let I_n (x)= ∫_0 ^∞ ((t sin(t))/((t^2 +x^2 )^n ))dt 1) find a relation between I_(n+1) and I_n 2) calculate I_2 (x) and I_3 (x) 3) calculate ∫


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Question Number 36187 by prof Abdo imad last updated on 30/May/18

$l e t I_{n} (x) = \int_{0}^{\infty} \frac{t s i n (t)}{{(t^{2} + x^{2})}^{n}} d t$ $1) f i n d a r e l a t i o n b e t w e e n I_{n + 1} a n d I_{n}$ $2) c a l c u l a t e I_{2} (x) a n d I_{3} (x)$ $3) c a l c u l a t e \int_{0}^{\infty} \frac{t s i n (t)}{{(2 + t^{2})}^{2}} d t$
Commented by maxmathsup by imad last updated on 15/Aug/18
$3) let A = ∫_0 ^∞ ((t sint)/((2+t^2 )^2 )) dt changement t =(√2)x give A =∫_0 ^∞ (((√2)xsin((√2)x))/(4(1+x^2 ))) (√2)dx =(1/2) ∫_0 ^∞ ((xsin(x(√2)))/((x^2 +1)^2 ))dx =(1/4) ∫_(−∞) ^(+∞) ((x sin(x(√2)))/((x^2 +1)^2 ))dx ⇒ 4A = Im( ∫_(−∞) ^(+∞) ((x e^(ix(√2)) )/((x^2 +1)^2 ))) let considere the complex function ϕ(z) = ((z e^(iz(√2)) )/((z^2 +1)^2 )) ⇒ϕ(z) = ((z e^(iz(√2)) )/((z−i)^2 (z+i)^2 )) so the poles of ϕ are i and −i (doubles) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) but Res(ϕ,i) =lim_(z→i) (1/((2−1)! )) {(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) { ((z e^(iz(√2)) )/((z+i)^2 ))}^((1)) =lim_(z→i) (((e^(iz(√2)) +i(√2)z e^(iz(√2)) )(z+i)^2 −2(z+i)ze^(iz(√2)) )/((z+i)^4 )) =lim_(z→i) (((z+i) (e^(iz(√2)) +i(√2)z e^(iz(√2)) )−2 z e^(iz(√2)) )/((z+i)^3 )) =(((2i)( e^(−(√2)) −(√2)e^(−(√2)) )−2i e^(−(√2)) )/((2i)^3 )) ((−2(√2)i e^(−(√2)) )/(−8i)) = ((√2)/4) e^(−(√2)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ((√2)/4) e^(−(√2)) =((iπ(√2))/2) e^(−(√2)) 4A =Im( ∫_(−∞) ^(+∞) ϕ(z)dz ) =((π(√2))/2) e^(−(√2)) ⇒ A =((π(√2))/8) e^(−(√2)) .$
$3) l e t A = \int_{0}^{\infty} \frac{t s i n t}{{(2 + t^{2})}^{2}} d t c h a n g e m e n t t = \sqrt{2} x g i v e$ $A = \int_{0}^{\infty} \frac{\sqrt{2} x s i n (\sqrt{2} x)}{4 (1 + x^{2})} \sqrt{2} d x = \frac{1}{2} \int_{0}^{\infty} \frac{x s i n (x \sqrt{2})}{{(x^{2} + 1)}^{2}} d x$ $= \frac{1}{4} \int_{- \infty}^{+ \infty} \frac{x s i n (x \sqrt{2})}{{(x^{2} + 1)}^{2}} d x \Rightarrow$ $4 A = I m (\int_{- \infty}^{+ \infty} \frac{x e^{i x \sqrt{2}}}{{(x^{2} + 1)}^{2}}) l e t c o n s i d e r e t h e c o m p l e x f u n c t i o n$ $φ (z) = \frac{z e^{i z \sqrt{2}}}{{(z^{2} + 1)}^{2}} \Rightarrow φ (z) = \frac{z e^{i z \sqrt{2}}}{{(z - i)}^{2} {(z + i)}^{2}} s o t h e p o l e s o f φ a r e i a n d - i$ $(d o u b l e s) r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i) b u t$ $R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to i} {\frac{z e^{i z \sqrt{2}}}{{(z + i)}^{2}}}^{(1)} = {l i m}_{z \to i} \frac{(e^{i z \sqrt{2}} + i \sqrt{2} z e^{i z \sqrt{2}}) {(z + i)}^{2} - 2 (z + i) {z e}^{i z \sqrt{2}}}{{(z + i)}^{4}}$ $= {l i m}_{z \to i} \frac{(z + i) (e^{i z \sqrt{2}} + i \sqrt{2} z e^{i z \sqrt{2}}) - 2 z e^{i z \sqrt{2}}}{{(z + i)}^{3}}$ $= \frac{(2 i) (e^{- \sqrt{2}} - \sqrt{2} e^{- \sqrt{2}}) - 2 i e^{- \sqrt{2}}}{{(2 i)}^{3}}$ $\frac{- 2 \sqrt{2} i e^{- \sqrt{2}}}{- 8 i} = \frac{\sqrt{2}}{4} e^{- \sqrt{2}} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{\sqrt{2}}{4} e^{- \sqrt{2}} = \frac{i π \sqrt{2}}{2} e^{- \sqrt{2}}$ $4 A = I m (\int_{- \infty}^{+ \infty} φ (z) d z) = \frac{π \sqrt{2}}{2} e^{- \sqrt{2}} \Rightarrow A = \frac{π \sqrt{2}}{8} e^{- \sqrt{2}} .$
Commented by maxmathsup by imad last updated on 19/Aug/18
$2) we have I_2 (x)= ∫_0 ^∞ ((t sint)/((t^2 +x^2 )^2 )) dt =(1/2) ∫_(−∞) ^(+∞) ((tsint)/((t^2 +x^2 )^2 ))dt ⇒ 2I_2 (x) =Im( ∫_(−∞) ^(+∞) ((t e^(it) )/((t^2 +x^2 )^2 ))dt)=Im(A(x)) changement t=xu give A(x) = ∫_(−∞) ^(+∞) ((xu e^(ixu) )/(x^4 (1+u^2 )^2 )) x du = (1/x^2 ) ∫_(−∞) ^(+∞) ((u e^(ixu) )/((u^2 +1)^2 )) du (we suppose x>0) let consider the complex function ϕ(z) =((z e^(ixz) )/((z^2 +1)^2 )) ϕ(z) =((z e^(ixz) )/((z−i)^2 (z+i)^2 )) the poles of ϕ are i and −i(doubles) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) but Res(ϕ,i) =lim_(z→i) (1/((2−1!)) { (z−i)^2 ϕ(z)}^((1)) =lim_(z→i) { ((z e^(ixz) )/((z+i)^2 ))}^((1)) =lim_(z→i) (((e^(ixz) +ixz e^(ixz) )(z+i)^2 −2(z+i)z e^(ixz) )/((z+i)^4 )) =lim_(z→i) (((1+ixz)e^(ixz) (z+i)−2z e^(ixz) )/((z+i)^3 )) =(((1−x)e^(−x) (2i)−2i e^(−x) )/((2i)^3 )) = ((2i(1−x)e^(−x) −2i e^(−x) )/(−8i)) =(((2−2x−2)e^(−x) )/(−8)) = ((x e^(−x) )/4) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ((x e^(−x) )/4) =((iπ)/2) x e^(−x) ⇒ 2I_2 (x) = ((πx)/2) e^(−x) ⇒ I_2 (x) =((πx)/4) e^(−x) . if x<0 x =−α with α>0 ⇒I_2 (x)=I_2 (−α) =∫_0 ^∞ ((tsint)/((t^2 +α^2 )^2 )) =I_2 (α) =((πα)/4) e^(−α) =−((πx)/4) e^x .$
$2) w e h a v e I_{2} (x) = \int_{0}^{\infty} \frac{t s i n t}{{(t^{2} + x^{2})}^{2}} d t = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{t s i n t}{{(t^{2} + x^{2})}^{2}} d t \Rightarrow$ $2 I_{2} (x) = I m (\int_{- \infty}^{+ \infty} \frac{t e^{i t}}{{(t^{2} + x^{2})}^{2}} d t) = I m (A (x)) c h a n g e m e n t t = x u g i v e$ $A (x) = \int_{- \infty}^{+ \infty} \frac{x u e^{i x u}}{x^{4} {(1 + u^{2})}^{2}} x d u = \frac{1}{x^{2}} \int_{- \infty}^{+ \infty} \frac{u e^{i x u}}{{(u^{2} + 1)}^{2}} d u (w e s u p p o s e x > 0)$ $l e t c o n s i d e r t h e c o m p l e x f u n c t i o n φ (z) = \frac{z e^{i x z}}{{(z^{2} + 1)}^{2}}$ $φ (z) = \frac{z e^{i x z}}{{(z - i)}^{2} {(z + i)}^{2}} t h e p o l e s o f φ a r e i a n d - i (d o u b l e s) \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i) b u t$ $R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(2 - 1!} {{(z - i)}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to i} {\frac{z e^{i x z}}{{(z + i)}^{2}}}^{(1)} = {l i m}_{z \to i} \frac{(e^{i x z} + i x z e^{i x z}) {(z + i)}^{2} - 2 (z + i) z e^{i x z}}{{(z + i)}^{4}}$ $= {l i m}_{z \to i} \frac{(1 + i x z) e^{i x z} (z + i) - 2 z e^{i x z}}{{(z + i)}^{3}}$ $= \frac{(1 - x) e^{- x} (2 i) - 2 i e^{- x}}{{(2 i)}^{3}} = \frac{2 i (1 - x) e^{- x} - 2 i e^{- x}}{- 8 i}$ $= \frac{(2 - 2 x - 2) e^{- x}}{- 8} = \frac{x e^{- x}}{4} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{x e^{- x}}{4} = \frac{i π}{2} x e^{- x} \Rightarrow$ $2 I_{2} (x) = \frac{π x}{2} e^{- x} \Rightarrow I_{2} (x) = \frac{π x}{4} e^{- x} .$ $i f x < 0 x = - α w i t h α > 0 \Rightarrow I_{2} (x) = I_{2} (- α) = \int_{0}^{\infty} \frac{t s i n t}{{(t^{2} + α^{2})}^{2}} = I_{2} (α)$ $= \frac{π α}{4} e^{- α} = - \frac{π x}{4} e^{x} .$
Commented by maxmathsup by imad last updated on 19/Aug/18

$e r r o r a t t h e f i n a l l i n e w e h a v e A (x) = \frac{1}{x^{2}} \int_{- \infty}^{+ \infty} φ (z) d z$ $= \frac{1}{x^{2}} \frac{i π}{2} x e^{- x} = \frac{i π}{2 x} e^{- x} \Rightarrow I_{2} (x) = \frac{1}{2} I m (A (x)) = \frac{1}{2} \frac{π}{2 x} e^{- x} \Rightarrow$ $I_{2} (x) = \frac{π}{4 x} e^{- x} w i t h x > 0 a n d i f x < 0$ $I_{2} (x) = - \frac{π}{4 x} e^{x} .$
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