let f(x) =∫_0 ^∞ ((sin(t^2 ))/((x^2 +t^2 )^2 ))dt with x>0 1)determine a explicit form for f(x) 2) find also g(x) =∫_0 ^∞ ((sin(t^2 ))/((x^2 +t^2 )^3 ))dt 3) give f^((n)) (x) at form of integral and calculate f^((n)) (1). 4) find the valueof ∫_0 ^∞ ((sin(t^2 ))/((1+t^2 )^2 )) dt and ∫


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 67744 by mathmax by abdo last updated on 31/Aug/19

$l e t f (x) = \int_{0}^{\infty} \frac{s i n (t^{2})}{{(x^{2} + t^{2})}^{2}} d t w i t h x > 0$ $1) d e t e r m i n e a e x p l i c i t f o r m f o r f (x)$ $2) f i n d a l s o g (x) = \int_{0}^{\infty} \frac{s i n (t^{2})}{{(x^{2} + t^{2})}^{3}} d t$ $3) g i v e f^{(n)} (x) a t f o r m o f i n t e g r a l a n d c a l c u l a t e f^{(n)} (1) .$ $4) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{s i n (t^{2})}{{(1 + t^{2})}^{2}} d t a n d \int_{0}^{\infty} \frac{s i n (t^{2})}{{(1 + t^{2})}^{3}} d t$
Commented by mathmax by abdo last updated on 31/Aug/19
$1) we have f(x)=∫_0 ^∞ ((sin(t^2 ))/((x^2 +t^2 )^2 ))dt cha7gement t=xu give f(x) =∫_0 ^∞ ((sin(x^2 u^2 ))/(x^4 (1+u^2 )^2 )) xdu =(1/x^3 ) ∫_0 ^∞ ((sin(x^2 u^2 ))/((u^2 +1)^2 ))du =(1/(2x^3 ))∫_(−∞) ^(+∞) ((sin(x^2 u^2 ))/((u^2 +1)^2 ))du =(1/(2x^3 )) Im(∫_(−∞) ^(+∞) (e^(ix^2 u^2 ) /((u^2 +1)^2 ))du)let W(z)=(e^(ix^2 z^2 ) /((z^2 +1)^2 )) ⇒W(z) =(e^(ix^2 z^2 ) /((z−i)^2 (z+i)^2 )) so the poles of W are i and −i(doubles) residus theorem give ∫_(−∞) ^(+∞) W(z)dz =2iπ Res(W,i) and Res(W,i) =lim_(z→i) (1/((2−1)!)){(z−i)^2 W(z)}^((1)) =lim_(z→i) { (e^(ix^2 z^2 ) /((z+i)^2 ))}^((1)) =lim_(z→i) ((2ix^2 z e^(ix^2 z^2 ) (z+i)^2 −2(z+i)e^(ix^2 z^2 ) )/((z+i)^4 )) =lim_(z→i) ((2ix^2 z e^(ix^2 z^2 ) (z+i)−2 e^(ix^2 z^2 ) )/((z+i)^3 )) =lim_(z→i) ((2ix^2 z(z+i)−2)/((z+i)^3 )) e^(ix^2 z^2 ) =((((2i)(−2x^2 )−2)e^(−ix^2 ) )/((2i)^3 )) =(((−4ix^2 −2) e^(−ix^2 ) )/(−8i)) =(((2ix^2 −1)e^(−ix^2 ) )/(4i)) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ×(((2ix^2 −1)e^(−ix^2 ) )/(4i)) =(π/2)(2ix^2 −1)e^(−ix^2 ) =(π/2)(2ix^2 −1)(cos(x^2 )−isin(x^2 )) =(π/2){2ix^2 cos(x^2 ) +2x^2 sin(x^2 )−cos(x^2 )+isin(x^2 )} =(π/2){2x^2 sin(x^2 )−cos(x^2 ) +i(2x^2 cos(x^2 )+sin(x^2 )} ⇒ f(x)=((2x^2 cos(x^2 )+sin(x^2 ))/(2x^3 )) .$
$1) w e h a v e f (x) = \int_{0}^{\infty} \frac{s i n (t^{2})}{{(x^{2} + t^{2})}^{2}} d t c h a 7 g e m e n t t = x u g i v e$ $f (x) = \int_{0}^{\infty} \frac{s i n (x^{2} u^{2})}{x^{4} {(1 + u^{2})}^{2}} x d u = \frac{1}{x^{3}} \int_{0}^{\infty} \frac{s i n (x^{2} u^{2})}{{(u^{2} + 1)}^{2}} d u$ $= \frac{1}{2 x^{3}} \int_{- \infty}^{+ \infty} \frac{s i n (x^{2} u^{2})}{{(u^{2} + 1)}^{2}} d u = \frac{1}{2 x^{3}} I m (\int_{- \infty}^{+ \infty} \frac{e^{{i x}^{2} u^{2}}}{{(u^{2} + 1)}^{2}} d u) l e t$ $W (z) = \frac{e^{{i x}^{2} z^{2}}}{{(z^{2} + 1)}^{2}} \Rightarrow W (z) = \frac{e^{{i x}^{2} z^{2}}}{{(z - i)}^{2} {(z + i)}^{2}} s o t h e p o l e s o f W a r e i$ $a n d - i (d o u b l e s) r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π R e s (W, i) a n d$ $R e s (W, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} W (z)}^{(1)}$ $= {l i m}_{z \to i} {\frac{e^{{i x}^{2} z^{2}}}{{(z + i)}^{2}}}^{(1)} = {l i m}_{z \to i} \frac{2 {i x}^{2} z e^{{i x}^{2} z^{2}} {(z + i)}^{2} - 2 (z + i) e^{{i x}^{2} z^{2}}}{{(z + i)}^{4}}$ $= {l i m}_{z \to i} \frac{2 {i x}^{2} z e^{{i x}^{2} z^{2}} (z + i) - 2 e^{{i x}^{2} z^{2}}}{{(z + i)}^{3}}$ $= {l i m}_{z \to i} \frac{2 {i x}^{2} z (z + i) - 2}{{(z + i)}^{3}} e^{{i x}^{2} z^{2}} = \frac{((2 i) (- 2 x^{2}) - 2) e^{- {i x}^{2}}}{{(2 i)}^{3}}$ $= \frac{(- 4 {i x}^{2} - 2) e^{- {i x}^{2}}}{- 8 i} = \frac{(2 {i x}^{2} - 1) e^{- {i x}^{2}}}{4 i} \Rightarrow$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π \times \frac{(2 {i x}^{2} - 1) e^{- {i x}^{2}}}{4 i} = \frac{π}{2} (2 {i x}^{2} - 1) e^{- {i x}^{2}}$ $= \frac{π}{2} (2 {i x}^{2} - 1) (c o s (x^{2}) - i s i n (x^{2}))$ $= \frac{π}{2} {2 {i x}^{2} c o s (x^{2}) + 2 x^{2} s i n (x^{2}) - c o s (x^{2}) + i s i n (x^{2})}$ $= \frac{π}{2} {2 x^{2} s i n (x^{2}) - c o s (x^{2}) + i (2 x^{2} c o s (x^{2}) + s i n (x^{2})} \Rightarrow$ $f (x) = \frac{2 x^{2} c o s (x^{2}) + s i n (x^{2})}{2 x^{3}} .$
Commented by mathmax by abdo last updated on 31/Aug/19
$sorry f(x) =(π/(4x^3 )){2x^2 cos(x^2 )+sin(x^2 )}$
$s o r r y f (x) = \frac{π}{4 x^{3}} {2 x^{2} c o s (x^{2}) + s i n (x^{2})}$
Commented by mathmax by abdo last updated on 31/Aug/19
$2) we have f^′ (x) =−∫_0 ^∞ ((−2(2x)(x^2 +t^2 ))/((x^2 +t^2 )^4 ))sin(t^2 )dt =4x∫_0 ^∞ ((sin(t^2 ))/((x^2 +t^2 )^3 )) =4x g(x) ⇒g(x) =(1/(4x))f^′ (x) we have f(x)=(π/(4x^3 )){2x^2 cos(x^2 )+sin(x^2 )} ⇒ f^′ (x) =(π/4)(((−3x^2 )/x^6 )){2x^2 cos(x^2 )+sin(x^2 )} +(π/(4x^3 )){ 4xcos(x^2 )−4x^3 sin(x^2 )+2xcos(x^2 )}=...$
$2) w e h a v e f^{^{'}} (x) = - \int_{0}^{\infty} \frac{- 2 (2 x) (x^{2} + t^{2})}{{(x^{2} + t^{2})}^{4}} s i n (t^{2}) d t = 4 x \int_{0}^{\infty} \frac{s i n (t^{2})}{{(x^{2} + t^{2})}^{3}}$ $= 4 x g (x) \Rightarrow g (x) = \frac{1}{4 x} f^{^{'}} (x) w e h a v e$ $f (x) = \frac{π}{4 x^{3}} {2 x^{2} c o s (x^{2}) + s i n (x^{2})} \Rightarrow$ $f^{^{'}} (x) = \frac{π}{4} (\frac{- 3 x^{2}}{x^{6}}) {2 x^{2} c o s (x^{2}) + s i n (x^{2})}$ $+ \frac{π}{4 x^{3}} {4 x c o s (x^{2}) - 4 x^{3} s i n (x^{2}) + 2 x c o s (x^{2})} = . . .$
Commented by mathmax by abdo last updated on 31/Aug/19
$4) ∫_0 ^∞ ((sin(t^2 ))/((1+t^2 )^2 ))dt =f(1)=(π/4){2cos(1) +sin(1)} =(π/2)cos(1)+(π/4)sin(1). ∫_0 ^∞ ((sin(t^2 ))/((1+t^2 )^3 ))dt =g(1) rest to calculste g(1)...$
$4) \int_{0}^{\infty} \frac{s i n (t^{2})}{{(1 + t^{2})}^{2}} d t = f (1) = \frac{π}{4} {2 c o s (1) + s i n (1)}$ $= \frac{π}{2} c o s (1) + \frac{π}{4} s i n (1) .$ $\int_{0}^{\infty} \frac{s i n (t^{2})}{{(1 + t^{2})}^{3}} d t = g (1) r e s t t o c a l c u l s t e g (1) . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum