let f(x)=∫_0 ^∞ ((tsin(tx))/(1+t^4 ))dt with x>0 1) find a explicit form of f(x) 2) find the value of ∫


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Question Number 53785 by maxmathsup by imad last updated on 25/Jan/19

$l e t f (x) = \int_{0}^{\infty} \frac{t s i n (t x)}{1 + t^{4}} d t w i t h x > 0$ $1) f i n d a e x p l i c i t f o r m o f f (x)$ $2) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{t s i n (2 t)}{1 + t^{4}} d t .$
Commented bymaxmathsup by imad last updated on 26/Jan/19
$1) we have 2f(x)=∫_(−∞) ^(+∞) ((t sin(tx))/(t^4 +1))dt=Im(∫_(−∞) ^(+∞) ((t e^(itx) )/(t^4 +1))dt) let consider the complex function ϕ(z) = ((z e^(ixz) )/(z^4 +1)) ⇒ϕ(z) =((z e^(ixz) )/((z^2 −i)(z^2 +i))) =((z e^(ixz) )/((z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) the poles of ϕ are +^− e^((iπ)/4) and +^− e^(−((iπ)/4)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ , e^((iπ)/4) ) +Res(ϕ,−e^(−((iπ)/4)) )} Res(ϕ, e^((iπ)/4) ) = ((e^((iπ)/4) e^(ix((1/(√2))+(i/(√2)))) )/(2e^((iπ)/4) (2i))) =(1/(4i)) e^((ix)/(√2)) .e^(−(t/(√2))) =(e^(−(x/(√2))) /(4i)) e^((ix)/(√2)) Res(ϕ,−e^(−((iπ)/4)) ) =((−e^(−((iπ)/4)) e^(ix(−(1/(√2))+(i/(√2)))) )/((−2ie^(−((iπ)/4)) )(−2i))) =−(1/(4i)) e^(−(x/(√2))) e^(−((ix)/(√2))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ e^(−(x/(√2))) { (1/(4i)) e^((ix)/(√2)) −(1/(4i)) e^(−((ix)/2)) } =(π/2) e^(−(x/(√2))) { 2i sin((x/(√2)))} =iπ e^(−(x/(√2))) sin((x/(√2))) ⇒2f(x)=π e^(−(x/(√(2 )))) sin((x/(√2))) ⇒ ★f(x) =(π/2) e^(−(x/(√2))) sin((x/(√2)))★ 2) this integral is a spacial case ∫_0 ^(+∞) ((tsin(2t))/(1+t^4 )) dt =f(2) = (π/2) e^(−(√2)) sin((√2)) .$
$1) w e h a v e 2 f (x) = \int_{- \infty}^{+ \infty} \frac{t s i n (t x)}{t^{4} + 1} d t = I m (\int_{- \infty}^{+ \infty} \frac{t e^{i t x}}{t^{4} + 1} d t) l e t c o n s i d e r t h e c o m p l e x f u n c t i o n$ $φ (z) = \frac{z e^{i x z}}{z^{4} + 1} \Rightarrow φ (z) = \frac{z e^{i x z}}{(z^{2} - i) (z^{2} + i)} = \frac{z e^{i x z}}{(z - e^{\frac{i π}{4}}) (z + e^{\frac{i π}{4}}) (z - e^{- \frac{i π}{4}}) (z + e^{- \frac{i π}{4}})}$ $t h e p o l e s o f φ a r e \bar{+} e^{\frac{i π}{4}} a n d \bar{+} e^{- \frac{i π}{4}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, e^{\frac{i π}{4}}) + R e s (φ, - e^{- \frac{i π}{4}})}$ $R e s (φ, e^{\frac{i π}{4}}) = \frac{e^{\frac{i π}{4}} e^{i x (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}})}}{2 e^{\frac{i π}{4}} (2 i)} = \frac{1}{4 i} e^{\frac{i x}{\sqrt{2}}} . e^{- \frac{t}{\sqrt{2}}} = \frac{e^{- \frac{x}{\sqrt{2}}}}{4 i} e^{\frac{i x}{\sqrt{2}}}$ $R e s (φ, - e^{- \frac{i π}{4}}) = \frac{- e^{- \frac{i π}{4}} e^{i x (- \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}})}}{(- 2 {i e}^{- \frac{i π}{4}}) (- 2 i)} = - \frac{1}{4 i} e^{- \frac{x}{\sqrt{2}}} e^{- \frac{i x}{\sqrt{2}}} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π e^{- \frac{x}{\sqrt{2}}} {\frac{1}{4 i} e^{\frac{i x}{\sqrt{2}}} - \frac{1}{4 i} e^{- \frac{i x}{2}}}$ $= \frac{π}{2} e^{- \frac{x}{\sqrt{2}}} {2 i s i n (\frac{x}{\sqrt{2}})} = i π e^{- \frac{x}{\sqrt{2}}} s i n (\frac{x}{\sqrt{2}}) \Rightarrow 2 f (x) = π e^{- \frac{x}{\sqrt{2}}} s i n (\frac{x}{\sqrt{2}}) \Rightarrow$ $★ f (x) = \frac{π}{2} e^{- \frac{x}{\sqrt{2}}} s i n (\frac{x}{\sqrt{2}}) ★$ $2) t h i s i n t e g r a l i s a s p a c i a l c a s e$ $\int_{0}^{+ \infty} \frac{t s i n (2 t)}{1 + t^{4}} d t = f (2) = \frac{π}{2} e^{- \sqrt{2}} s i n (\sqrt{2}) .$
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