find f(x) =∫_0 ^(+∞) ((t arctan(xt))/(1+t^4 )) dt


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Question Number 49968 by maxmathsup by imad last updated on 12/Dec/18

$f i n d f (x) = \int_{0}^{+ \infty} \frac{t a r c t a n (x t)}{1 + t^{4}} d t$
Commented by mathmax by abdo last updated on 03/Nov/19
$we have 2f(x)=∫_(−∞) ^(+∞) ((tarctan(xt))/(t^4 +1))dt let W(z)=((z arctan(xz))/(z^4 +1)) ⇒f(z)=((z arctan(xz))/((z^2 −i)(z^2 +i))) =((z arctan(xz))/((z−(√i))(z+(√i))(z−(√(−i)))(z+(√(−i))))) =((z arctan(xz))/((z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^((iπ)/4) ))) ∫_(−∞) ^(+∞) W(z)dz =2iπ{ Res(W,e^((iπ)/4) ) +Res(W,−e^(−((iπ)/4)) )} Res(W,e^((iπ)/4) ) =lim_(z→e^((iπ)/4) ) (z−e^((iπ)/4) )W(z)=((e^((iπ)/4) arctan(xe^((iπ)/4) ))/(2e^((iπ)/4) (2i))) =(1/(4i)) arctan(x e^((iπ)/4) ) Res(W,−e^(−((iπ)/4)) ) =lim_(z→−e^(−((iπ)/4)) ) (z+e^(−((iπ)/4)) )W(z) =((−e^(−((iπ)/4)) arctan(−xe^(−((iπ)/4)) ))/((−2e^(−((iπ)/4)) )(−2i))) =(1/(4i)) arctan(x e^(−((iπ)/4)) ) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ{(1/(4i))arctan(xe^((iπ)/4) ) +(1/(4i)) arctan(x e^(−((iπ)/4)) )} =(π/2){ arctan(xe^((iπ)/4) )+arctan(xe^(−((iπ)/4)) )} =2f(x) ⇒ f(x)=(π/4){ arctan(xe^((iπ)/4) )+arctan(xe^(−((iπ)/4)) )} rest to find arctan(z) +arctan(z^− )...be continued..$
$w e h a v e 2 f (x) = \int_{- \infty}^{+ \infty} \frac{t a r c t a n (x t)}{t^{4} + 1} d t l e t W (z) = \frac{z a r c t a n (x z)}{z^{4} + 1}$ $\Rightarrow f (z) = \frac{z a r c t a n (x z)}{(z^{2} - i) (z^{2} + i)} = \frac{z a r c t a n (x z)}{(z - \sqrt{i}) (z + \sqrt{i}) (z - \sqrt{- i}) (z + \sqrt{- i})}$ $= \frac{z a r c t a n (x z)}{(z - e^{\frac{i π}{4}}) (z + e^{\frac{i π}{4}}) (z - e^{- \frac{i π}{4}}) (z + e^{\frac{i π}{4}})}$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π {R e s (W, e^{\frac{i π}{4}}) + R e s (W, - e^{- \frac{i π}{4}})}$ $R e s (W, e^{\frac{i π}{4}}) = {l i m}_{z \to e^{\frac{i π}{4}}} (z - e^{\frac{i π}{4}}) W (z) = \frac{e^{\frac{i π}{4}} a r c t a n ({x e}^{\frac{i π}{4}})}{2 e^{\frac{i π}{4}} (2 i)}$ $= \frac{1}{4 i} a r c t a n (x e^{\frac{i π}{4}})$ $R e s (W, - e^{- \frac{i π}{4}}) = {l i m}_{z \to - e^{- \frac{i π}{4}}} (z + e^{- \frac{i π}{4}}) W (z)$ $= \frac{- e^{- \frac{i π}{4}} a r c t a n (- {x e}^{- \frac{i π}{4}})}{(- 2 e^{- \frac{i π}{4}}) (- 2 i)} = \frac{1}{4 i} a r c t a n (x e^{- \frac{i π}{4}}) \Rightarrow$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π {\frac{1}{4 i} a r c t a n ({x e}^{\frac{i π}{4}}) + \frac{1}{4 i} a r c t a n (x e^{- \frac{i π}{4}})}$ $= \frac{π}{2} {a r c t a n ({x e}^{\frac{i π}{4}}) + a r c t a n ({x e}^{- \frac{i π}{4}})} = 2 f (x) \Rightarrow$ $f (x) = \frac{π}{4} {a r c t a n ({x e}^{\frac{i π}{4}}) + a r c t a n ({x e}^{- \frac{i π}{4}})}$ $r e s t t o f i n d a r c t a n (z) + a r c t a n (\bar{z}) . . . b e c o n t i n u e d . .$
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