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-0-dx-x-8-1-

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) 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}} .