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Question Number 43935 by abdo.msup.com last updated on 18/Sep/18

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

2) c a l c u l a t e f^{^{'}} (a) .

Commented by maxmathsup by imad last updated on 19/Sep/18

w e h a v e f (a) = \int_{0}^{\infty} \frac{{s i n}^{2} (a x)}{x^{2}} d x =_{a x = t} \int_{0}^{\infty} \frac{{s i n}^{2} t}{{(\frac{t}{a})}^{2}} \frac{d t}{a}

= a \int_{0}^{\infty} \frac{{s i n}^{2} t}{t^{2}} d t b y p a r t s u^{^{'}} = \frac{1}{t^{2}} a n d v = {s i n}^{2} t \Rightarrow

f (a) = a {{[- \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}

= a {2 \int_{0}^{\infty} \frac{s i n t c o s t}{t} d t} = a \int_{0}^{\infty} \frac{s i n (2 t)}{t} d t =_{2 t = u} a \int_{0}^{\infty} \frac{s i n (u)}{\frac{u}{2}} \frac{d u}{2}

= a \int_{0}^{\infty} \frac{s i n (u)}{u} d u = a \frac{π}{2} \Rightarrow f (a) = \frac{π a}{2} .

2) w e h a v e f (a) = \frac{π a}{2} \Rightarrow f^{^{'}} (a) = 0 .