let f(a)=∫_0 ^∞ ((cos(x^2 ) +sin(x^2 ))/((x^2 +a^2 )^2 )) dx with a>0 1) calculate f(a) 2) find the values of ∫_0 ^∞ ((cos(x^2 )+sin(x^2 ))/((x^2 +1)^2 ))dx and ∫_0 ^∞ ((cos(x^2 )+sin(x^2 ))/((x^2 +3)^2 ))dx


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Question Number 64970 by mathmax by abdo last updated on 23/Jul/19

$l e t f (a) = \int_{0}^{\infty} \frac{c o s (x^{2}) + s i n (x^{2})}{{(x^{2} + a^{2})}^{2}} d x w i t h a > 0$ $1) c a l c u l a t e f (a)$ $2) f i n d t h e v a l u e s o f \int_{0}^{\infty} \frac{c o s (x^{2}) + s i n (x^{2})}{{(x^{2} + 1)}^{2}} d x a n d$ $\int_{0}^{\infty} \frac{c o s (x^{2}) + s i n (x^{2})}{{(x^{2} + 3)}^{2}} d x$
Commented by~ À ® @ 237 ~ last updated on 23/Jul/19

$w e a l w a y s h a v e c o s (x^{2}) + s i n (x^{2}) = 1$ $s o f (a) = \int_{0}^{\infty} \frac{1}{{(x^{2} + a^{2})}^{2}} d x$ $l e t c h a n g e x = a . t a n t d x = a (1 + {t a n}^{2} t) d t$ $f (a) = \int_{0}^{\frac{π}{2}} \frac{a (1 + {t a n}^{2} t) d t}{{(a^{2} {t a n}^{2} t + a^{2})}^{2}}$ $= \frac{1}{a^{3}} \int_{0}^{\frac{π}{2}} \frac{1}{(1 + {t a n}^{2} t)} d t$ $= \frac{1}{a^{3}} \int_{0}^{\frac{π}{2}} {c o s}^{2} t d t$ $k n o w i n g t h a t {c o s}^{2} t = \frac{1 + c o s 2 t}{2} w e f i n a l l y g o t$ $f (a) = \frac{1}{a^{3}} {[\frac{t}{2} + \frac{1}{4} s i n 2 t]}_{0}^{\frac{π}{2}}$ $= \frac{π}{4 a^{3}}$ $t h e n f (1) = \frac{π}{4} a n d f (\sqrt{3}) = \frac{π}{12 \sqrt{3}}$
Commented bymathmax by abdo last updated on 23/Jul/19

$t h a n k y o u s i r .$
Commented bymathmax by abdo last updated on 23/Jul/19

$r e a l l y i t s c o s (x^{2}) - s i n (x^{2}) n o t + b u t n e v e r m i n d i w i l l p o s t$ $a n o t h e r q u e s t i o n . . .$
Commented byMJS last updated on 24/Jul/19

${(\cos x)}^{2} + {(\sin x)}^{2} = 1$ $but$ $\cos (x^{2}) + \sin (x^{2}) = \sqrt{2} \sin (x^{2} + \frac{π}{4})$ $which is n o t always = 1$
Commented bymathmax by abdo last updated on 24/Jul/19

$s i r \sim 237 y o u a n s w e r i s n o t c o r r e c t . . . .$
Commented bymathmax by abdo last updated on 24/Jul/19

$y o u a r e r i g h t s i r i h a v e c o m m i t e d a e r r o r i d e l e t t h i s p o s t$ $a n d g i v e t h e r i g h t a n s w e r . . .$
Commented bymathmax by abdo last updated on 24/Jul/19
$1)we have cos(x^2 )+sin(x^2 ) =(√2)cos(x^2 −(π/4)) ⇒ f(a) =(√2)∫_0 ^∞ ((cos(x^2 −(π/4)))/((x^2 +a^2 )^2 ))dx ⇒2f(a) =(√2)∫_(−∞) ^(+∞) ((cos(x^2 −(π/4)))/((x^2 +a^2 )^2 ))dx ⇒(√2)f(a) =Re(∫_(−∞) ^(+∞) (e^(i(x^2 −(π/4))) /((x^2 +a^2 )^2 ))ex) let ϕ(z) =(e^(i(z^2 −(π/4))) /((z^2 +a^2 )^2 )) ⇒ ϕ(z) =(e^(i(z^2 −(π/4))) /((z−ia)^2 (z+ia)^2 )) the poles of ϕ are +^− ia (a>0) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,ia) Res(ϕ,ia) =lim_(z→ia) (z−ia)^2 {(z−ia)^2 ϕ(z)}^((1)) =lim_(z→ia) {(e^(i(z^2 −(π/4))) /((z+ia)^2 ))}^((1)) =e^(−((iπ)/4)) lim_(z→ia) {(e^(iz^2 ) /((z+ia)^2 ))}^((1)) =e^(−((iπ)/4)) lim_(z→ia) ((2iz e^(iz^2 ) (z+ia)^2 −2(z+ia)e^(iz^2 ) )/((z+ia)^4 )) =e^(−((iπ)/4)) lim_(z→ia) (((2iz(z+ia)−2)e^(iz^2 ) )/((z+ia)^3 )) =e^(−i(π/4)) ((2i(ia)(2ia)−2)/((2ia)^3 )) e^(i(ia)^2 ) = e^(−((iπ)/4)) ((−4ia^2 −2)/(−8ia^3 )) e^(−ia^2 ) =((2ia^2 −1)/(4ia^3 )) e^(−i((π/4)+a^2 )) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ((2ia^2 −1)/(4ia^3 )) e^(−i((π/4)+a^2 )) =−(π/(2a^3 ))(1−2ia^2 )( cos((π/4)+a^2 )−isin((π/4)+a^2 )) =−(π/(2a^3 )){cos((π/4)+a^2 )−isin((π/4)+a^2 )−2ia^2 cos((π/4)+a^2 )−2a^2 sin((π/4)+a^2 )} (√2)f(a) =−(π/(2a^3 ))(cos((π/4)+a^2 )−2a^2 sin((π/4) +a^2 )) =−(π/(2a^3 ))cos((π/4)+a^2 ) +(π/a) sin((π/4) +a^2 ) ⇒ f(a) =(π/(a(√2)))sin((π/4) +a^2 )−(π/(2(√2)a^3 )) cos((π/4) +a^2 )$
$1) w e h a v e c o s (x^{2}) + s i n (x^{2}) = \sqrt{2} c o s (x^{2} - \frac{π}{4}) \Rightarrow$ $f (a) = \sqrt{2} \int_{0}^{\infty} \frac{c o s (x^{2} - \frac{π}{4})}{{(x^{2} + a^{2})}^{2}} d x \Rightarrow 2 f (a) = \sqrt{2} \int_{- \infty}^{+ \infty} \frac{c o s (x^{2} - \frac{π}{4})}{{(x^{2} + a^{2})}^{2}} d x$ $\Rightarrow \sqrt{2} f (a) = R e (\int_{- \infty}^{+ \infty} \frac{e^{i (x^{2} - \frac{π}{4})}}{{(x^{2} + a^{2})}^{2}} e x) l e t φ (z) = \frac{e^{i (z^{2} - \frac{π}{4})}}{{(z^{2} + a^{2})}^{2}} \Rightarrow$ $φ (z) = \frac{e^{i (z^{2} - \frac{π}{4})}}{{(z - i a)}^{2} {(z + i a)}^{2}} t h e p o l e s o f φ a r e \bar{+} i a (a > 0) r e s i d u s$ $t h e o r e m g i v e \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i a)$ $R e s (φ, i a) = {l i m}_{z \to i a} {(z - i a)}^{2} {{(z - i a)}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to i a} {\frac{e^{i (z^{2} - \frac{π}{4})}}{{(z + i a)}^{2}}}^{(1)} = e^{- \frac{i π}{4}} {l i m}_{z \to i a} {\frac{e^{{i z}^{2}}}{{(z + i a)}^{2}}}^{(1)}$ $= e^{- \frac{i π}{4}} {l i m}_{z \to i a} \frac{2 i z e^{{i z}^{2}} {(z + i a)}^{2} - 2 (z + i a) e^{{i z}^{2}}}{{(z + i a)}^{4}}$ $= e^{- \frac{i π}{4}} {l i m}_{z \to i a} \frac{(2 i z (z + i a) - 2) e^{{i z}^{2}}}{{(z + i a)}^{3}}$ $= e^{- i \frac{π}{4}} \frac{2 i (i a) (2 i a) - 2}{{(2 i a)}^{3}} e^{i {(i a)}^{2}} = e^{- \frac{i π}{4}} \frac{- 4 {i a}^{2} - 2}{- 8 {i a}^{3}} e^{- {i a}^{2}}$ $= \frac{2 {i a}^{2} - 1}{4 {i a}^{3}} e^{- i (\frac{π}{4} + a^{2})} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{2 {i a}^{2} - 1}{4 {i a}^{3}} e^{- i (\frac{π}{4} + a^{2})}$ $= - \frac{π}{2 a^{3}} (1 - 2 {i a}^{2}) (c o s (\frac{π}{4} + a^{2}) - i s i n (\frac{π}{4} + a^{2}))$ $= - \frac{π}{2 a^{3}} {c o s (\frac{π}{4} + a^{2}) - i s i n (\frac{π}{4} + a^{2}) - 2 {i a}^{2} c o s (\frac{π}{4} + a^{2}) - 2 a^{2} s i n (\frac{π}{4} + a^{2})}$ $\sqrt{2} f (a) = - \frac{π}{2 a^{3}} (c o s (\frac{π}{4} + a^{2}) - 2 a^{2} s i n (\frac{π}{4} + a^{2}))$ $= - \frac{π}{2 a^{3}} c o s (\frac{π}{4} + a^{2}) + \frac{π}{a} s i n (\frac{π}{4} + a^{2}) \Rightarrow$ $f (a) = \frac{π}{a \sqrt{2}} s i n (\frac{π}{4} + a^{2}) - \frac{π}{2 \sqrt{2} a^{3}} c o s (\frac{π}{4} + a^{2})$
Commented bymathmax by abdo last updated on 24/Jul/19

$2) \int_{0}^{\infty} \frac{c o s (x^{2}) + s i n (x^{2})}{{(x^{2} + 1)}^{2}} d x = f (1) = \frac{π}{\sqrt{2}} s i n (\frac{π}{4} + 1) - \frac{π}{2 \sqrt{2}} c o s (\frac{π}{4} + 1)$ $\int_{0}^{\infty} \frac{c o s (x^{2}) + s i n (x^{2})}{{(x^{2} + 3)}^{2}} = f (\sqrt{3}) = \frac{π}{\sqrt{6}} s i n (\frac{π}{4} + 3) - \frac{π}{2 \sqrt{2} 3 \sqrt{3}} c o s (\frac{π}{4} + 3)$ $= \frac{π}{\sqrt{6}} s i n (3 + \frac{π}{4}) - \frac{π}{6 \sqrt{6}} c o s (3 + \frac{π}{4}) .$
Commented by~ À ® @ 237 ~ last updated on 25/Jul/19

$y e s y o u a r e r i g h t . I d i d n o t m i n d t h e d i f f e r e n c e a t t h i s t i m e . S o r r y$
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