let f_n (a)=∫_(−∞) ^∞ ((sin(x^n ))/(x^2 +a^2 )) dx with a positif real not 0 and n from N 1) find a explicit form of f(a) 2) calculate g_n (a) =∫_(−∞) ^(+∞) ((sin(x^n ))/((x^2 +a^2 )^2 ))dx 3) calculate ∫_(−∞) ^(+∞) ((sin(x^3 ))/(x^2 +4)) dx and ∫_(−∞) ^(+∞) ((sin(x^2 ))/(x^2 +9))dx 4) calculate ∫_(−∞) ^(+∞) ((sin(x^3 ))/((x^2 +4)^2 ))dx .


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Question Number 56699 by maxmathsup by imad last updated on 21/Mar/19

$l e t f_{n} (a) = \int_{- \infty}^{\infty} \frac{s i n (x^{n})}{x^{2} + a^{2}} d x w i t h a p o s i t i f r e a l n o t 0 a n d n f r o m N$ $1) f i n d a e x p l i c i t f o r m o f f (a)$ $2) c a l c u l a t e g_{n} (a) = \int_{- \infty}^{+ \infty} \frac{s i n (x^{n})}{{(x^{2} + a^{2})}^{2}} d x$ $3) c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{s i n (x^{3})}{x^{2} + 4} d x a n d \int_{- \infty}^{+ \infty} \frac{s i n (x^{2})}{x^{2} + 9} d x$ $4) c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{s i n (x^{3})}{{(x^{2} + 4)}^{2}} d x .$
Commented by maxmathsup by imad last updated on 23/Mar/19
$1)we have f_n (a) =Im( ∫_(−∞) ^(+∞) (e^(ix^n ) /(x^2 +a^2 ))dx) let consider the complex function ϕ(z) =(e^(iz^n ) /(z^2 +a^2 )) ⇒ϕ(z)=(e^(iz^n ) /((z−ia)(z+ia))) so the poles of ϕ are ia and −ia residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,ia) Res(ϕ,ia) =lim_(z→ia) (z−ia)ϕ(z)= (e^(i(ia)^n ) /(2ia)) =(e^(i^(n+1) a^n ) /(2ia)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (e^(i^(n+1) a^n ) /(2ia)) =(π/a) e^(i^(n+1) a^n ) if n is even n=2p ⇒∫_(−∞) ^(+∞) ϕ(z)dz =(π/a) e^(i^(2p+1) a^n ) =(π/a) e^(i(−1)^p a^(2p) ) =(π/a){cos((−1)^p a^(2p) +isin((−1)^p a^(2p) } ⇒f_n (a) =(π/a) sin((−1)^p a^(2p) ) (n=2p) if n is odd ⇒n=2p+1 ⇒∫_(−∞) ^(+∞) ϕ(z)dz =(π/a) e^(i^(2p+2) a^(2p+1) ) =(π/a) e^((−1)^(p+1) a^(2p+1) ) ⇒ f_n (a) =0 2) we have by derivation f_n ^′ (a) = ∫_(−∞) ^(+∞) (∂/∂a)(((sin(x^n ))/(x^2 +a^2 )))dx =−∫_(−∞) ^(+∞) ((2a sin(x^n ))/((x^2 +a^2 )^2 )) dx =−2a g_n (a) ⇒g_n (a)=−(1/(2a)) f_n ^′ (a) g_(2p) (a) =−(1/(2a)) f_(2p) ^′ (a) =−(1/(2a)) {−(π/a^2 ) sin{(−1)^p a^(2p) }+(π/a)(2p)(−1)^p a^(2p−1) cos{(−1)^p a^(2p) } =(π/(2a^3 )) sin{(−1)^p a^(2p) } +((pπ)/a^2 ) (−1)^(p+1) a^(2p−1) cos{(−1)^p a^(2p) } g_(2p+1) (a) =0 due to f_(2p+1) (a)=0 . 3) ∫_(−∞) ^(+∞) ((sin(x^3 ))/(x^2 +4)) dx =f_3 (2)=0 because 3 is odd. ∫_(−∞) ^(+∞) ((sin(x^2 ))/(x^2 +9)) =f_2 (3) =(π/3) sin(−9) =−(π/3) sin(9) .$
$1) w e h a v e f_{n} (a) = I m (\int_{- \infty}^{+ \infty} \frac{e^{{i x}^{n}}}{x^{2} + a^{2}} d x) l e t c o n s i d e r t h e c o m p l e x f u n c t i o n$ $φ (z) = \frac{e^{{i z}^{n}}}{z^{2} + a^{2}} \Rightarrow φ (z) = \frac{e^{{i z}^{n}}}{(z - i a) (z + i a)} s o t h e p o l e s o f φ a r e i a a n d - i a$ $r e s i d u s t h e o r e m g i v e \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i a)$ $R e s (φ, i a) = {l i m}_{z \to i a} (z - i a) φ (z) = \frac{e^{i {(i a)}^{n}}}{2 i a} = \frac{e^{i^{n + 1} a^{n}}}{2 i a} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{e^{i^{n + 1} a^{n}}}{2 i a} = \frac{π}{a} e^{i^{n + 1} a^{n}}$ $i f n i s e v e n n = 2 p \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = \frac{π}{a} e^{i^{2 p + 1} a^{n}} = \frac{π}{a} e^{i {(- 1)}^{p} a^{2 p}}$ $= \frac{π}{a} {c o s ({(- 1)}^{p} a^{2 p} + i s i n ({(- 1)}^{p} a^{2 p}} \Rightarrow f_{n} (a) = \frac{π}{a} s i n ({(- 1)}^{p} a^{2 p}) (n = 2 p)$ $i f n i s o d d \Rightarrow n = 2 p + 1 \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = \frac{π}{a} e^{i^{2 p + 2} a^{2 p + 1}} = \frac{π}{a} e^{{(- 1)}^{p + 1} a^{2 p + 1}} \Rightarrow$ $f_{n} (a) = 0$ $2) w e h a v e b y d e r i v a t i o n f_{n}^{^{'}} (a) = \int_{- \infty}^{+ \infty} \frac{\partial}{\partial a} (\frac{s i n (x^{n})}{x^{2} + a^{2}}) d x$ $= - \int_{- \infty}^{+ \infty} \frac{2 a s i n (x^{n})}{{(x^{2} + a^{2})}^{2}} d x = - 2 a g_{n} (a) \Rightarrow g_{n} (a) = - \frac{1}{2 a} f_{n}^{^{'}} (a)$ $g_{2 p} (a) = - \frac{1}{2 a} f_{2 p}^{^{'}} (a) = - \frac{1}{2 a} {- \frac{π}{a^{2}} s i n {{(- 1)}^{p} a^{2 p}} + \frac{π}{a} (2 p) {(- 1)}^{p} a^{2 p - 1} c o s {{(- 1)}^{p} a^{2 p}}$ $= \frac{π}{2 a^{3}} s i n {{(- 1)}^{p} a^{2 p}} + \frac{p π}{a^{2}} {(- 1)}^{p + 1} a^{2 p - 1} c o s {{(- 1)}^{p} a^{2 p}}$ $g_{2 p + 1} (a) = 0 d u e t o f_{2 p + 1} (a) = 0 .$ $3) \int_{- \infty}^{+ \infty} \frac{s i n (x^{3})}{x^{2} + 4} d x = f_{3} (2) = 0 b e c a u s e 3 i s o d d .$ $\int_{- \infty}^{+ \infty} \frac{s i n (x^{2})}{x^{2} + 9} = f_{2} (3) = \frac{π}{3} s i n (- 9) = - \frac{π}{3} s i n (9) .$
Commented by maxmathsup by imad last updated on 23/Mar/19

$4) \int_{- \infty}^{+ \infty} \frac{s i n (x^{3})}{{(x^{2} + 4)}^{2}} d x = g_{3} (2) = 0 b e c a u s e 3 i s o d d .$
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