docs: maths: minor improvements

src/content/docs/maths/real-analysis/15-continuity-theorems.md

@@ -23,8 +23,10 @@ $f(a)<u<f(b)$: $\exists c \in (a,b)$ such that $f(c)=u$.
 :::note[Proof Hint]
 
 - Define $g(x)=f(x)-u$
-- Define $A=\{ x \in [a,b) \,|\,g(x)\gt 0 \}$
-- Show that $\sup A$ ($=c$) exists. Assume and contradict these cases:
+- Start with $f(b)\lt u \lt f(a)$
+- Define $A=\{ x \in [a,b] \,|\,g(x)\gt 0 \}$
+- Show that $\sup A$ ($=c$) exists.
+- Use the continuity definition. Assume and contradict these cases:
   - $c=a$ (use $2\epsilon = g(a)$)
   - $c=b$ (use $2\epsilon = -g(b)$)
   - $c\in(a,b)$ then contradict:

src/content/docs/maths/real-analysis/32-uniformly-cauchy.md

@@ -17,10 +17,11 @@ u_n(x)-u_m(x)
 \rvert \lt \epsilon
 ```
 
-If $u_n(x)$ is a sequence of real-valued functions, then:
+If $u_n(x)$ is a sequence of real-valued functions, then: $ $
 
 ```math
-u_n(x)\text{ converges uniformly} \iff
+u_n(x)\text{ converges uniformly}
+\iff
 u_n(x)\text{ is uniformly Cauchy}
 ```
 

src/content/docs/maths/riemann-integration/16-improper-riemann-integrals.md

@@ -21,17 +21,6 @@ Suppose $f:(a,b]\to\mathbb{R}$ is integrable on $[c,b]\;\forall c\in (a,b)$.
 Can be similarly defined on the other endpoint. The above integral converges
 **iff** the limit exists and finite. Otherwise diverges.
 
-#### Examples
-
-```math
-\int_0^1 \frac{1}{x^p}\,\text{d}x =
-\lim_\limits{\epsilon \to 0^+}
-\int_\epsilon^1 \frac{1}{x^p}\,\text{d}x
-```
-
-The above integral converges (to $\frac{1}{p-1}$) **iff** $0 \lt p \lt 1$. When
-$p\ge 1$, it diverges to $\infty$.
-
 ### Type 2
 
 A function defined on unbounded interval (including $\infty$). $ $
@@ -45,17 +34,6 @@ Suppose $f:[a,\infty)\to\mathbb{R}$ is integrable on $[a,r] \forall r > a$.
 Can be similarly defined on the other endpoint. The above integral converges
 **iff** the limit exists and finite. Otherwise diverges.
 
-#### Examples
-
-```math
-\int_1^\infty \frac{1}{x^p}\,\text{d}x =
-\lim_\limits{r \to \infty}
-\int_1^r \frac{1}{x^p}\,\text{d}x
-```
-
-The above integral converges (to $\frac{1}{p-1}$) **iff** $p \gt 1$. When
-$0 \lt p \le 1$, it diverges to $\infty$.
-
 ### Type 3
 
 A function that is undefined at finite number of points. The integral can be
@@ -79,3 +57,29 @@ Convergence of improper integrals is similar to the convergence of
 \implies
 \int_a^b f \;\text{converges}
 ```
+
+## Common integrals
+
+```math
+\int_0^1 \frac{1}{x^p}\,\text{d}x
+\;\;\;
+\;\;\;
+\;\;\;
+\int_1^\infty \frac{1}{x^p}\,\text{d}x
+```
+
+The above integrals converge **iff** $p$ is in the integrating (open) interval.
+Converges to $\frac{1}{p-1}$ in that case.
+
+```math
+\int_0^1 \frac{\sin^2 x}{x^2}\,\text{d}x
+\;\;\;
+\;\;\;
+\;\;\;
+\int_1^\infty \frac{\sin^2 x}{x^2}\,\text{d}x
+```
+
+Both of the above integrals converges. Direct comparison test can be used.
+
+- For the 1st integral, $\sqrt{x^{-1}}$ can be used
+- For the 2nd integral, $x^{-2}$ can be used