0^R \frac{r}{\sqrt{r^2+z^2}} \text{d}r
&\xlongequal{r=|z|\tan \phi} \int_0^{\arctan \frac{R}{z}} \frac{\tan \phi}{1+\tan^2 \phi} \text{d} \phi \\
&= \int_0^{\arctan \frac{R}{z}} \sin \phi \text{d} \phi \\
&= -\cos \phi \Big|_0^{\arctan \frac{R}{z}} \\
&= 1 - \frac{|z|}{\sqrt{R^2+z^2}}
\end{aligned}
$${#eq-1}
# Exponential
$$
\begin{aligned}
\int_0^r r^k \exp (-\frac{r}{a})
&= \int_0^r r^k \d (\exp (-\frac{r}{a})) \cdot (-a) \\
&= -a(r^k \exp(-\frac{r}{a}) \Big|_0^r - \int_0^a k r^{k-1} \exp(-\frac{r}{a})\d r) \\
&= ...
\end{aligned}
$${#eq-exp-part}
$$
\int (\frac{1}{u^2}+\frac{2}{u})\exp(-2u) \d u = \frac{\exp(-2u)}{u}
$${#eq-exp-coincidence}