1)findF(a)= ∫_0 ^∞ ((cos(ln(2+x^2 )))/(a^2 +x^2 ))dx witha>0 2) find the value of ∫


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Question Number 57490 by Abdo msup. last updated on 05/Apr/19

$1) f i n d F (a) = \int_{0}^{\infty} \frac{c o s (l n (2 + x^{2}))}{a^{2} + x^{2}} d x w i t h a > 0$ $2) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{c o s (l n (2 + x^{2}))}{4 + x^{2}} d x .$
Commented bymaxmathsup by imad last updated on 06/Apr/19
$1) we have 2F(a) =∫_(−∞) ^(+∞) ((cos(ln(2+x^2 )))/(a^2 +x^2 )) dx =Re(∫_(−∞) ^(+∞) (e^(iln(2+x^2 )) /(x^2 +a^2 )) dx) let ϕ(z) = (e^(iln(2+z^2 )) /(z^2 +a^2 )) ⇒ϕ(z) =(e^(inln(2+z^2 )) /((z−ia)(z+ia))) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,ia) Res(ϕ,ia) =lim_(z→ia) (z−ia)ϕ(z) =(e^(iln(2 +(ia)^2 )) /(2ia)) =(e^(iln(2−a^2 )) /(2ia)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (e^(iln(2−a^2 )) /(2ia)) =(π/a){cos(ln(2−a^2 ))+isin(ln(2−a^2 ))} ⇒ 2F(a) =(π/a)cos{ln(2−a^2 )} ⇒ ★F(a) =(π/(2a)) cos{ln(2−a^2 } ★$
$1) w e h a v e 2 F (a) = \int_{- \infty}^{+ \infty} \frac{c o s (l n (2 + x^{2}))}{a^{2} + x^{2}} d x = R e (\int_{- \infty}^{+ \infty} \frac{e^{i l n (2 + x^{2})}}{x^{2} + a^{2}} d x)$ $l e t φ (z) = \frac{e^{i l n (2 + z^{2})}}{z^{2} + a^{2}} \Rightarrow φ (z) = \frac{e^{i n l n (2 + z^{2})}}{(z - i a) (z + i a)} r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i a)$ $R e s (φ, i a) = {l i m}_{z \to i a} (z - i a) φ (z) = \frac{e^{i l n (2 + {(i a)}^{2})}}{2 i a} = \frac{e^{i l n (2 - a^{2})}}{2 i a} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{e^{i l n (2 - a^{2})}}{2 i a} = \frac{π}{a} {c o s (l n (2 - a^{2})) + i s i n (l n (2 - a^{2}))} \Rightarrow$ $2 F (a) = \frac{π}{a} c o s {l n (2 - a^{2})} \Rightarrow ★ F (a) = \frac{π}{2 a} c o s {l n (2 - a^{2}} ★$
Commented bymaxmathsup by imad last updated on 06/Apr/19
$2) let I =∫_0 ^∞ ((cos{ln(2+x^2 )})/(4+x^2 ))dx ⇒2I =∫_(−∞) ^(+∞) ((cos{ln(2+x^2 )})/(x^2 +4))dx =Re(∫_(−∞) ^(+∞) (e^(i ln(2+x^2 )) /(x^2 +4))dx} let w(z) =(e^(iln(2+z^2 )) /(z^2 +4)) the poles of ϕ are 2i and −2i ∫_(−∞) ^(+∞) w(z)dz =2iπ Res(ϕ,2i) but w(z) =(e^(iln(2+z^2 )) /((z−2i)(z+2i))) ⇒ Res(w,2i) = (e^(iln(2−4)) /(4i)) =(e^(iln(−2)) /(4i)) =(e^(i{ln(2)+ln(−1)}) /(4i)) =(e^(i{ln(2)+iπ}) /(4i)) =(e^(−π) /(4i)) e^(iln(2)) ⇒ ∫_(−∞) ^(+∞) w(z)dz =2iπ (e^(−π) /(4i)) e^(iln(2)) =(e^(−π) /2){ cos(ln(2))+isin(ln(2)}⇒ 2I =(e^(−π) /2) cos{ln(2)} ⇒ ★I =(e^(−π) /4) cos{ln(2)}★$
$2) l e t I = \int_{0}^{\infty} \frac{c o s {l n (2 + x^{2})}}{4 + x^{2}} d x \Rightarrow 2 I = \int_{- \infty}^{+ \infty} \frac{c o s {l n (2 + x^{2})}}{x^{2} + 4} d x$ $= R e (\int_{- \infty}^{+ \infty} \frac{e^{i l n (2 + x^{2})}}{x^{2} + 4} d x} l e t w (z) = \frac{e^{i l n (2 + z^{2})}}{z^{2} + 4} t h e p o l e s o f φ a r e 2 i a n d - 2 i$ $\int_{- \infty}^{+ \infty} w (z) d z = 2 i π R e s (φ, 2 i) b u t w (z) = \frac{e^{i l n (2 + z^{2})}}{(z - 2 i) (z + 2 i)} \Rightarrow$ $R e s (w, 2 i) = \frac{e^{i l n (2 - 4)}}{4 i} = \frac{e^{i l n (- 2)}}{4 i} = \frac{e^{i {l n (2) + l n (- 1)}}}{4 i} = \frac{e^{i {l n (2) + i π}}}{4 i} = \frac{e^{- π}}{4 i} e^{i l n (2)} \Rightarrow$ $\int_{- \infty}^{+ \infty} w (z) d z = 2 i π \frac{e^{- π}}{4 i} e^{i l n (2)} = \frac{e^{- π}}{2} {c o s (l n (2)) + i s i n (l n (2)} \Rightarrow$ $2 I = \frac{e^{- π}}{2} c o s {l n (2)} \Rightarrow ★ I = \frac{e^{- π}}{4} c o s {l n (2)} ★$
	Answered by einsteindrmaths@hotmail.fr last updated on 05/Apr/19
	$a)f(a)=(1/2)∫_(−∞) ^∞ ((cosln (2+x^2 ))/(a^2 +x^2 ))dx letf(z)=((cos(ln(2+z^2 )))/(a^2 +z^2 )) withe z≠{−i(√2) .i(√2) } res(f(z).ia)=lim_(z→ia) ((cos(ln(z^2 +2)))/((z+ia)))=((cos(ln(2−a^2 )))/(2ia)) we get f(a)=(1/2)2iπ×((cos(ln(2−a^2 )))/(2ia))=((πcos(ln(2−a^2 )))/(2a)) for a≠i(√2) 2)f(2)=π((cos(ln(−2)))/4)=(π/4)[cos( ln2).cosh(π)+sin (ln2) sh(π) ]$
	$a) f (a) = \frac{1}{2} \underset{- \infty}{\int^{\infty}} \frac{c o s \ln (2 + x^{2})}{a^{2} + x^{2}} d x$ $l e t f (z) = \frac{c o s (l n (2 + z^{2}))}{a^{2} + z^{2}} w i t h e z \neq {- i \sqrt{2} . i \sqrt{2}}$ $r e s (f (z) . i a) = {l i m}_{z \to i a} \frac{c o s (l n (z^{2} + 2))}{(z + i a)} = \frac{c o s (l n (2 - a^{2}))}{2 i a}$ $w e g e t f (a) = \frac{1}{2} 2 i π \times \frac{c o s (l n (2 - a^{2}))}{2 i a} = \frac{π c o s (l n (2 - a^{2}))}{2 a} f o r a \neq i \sqrt{2}$ $2) f (2) = π \frac{c o s (l n (- 2))}{4} = \frac{π}{4} [\cos (l n 2) . \cosh (π) + s i n (l n 2) s h (π)]$
	Commented bymaxmathsup by imad last updated on 05/Apr/19

	$s i r e i n s t e i n h o w y o u g e t l n (- 2) t h i s n u m b e r i s c o m p l e x a n d f (2) i s r e a l . . .$
	Commented byeinsteindrmaths@hotmail.fr last updated on 06/Apr/19

	$l n (z) = l n ∣ z ∣ + i a r g (z)$ $l n (- 2) = l n (2) + i π$ $F (a) = π \frac{c o s (l n (2 - a^{2}))}{2 a}$ $F (2) = \frac{π c o s (l n 2 + i π)}{4} = \frac{π c o s (l n 2) c o s (i π) - π s i n (l n 2) s i n (i π)}{4}$ $w i t h e s i n (i π) = - s h (π)$ $a n d c o s (i π) = c h (π)$ $w e g e t \frac{π (c o s (l n 2) \times c h (π) + s i n (l n 2) s h (π))}{4}$
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