let f(x) =∫_0 ^(+∞) ((sin(xt^2 −1))/(t^4 +1)) dt 1) find a explicit form of f(x) 2) let g(x) =∫_0 ^∞ ((t^2 cos(xt^2 −1))/(t^4 +1)) dt find a explicit form of g(x) 3) calculate ∫_0 ^∞ ((sin(2t^2 −1))/(t^4 +1)) 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 57237 by maxmathsup by imad last updated on 31/Mar/19

$l e t f (x) = \int_{0}^{+ \infty} \frac{s i n ({x t}^{2} - 1)}{t^{4} + 1} d t$ $1) f i n d a e x p l i c i t f o r m o f f (x)$ $2) l e t g (x) = \int_{0}^{\infty} \frac{t^{2} c o s ({x t}^{2} - 1)}{t^{4} + 1} d t$ $f i n d a e x p l i c i t f o r m o f g (x)$ $3) c a l c u l a t e \int_{0}^{\infty} \frac{s i n (2 t^{2} - 1)}{t^{4} + 1} d t a n d \int_{0}^{\infty} \frac{t^{2} c o s (3 t^{2} - 1)}{t^{4} + 1} d t .$
Commented by maxmathsup by imad last updated on 05/Apr/19
$1) we have 2f(x)=∫_(−∞) ^(+∞) ((sin(xt^2 −1))/(t^4 +1))dt =Im (∫_(−∞) ^(+∞) (e^(i(xt^2 −1)) /(t^4 +1))dt) let ϕ(z) =(e^(i(xz^2 −1)) /(z^4 +1)) ⇒ϕ(z) =(e^(i(xz^2 −1)) /((z^2 −i)(z^2 +i))) =(e^(i(xz^2 −1)) /((z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) so the poles of ϕ are +^− e^((iπ)/4) and +^− e^(−((iπ)/4)) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ {Res(ϕ,e^((iπ)/4) )+Res(ϕ,−e^(−((iπ)/4)) )} Res(ϕ,e^((iπ)/4) ) =lim_(z→e^((iπ)/4) ) (z−e^((iπ)/4) )ϕ(z) = (e^(i(x(i)−1)) /(2e^((iπ)/4) (2i))) =((e^(−x) e^(−i) )/(4i e^((iπ)/4) )) =(e^(−x) /(4i)) e^(−(1+(π/4))i) Res(ϕ,−e^(−((iπ)/4)) ) =lim_(z→−e^(−((iπ)/4)) ) (z+e^(−((iπ)/4)) )ϕ(z) = (e^(i(−xi −1)) /(−2 e^(−((iπ)/4)) (−2i))) = (e^x /(4i)) e^(−i) e^((iπ)/4) =(e^x /(4i)) e^((−1+(π/4))i) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (1/(4i)){ e^(−i) { e^(−x) e^(−((iπ)/4)) +e^x e^((iπ)/4) }} =(π/2) e^(−i) {e^(−x) ( (1/(√2)) −(i/(√2))) +e^x ((1/(√2)) +(i/(√2)))} =(π/2) e^(−i) { (1/(√2))(e^x +e^(−x) )+(i/(√2))(e^x −e^(−x) )} =(π/(2(√2)))(cos1 −isin1){ 2ch(x)+2ish(x)} =(π/(√2))(cos(1)−isin(1))(ch(x)+ish(x)) =(π/(√2)){cos(1)ch(x) +icos(1)sh(x)−isin(1)ch(x) +sin(1)sh(x)} ⇒ f(x) =(π/(2(√2))){cos(1)sh(x) −sin(1)ch(x)} 2) we have f^′ (x)=∫_0 ^∞ ((t^2 cos(xt^2 −1))/(t^4 +1))dt =g(x) ⇒ g(x) =(π/(2(√2))){cos(1)ch(x)−sin(1)sh(x)} .$
$1) w e h a v e 2 f (x) = \int_{- \infty}^{+ \infty} \frac{s i n ({x t}^{2} - 1)}{t^{4} + 1} d t = I m (\int_{- \infty}^{+ \infty} \frac{e^{i ({x t}^{2} - 1)}}{t^{4} + 1} d t)$ $l e t φ (z) = \frac{e^{i ({x z}^{2} - 1)}}{z^{4} + 1} \Rightarrow φ (z) = \frac{e^{i ({x z}^{2} - 1)}}{(z^{2} - i) (z^{2} + i)} = \frac{e^{i ({x z}^{2} - 1)}}{(z - e^{\frac{i π}{4}}) (z + e^{\frac{i π}{4}}) (z - e^{- \frac{i π}{4}}) (z + e^{- \frac{i π}{4}})}$ $s o t h e p o l e s o f φ a r e \bar{+} e^{\frac{i π}{4}} a n d \bar{+} e^{- \frac{i π}{4}} r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, e^{\frac{i π}{4}}) + R e s (φ, - e^{- \frac{i π}{4}})}$ $R e s (φ, e^{\frac{i π}{4}}) = {l i m}_{z \to e^{\frac{i π}{4}}} (z - e^{\frac{i π}{4}}) φ (z) = \frac{e^{i (x (i) - 1)}}{2 e^{\frac{i π}{4}} (2 i)} = \frac{e^{- x} e^{- i}}{4 i e^{\frac{i π}{4}}} = \frac{e^{- x}}{4 i} e^{- (1 + \frac{π}{4}) i}$ $R e s (φ, - e^{- \frac{i π}{4}}) = {l i m}_{z \to - e^{- \frac{i π}{4}}} (z + e^{- \frac{i π}{4}}) φ (z) = \frac{e^{i (- x i - 1)}}{- 2 e^{- \frac{i π}{4}} (- 2 i)}$ $= \frac{e^{x}}{4 i} e^{- i} e^{\frac{i π}{4}} = \frac{e^{x}}{4 i} e^{(- 1 + \frac{π}{4}) i} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{1}{4 i} {e^{- i} {e^{- x} e^{- \frac{i π}{4}} + e^{x} e^{\frac{i π}{4}}}}$ $= \frac{π}{2} e^{- i} {e^{- x} (\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}) + e^{x} (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}})}$ $= \frac{π}{2} e^{- i} {\frac{1}{\sqrt{2}} (e^{x} + e^{- x}) + \frac{i}{\sqrt{2}} (e^{x} - e^{- x})}$ $= \frac{π}{2 \sqrt{2}} (c o s 1 - i s i n 1) {2 c h (x) + 2 i s h (x)}$ $= \frac{π}{\sqrt{2}} (c o s (1) - i s i n (1)) (c h (x) + i s h (x))$ $= \frac{π}{\sqrt{2}} {c o s (1) c h (x) + i c o s (1) s h (x) - i s i n (1) c h (x) + s i n (1) s h (x)} \Rightarrow$ $f (x) = \frac{π}{2 \sqrt{2}} {c o s (1) s h (x) - s i n (1) c h (x)}$ $2) w e h a v e f^{^{'}} (x) = \int_{0}^{\infty} \frac{t^{2} c o s ({x t}^{2} - 1)}{t^{4} + 1} d t = g (x) \Rightarrow$ $g (x) = \frac{π}{2 \sqrt{2}} {c o s (1) c h (x) - s i n (1) s h (x)} .$
Commented by maxmathsup by imad last updated on 05/Apr/19
$3) ∫_0 ^∞ ((sin(2t^2 −1))/(t^4 +1)) dt =f(2) =(π/(2(√2))){cos(1)sh(2)−sin(1)ch(2)} ∫_0 ^∞ ((t^2 sin(3t^2 −1))/(t^4 +1)) dt =g(3) =(π/(2(√2))){cos(1)ch(3)−sin(1)sh(3)} .$
$3) \int_{0}^{\infty} \frac{s i n (2 t^{2} - 1)}{t^{4} + 1} d t = f (2) = \frac{π}{2 \sqrt{2}} {c o s (1) s h (2) - s i n (1) c h (2)}$ $\int_{0}^{\infty} \frac{t^{2} s i n (3 t^{2} - 1)}{t^{4} + 1} d t = g (3) = \frac{π}{2 \sqrt{2}} {c o s (1) c h (3) - s i n (1) s h (3)} .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum