let I = ∫_0 ^∞ (dx/(x^4 −i)) and J = ∫_0 ^∞ (dx/(x^4 +i)) 1) find values of I and J 2) calculate I +J 3) calculate ∫


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Question Number 43808 by maxmathsup by imad last updated on 15/Sep/18

$l e t I = \int_{0}^{\infty} \frac{d x}{x^{4} - i} a n d J = \int_{0}^{\infty} \frac{d x}{x^{4} + i}$ $1) f i n d v a l u e s o f I a n d J$ $2) c a l c u l a t e I + J$ $3) c a l c u l a t e \int_{0}^{\infty} \frac{d x}{x^{8} + 1}$
Commented by maxmathsup by imad last updated on 21/Sep/18
$1) we have i =(e^((iπ)/8) )^4 changement x =t e^((iπ)/8) give J= ∫_0 ^∞ ((e^((iπ)/8) dt)/(i t^4 +i)) =−i e^((iπ)/8) ∫_0 ^∞ (dt/(1+t^4 )) also cha7gement t =u^(1/4) ⇒ ∫_0 ^∞ (dt/(1+t^4 )) =∫_0 ^∞ (1/(1+u)) (1/4) u^((1/4)−1) du =(1/4) (π/(sin((π/4)))) =(π/4) (1/((√2)/2)) =(π/(2(√2))) ⇒ J =−((iπ)/(2(√2))) e^((iπ)/8) and I =J^− =((iπ)/(2(√2))) e^(−((iπ)/8)) 2) we have I +J =((iπ)/(2(√2))) e^(−((iπ)/8)) −((iπ)/(2(√2))) e^((iπ)/8) =−((iπ)/(2(√2))){e^((iπ)/8) −e^(−i(π/8)) } =−((iπ)/(2(√2)))(2i sin((π/8))) =(π/(√2)) ((√(2−(√2)))/2) =(π/(2(√2)))(√(2−(√2))). 3) we have I −J =∫_0 ^∞ ((1/(x^4 −i)) −(1/(x^4 +i)))dx =∫_0 ^∞ ((2i)/(x^8 +1)) dx ⇒∫_0 ^∞ (dx/(1+x^8 )) =(1/(2i)){I −J} but I −J = ((iπ)/(2(√2))) e^(−((iπ)/8)) +((iπ)/(2(√2))) e^((iπ)/8) =((iπ)/(2(√2))) (2cos((π/8))} =((iπ)/(√2)) ((√(2+(√2)))/2) ⇒ ∫_0 ^∞ (dx/(1+x^8 )) =(1/(2i)) ((iπ)/(2(√2)))(√(2+(√2)))=(π/(4(√2))) (√(2+(√2))). remark we have I +J =∫_0 ^∞ ((1/(x^4 −i)) +(1/(x^4 +i)))dx = ∫_0 ^∞ ((2x^4 )/(1+x^8 )) dx =(π/(2(√2)))(√(2−(√2))) ⇒∫_0 ^∞ (x^4 /(1+x^8 ))dx =(π/(4(√2)))(√(2−(√2))).$
$1) w e h a v e i = {(e^{\frac{i π}{8}})}^{4} c h a n g e m e n t x = t e^{\frac{i π}{8}} g i v e$ $J = \int_{0}^{\infty} \frac{e^{\frac{i π}{8}} d t}{i t^{4} + i} = - i e^{\frac{i π}{8}} \int_{0}^{\infty} \frac{d t}{1 + t^{4}} a l s o c h a 7 g e m e n t t = u^{\frac{1}{4}} \Rightarrow$ $\int_{0}^{\infty} \frac{d t}{1 + t^{4}} = \int_{0}^{\infty} \frac{1}{1 + u} \frac{1}{4} u^{\frac{1}{4} - 1} d u = \frac{1}{4} \frac{π}{s i n (\frac{π}{4})} = \frac{π}{4} \frac{1}{\frac{\sqrt{2}}{2}} = \frac{π}{2 \sqrt{2}} \Rightarrow$ $J = - \frac{i π}{2 \sqrt{2}} e^{\frac{i π}{8}} a n d I = \bar{J} = \frac{i π}{2 \sqrt{2}} e^{- \frac{i π}{8}}$ $2) w e h a v e I + J = \frac{i π}{2 \sqrt{2}} e^{- \frac{i π}{8}} - \frac{i π}{2 \sqrt{2}} e^{\frac{i π}{8}} = - \frac{i π}{2 \sqrt{2}} {e^{\frac{i π}{8}} - e^{- i \frac{π}{8}}}$ $= - \frac{i π}{2 \sqrt{2}} (2 i s i n (\frac{π}{8})) = \frac{π}{\sqrt{2}} \frac{\sqrt{2 - \sqrt{2}}}{2} = \frac{π}{2 \sqrt{2}} \sqrt{2 - \sqrt{2}} .$ $3) w e h a v e I - J = \int_{0}^{\infty} (\frac{1}{x^{4} - i} - \frac{1}{x^{4} + i}) d x$ $= \int_{0}^{\infty} \frac{2 i}{x^{8} + 1} d x \Rightarrow \int_{0}^{\infty} \frac{d x}{1 + x^{8}} = \frac{1}{2 i} {I - J} b u t$ $I - J = \frac{i π}{2 \sqrt{2}} e^{- \frac{i π}{8}} + \frac{i π}{2 \sqrt{2}} e^{\frac{i π}{8}} = \frac{i π}{2 \sqrt{2}} (2 c o s (\frac{π}{8})} = \frac{i π}{\sqrt{2}} \frac{\sqrt{2 + \sqrt{2}}}{2} \Rightarrow$ $\int_{0}^{\infty} \frac{d x}{1 + x^{8}} = \frac{1}{2 i} \frac{i π}{2 \sqrt{2}} \sqrt{2 + \sqrt{2}} = \frac{π}{4 \sqrt{2}} \sqrt{2 + \sqrt{2}} .$ $r e m a r k w e h a v e I + J = \int_{0}^{\infty} (\frac{1}{x^{4} - i} + \frac{1}{x^{4} + i}) d x$ $= \int_{0}^{\infty} \frac{2 x^{4}}{1 + x^{8}} d x = \frac{π}{2 \sqrt{2}} \sqrt{2 - \sqrt{2}} \Rightarrow \int_{0}^{\infty} \frac{x^{4}}{1 + x^{8}} d x = \frac{π}{4 \sqrt{2}} \sqrt{2 - \sqrt{2}} .$
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