Infi-2 6

Dirichlet's convergence test for integrals

\begin{matrix} Let f be a continuously differentiable, monotonically decreasing function \\ lim_{x \to \infty} f (x) = 0 \\ Let g be continuous \\ Let G (x) = \int_{a}^{x} g (t) d t be bounded \\ Then \int_{a}^{\infty} f (x) g (x) d x converges \end{matrix}

\begin{matrix} \int_{a}^{\infty} \frac{\cos (x)}{x} d x \\ f (x) = \frac{1}{x} \\ g (x) = \cos x = G (x) = \sin (x) - \sin (a) \\ ⟹ Integral converges \end{matrix}

\begin{matrix} \int_{a}^{\infty} \sin (x^{2}) d x \\ t = x^{2} ⟹ d t = 2 x d x ⟹ d x = \frac{d t}{2 \sqrt{t}} \\ \int_{a^{2}}^{\infty} \frac{\sin (t)}{2 \sqrt{t}} d t converges \\ But lim_{x \to \infty} \sin (x^{2}) \neq 0 \end{matrix}

\begin{matrix} \int_{- \infty}^{\infty} \frac{1}{x^{2} + 34} d x = \int_{- \infty}^{0} \frac{1}{x^{2} + 34} d x + \int_{0}^{\infty} \frac{1}{x^{2} + 34} d x \\ = lim_{b \to \infty} \frac{1}{\sqrt{34}} \arctan (\frac{x}{\sqrt{34}}) |_{x = - b}^{x = 0} + lim_{b \to \infty} \frac{1}{\sqrt{34}} \arctan (\frac{x}{\sqrt{34}}) |_{x = 0}^{x = b} = \frac{π}{\sqrt{34}} \end{matrix}

\begin{matrix} \int_{0}^{1} \cot x d x = lim_{a \to 0^{+}} \int_{a}^{1} \cot x d x = lim_{a \to 0^{+}} \ln | \sin x | |_{x = a}^{x = 1} = lim_{n \to \infty} \ln (\sin (1)) - \ln (\sin (a)) = \infty \end{matrix}

\begin{matrix} \forall n \geq 0 : \int_{0}^{\infty} x^{n} e^{- x} d x = n! \\ Proof (a little more complicated than the standard one) in Lecture 8 \end{matrix}

\begin{matrix} \int_{1}^{\infty} \sin (\frac{x + 20}{x^{3} - 2}) d x = \int_{1}^{3} \sin (\frac{x + 20}{x^{3} - 2}) d x + \int_{3}^{\infty} \sin (\frac{x + 20}{x^{3} - 2}) d x \\ \int_{3}^{\infty} \sin (\frac{x + 20}{x^{3} - 2}) d x converges ⟺ \int_{3}^{\infty} \frac{x + 20}{x^{3} - 2} d x converges \\ \int_{3}^{\infty} \frac{x + 20}{x^{3} - 2} d x converges ⟺ \int_{3}^{\infty} \frac{1}{x^{2}} d x converges \end{matrix}

\begin{matrix} \int_{0}^{1} \frac{\sin (\frac{1}{x})}{x} d x = {\begin{matrix} t = \frac{1}{x} \\ d t = - \frac{1}{x^{2}} d x \\ x = 0 ⟹ t = \infty \\ x = 1 ⟹ t = 1 \end{matrix}} = \int_{1}^{\infty} \frac{\sin (t)}{t} d t \\ ⟹ Converges by the Dirichlet’s test \\ | \frac{\sin t}{t} | \geq \frac{\sin^{2} t}{t} = \frac{1 - \cos (2 t)}{2 t} = \underset{Integral diverges}{\underset{⏟}{\frac{1}{2} (\frac{1}{t} - \frac{\cos (2 t)}{t})}} \end{matrix}

\begin{matrix} \int_{0}^{\infty} (- 1)^{⌊ x^{2} ⌋} d x = \sum_{n = 0}^{\infty} \int_{\sqrt{n}}^{\sqrt{n + 1}} (- 1)^{n} d x = \\ = \sum_{n = 0}^{\infty} (- 1)^{n} \frac{1}{\sqrt{n + 1} + \sqrt{n}} converges by the Leibniz test \end{matrix}

\begin{matrix} \int_{0}^{π / 2} \frac{1}{\sin^{a} (x^{2}) \cos^{b} (x)} d x = \\ = \int_{0}^{1} \frac{1}{\sin^{a} (x^{2}) \cos^{b} (x)} d x + \int_{1}^{π / 2} \frac{1}{\sin^{a} (x^{2}) \cos^{b} (x)} d x \\ lim_{x \to 0} \frac{\frac{1}{\cos^{b} x} \cdot \frac{1}{\sin^{a} (x^{2})}}{{(\frac{1}{x^{2}})}^{a}} = lim_{x \to 0} \frac{1}{\cos^{b} x} \cdot {(\frac{\frac{1}{\sin (x^{2})}}{\frac{1}{x^{2}}})}^{a} \underset{x \to 0}{\to} 1^{a} = 1 \\ ⟹ \int_{0}^{1} f (x) d x converges ⟺ 2 a < 1 ⟺ a < \frac{1}{2} \\ lim_{x \to \frac{π}{2}} \frac{\frac{1}{(π - x)^{b}}}{\frac{1}{\sin^{a} (x^{2}) \cos^{b} (x)}} = \frac{1}{\sin^{a} (\frac{π}{4})} \\ ⟹ \int_{1}^{π / 2} f (x) d x converges ⟺ b < 1 \end{matrix}