prove-that-0-sinx-x-dx-is-divergent-

Question Number 33170 by prof Abdo imad last updated on 11/Apr/18

p r o v e t h a t \int_{0}^{\infty} \frac{∣ s i n x ∣}{x} d x i s d i v e r g e n t .

Commented by prof Abdo imad last updated on 13/Apr/18

\int_{0}^{\infty} \frac{∣ s i n x ∣}{x} d x = {l i m}_{n \to + \infty} A_{n} w i t h

A_{n} = \int_{0}^{n π} \frac{∣ s i n x ∣}{x} d x b u t

A_{n} = \sum_{k = 0}^{n} \int_{k π}^{(k + 1) π} \frac{∣ s i n x ∣}{x} d x

=_{x = k π + t} \sum_{k = 0}^{n} \int_{0}^{π} \frac{s i n t}{k π + t} d t b u t

0 ⩽ t ⩽ π \Rightarrow k π ⩽ k π + t ⩽ (k + 1) π \Rightarrow

\frac{1}{(k + 1) π} ⩽ \frac{1}{k π + t} ⩽ \frac{1}{k π} \Rightarrow \frac{s i n t}{k π + t} ⩾ \frac{s i n t}{(k + 1) π} \forall t \in [0, π]

\Rightarrow \int_{0}^{π} \frac{s i n t}{k π + t} d t ⩾ \frac{1}{(k + 1) π} \int_{0}^{π} s i n t d t = \frac{2}{(k + 1) π} \Rightarrow

A_{n} ⩾ \frac{2}{π} \sum_{k = 0}^{n} \frac{1}{k + 1} \Rightarrow A_{n} ⩾ \frac{2}{π} \sum_{k = 1}^{n + 1} \frac{1}{k} \Rightarrow

A_{n} ⩾ \frac{2}{π} H_{n + 1}_{n \to + \infty} \to + \infty s o {l i m}_{n \to + \infty} A_{n} = + \infty

a n d t h e i n t e g r a l i s d i v e r g e n t .