find-0-1-cos-x-x-2-dx-with-gt-0-

Question Number 32994 by abdo imad last updated on 09/Apr/18

f i n d \int_{0}^{\infty} \frac{1 - c o s (λ x)}{x^{2}} d x w i t h λ > 0 .

Commented by abdo imad last updated on 15/Apr/18

c h a n g e m e n t λ x = u g i v e

I = \int_{0}^{\infty} \frac{1 - c o s u}{{(\frac{u}{λ})}^{2}} \frac{d u}{λ} = λ \int_{0}^{\infty} \frac{1 - c o s u}{u^{2}} d u b u t

\int_{0}^{\infty} \frac{1 - c o s u}{u^{2}} d u = 2 \int_{0}^{\infty} \frac{{s i n}^{2} (\frac{u}{2})}{u^{2}} d u

=_{\frac{u}{2} = t} 2 \int_{0}^{\infty} \frac{{s i n}^{2} t}{4 t^{2}} 2 d t = \int_{0}^{\infty} \frac{{s i n}^{2} t}{t^{2}} d t (b y p a r t s)

= {[- \frac{1}{t} {s i n}^{2} t]}_{0}^{+ \infty} - \int_{0}^{\infty} - \frac{1}{t} 2 s i n t c o s t d t

= \int_{0}^{\infty} \frac{s i n (2 t)}{t} d t =_{2 t = x} \int_{0}^{\infty} \frac{s i n (x)}{\frac{x}{2}} \frac{d x}{2} = \int_{0}^{+ \infty} \frac{s i n x}{x} d x

= \frac{π}{2} \Rightarrow I = \frac{λ π}{2} .