let f(x) = ∫_0 ^∞ ((x sinx)/(a^2 +x^4 ))dx with a>0 1) find a explicit form of f(a) 2) find g(a) = ∫_0 ^∞ ((xsinx)/((a^2 +x^4 )^2 ))dx 3)find the value of ∫_0 ^∞ ((x sinx)/(x^4 +1))dx 4) find the value of ∫


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Question Number 44466 by maxmathsup by imad last updated on 29/Sep/18

$l e t f (x) = \int_{0}^{\infty} \frac{x s i n x}{a^{2} + x^{4}} d x w i t h a > 0$ $1) f i n d a e x p l i c i t f o r m o f f (a)$ $2) f i n d g (a) = \int_{0}^{\infty} \frac{x s i n x}{{(a^{2} + x^{4})}^{2}} d x$ $3) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{x s i n x}{x^{4} + 1} d x$ $4) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{x s i n x}{{(x^{4} + 1)}^{2}} d x .$
Commented byabdo.msup.com last updated on 29/Sep/18

$f (a) = \int_{0}^{\infty} \frac{x s i n x}{a^{2} + x^{4}} d x w i t h a > 0 .$
Commented bymaxmathsup by imad last updated on 03/Oct/18
$1) we have f(a) =∫_0 ^∞ ((xsin(x))/(a^2 +x^4 ))dx ⇒2f(a) =∫_(−∞) ^(+∞) ((xsin(x))/(a^2 +x^4 ))dx =Im( ∫_(−∞) ^(+∞) ((x e^(ix) )/(a^2 +x^4 ))dx) changement x =(√a)t give I=∫_(−∞) ^(+∞) ((x e^(ix) )/(a^2 +x^4 ))dx = ∫_(−∞) ^(+∞) (((√a)t e^(i(√a)t) )/(a^2 +a^2 t^4 )) (√a)dt =(1/a) ∫_(−∞) ^(+∞) ((t e^(i(√a)t) )/(t^4 +1))dt ⇒ a I = ∫_(−∞) ^(+∞) ((t e^(i(√a)t) )/(t^4 +1))dt let consider the complex function ϕ(z) =((z e^(i(√a)z) )/(z^4 +1)) ⇒ϕ(z) =((z e^(i(√a)z) )/((z^2 −i)(z^2 +i))) =((z e^(i(√a)z) )/((z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^((iπ)/4) ))) so the poles of ϕ are +^− e^((iπ)/4) and +^− e^(−((iπ)/4)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,e^((iπ)/4) ) +Res(ϕ,−e^(−((iπ)/4)) )} but Res(ϕ, e^((iπ)/4) ) =lim_(z→e^((iπ)/4) ) (z−e^((iπ)/4) )ϕ(z) =((e^((iπ)/4) e^(i(√a)e^((iπ)/4) ) )/(2 e^((iπ)/4) (2i))) =(1/(4i)) e^(i(√a)((1/(√2))+(i/(√2)))) =(1/(4i)) e^(−((√a)/((√2) ))) e^(i((√a)/(√2))) Res(ϕ, −e^(−((iπ)/4)) ) =lim_(z→−e^(−((iπ)/4)) ) (z+e^(−((iπ)/4)) )ϕ(z) = ((e^(−((iπ)/4)) e^(i(√a) e^(−((iπ)/4)) ) )/((−2i)(−2e^(−((iπ)/4)) ))) =(1/(4i)) e^(i(√a){(1/(√2))−(i/(√2))}) =(1/(4i)) e^((√a)/(√2)) e^(i((√a)/(√(2 )))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ (1/(4i)) e^(−((√a)/(√2))) e^(i((√a)/(√2))) +(1/(4i)) e^((√a)/(√2)) e^(i((√a)/(√2))) } =(π/2) e^(i((√a)/(√2))) { e^((√a)/(√2)) +e^(−((√a)/(√2))) } =πch(((√a)/(√2))) {cos(((√a)/(√2)))+i sin(((√a)/(√2)))} ⇒ I =(π/a)ch(((√a)/(√2))){cos(((√a)/(√2)))+isin(((√a)/(√2)))} ⇒2f(a) =(π/a)ch(((√a)/(√2)))sin(((√a)/(√2))) ⇒ f(a) =(π/(2a)) ch(((√a)/(√2)))sin(((√a)/(√2))) with a>0 .$
$1) w e h a v e f (a) = \int_{0}^{\infty} \frac{x s i n (x)}{a^{2} + x^{4}} d x \Rightarrow 2 f (a) = \int_{- \infty}^{+ \infty} \frac{x s i n (x)}{a^{2} + x^{4}} d x$ $= I m (\int_{- \infty}^{+ \infty} \frac{x e^{i x}}{a^{2} + x^{4}} d x) c h a n g e m e n t x = \sqrt{a} t g i v e$ $I = \int_{- \infty}^{+ \infty} \frac{x e^{i x}}{a^{2} + x^{4}} d x = \int_{- \infty}^{+ \infty} \frac{\sqrt{a} t e^{i \sqrt{a} t}}{a^{2} + a^{2} t^{4}} \sqrt{a} d t = \frac{1}{a} \int_{- \infty}^{+ \infty} \frac{t e^{i \sqrt{a} t}}{t^{4} + 1} d t \Rightarrow$ $a I = \int_{- \infty}^{+ \infty} \frac{t e^{i \sqrt{a} t}}{t^{4} + 1} d t l e t c o n s i d e r t h e c o m p l e x f u n c t i o n$ $φ (z) = \frac{z e^{i \sqrt{a} z}}{z^{4} + 1} \Rightarrow φ (z) = \frac{z e^{i \sqrt{a} z}}{(z^{2} - i) (z^{2} + i)} = \frac{z e^{i \sqrt{a} z}}{(z - e^{\frac{i π}{4}}) (z + e^{\frac{i π}{4}}) (z - e^{- \frac{i π}{4}}) (z + e^{\frac{i π}{4}})}$ $s o t h e p o l e s o f φ a r e \bar{+} e^{\frac{i π}{4}} a n d \bar{+} e^{- \frac{i π}{4}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, e^{\frac{i π}{4}}) + R e s (φ, - e^{- \frac{i π}{4}})} b u t$ $R e s (φ, e^{\frac{i π}{4}}) = {l i m}_{z \to e^{\frac{i π}{4}}} (z - e^{\frac{i π}{4}}) φ (z) = \frac{e^{\frac{i π}{4}} e^{i \sqrt{a} e^{\frac{i π}{4}}}}{2 e^{\frac{i π}{4}} (2 i)} = \frac{1}{4 i} e^{i \sqrt{a} (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}})}$ $= \frac{1}{4 i} e^{- \frac{\sqrt{a}}{\sqrt{2}}} e^{i \frac{\sqrt{a}}{\sqrt{2}}}$ $R e s (φ, - e^{- \frac{i π}{4}}) = {l i m}_{z \to - e^{- \frac{i π}{4}}} (z + e^{- \frac{i π}{4}}) φ (z)$ $= \frac{e^{- \frac{i π}{4}} e^{i \sqrt{a} e^{- \frac{i π}{4}}}}{(- 2 i) (- 2 e^{- \frac{i π}{4}})} = \frac{1}{4 i} e^{i \sqrt{a} {\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}}} = \frac{1}{4 i} e^{\frac{\sqrt{a}}{\sqrt{2}}} e^{i \frac{\sqrt{a}}{\sqrt{2}}}$ $\Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {\frac{1}{4 i} e^{- \frac{\sqrt{a}}{\sqrt{2}}} e^{i \frac{\sqrt{a}}{\sqrt{2}}} + \frac{1}{4 i} e^{\frac{\sqrt{a}}{\sqrt{2}}} e^{i \frac{\sqrt{a}}{\sqrt{2}}}}$ $= \frac{π}{2} e^{i \frac{\sqrt{a}}{\sqrt{2}}} {e^{\frac{\sqrt{a}}{\sqrt{2}}} + e^{- \frac{\sqrt{a}}{\sqrt{2}}}} = π c h (\frac{\sqrt{a}}{\sqrt{2}}) {c o s (\frac{\sqrt{a}}{\sqrt{2}}) + i s i n (\frac{\sqrt{a}}{\sqrt{2}})} \Rightarrow$ $I = \frac{π}{a} c h (\frac{\sqrt{a}}{\sqrt{2}}) {c o s (\frac{\sqrt{a}}{\sqrt{2}}) + i s i n (\frac{\sqrt{a}}{\sqrt{2}})} \Rightarrow 2 f (a) = \frac{π}{a} c h (\frac{\sqrt{a}}{\sqrt{2}}) s i n (\frac{\sqrt{a}}{\sqrt{2}}) \Rightarrow$ $f (a) = \frac{π}{2 a} c h (\frac{\sqrt{a}}{\sqrt{2}}) s i n (\frac{\sqrt{a}}{\sqrt{2}}) w i t h a > 0 .$
Commented bymaxmathsup by imad last updated on 04/Oct/18

$2) w e h a v e f (a) = \int_{0}^{\infty} \frac{x s i n (x)}{a^{2} + x^{4}} d x \Rightarrow f^{^{'}} (a) = - \int_{0}^{\infty} \frac{2 a x s i n x}{{(a^{2} + x^{4})}^{2}} d x \Rightarrow$ $\int_{0}^{\infty} \frac{x s i n x}{{(a^{2} + x^{4})}^{2}} = - \frac{1}{2 a} f^{^{'}} (a) a n d f (a) i s k n o w n .$
Commented bymaxmathsup by imad last updated on 04/Oct/18

$3) \int_{0}^{\infty} \frac{x s i n (x)}{1 + x^{4}} d x = f (1) = \frac{π}{2} c h (\frac{1}{\sqrt{2}}) s i n (\frac{1}{\sqrt{2}}) .$
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