1-calculate-I-0-dx-x-2-i-and-J-0-dx-x-2-i-2-find-the-value-of-0-dx-x-4-1-

Question Number 43676 by maxmathsup by imad last updated on 13/Sep/18

1) c a l c u l a t e I = \int_{0}^{\infty} \frac{d x}{x^{2} - i} a n d J = \int_{0}^{\infty} \frac{d x}{x^{2} + i}

2) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{d x}{x^{4} + 1}

Commented by maxmathsup by imad last updated on 13/Sep/18

i^{2} = - 1

Commented by maxmathsup by imad last updated on 15/Sep/18

w e h a v e 2 I = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - i} l e t φ (z) = \frac{1}{z^{2} - i} w e h a v e

φ (z) = \frac{1}{(z - \sqrt{i}) (z + \sqrt{i})} = \frac{1}{(z - e^{\frac{i π}{4}}) (z + e^{\frac{i π}{4}})} r e s i d u s t h e o r e m g i v e

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, e^{\frac{i π}{4}}) = 2 i π . \frac{1}{2 e^{\frac{i π}{4}}} = i π e^{- \frac{i π}{4}} \Rightarrow

I = \frac{i π}{2} e^{- \frac{i π}{4}} a l s o w e h a v e 2 J = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + i} l e t ψ (z) = \frac{1}{z^{2} + i}

ψ (z) = \frac{1}{(z - \sqrt{- i}) (z + \sqrt{- i})} = \frac{1}{(z - e^{- \frac{i π}{4}}) (z + e^{- \frac{i π}{4}})} a n d

\int_{- \infty}^{+ \infty} ψ (z) d z = 2 i π R e s (ψ, - e^{- \frac{i π}{4}}) = 2 i π . \frac{1}{- 2 e^{- \frac{i π}{4}}} = - i π e^{\frac{i π}{4}} \Rightarrow J = - \frac{i π}{2} e^{\frac{i π}{4}}

a n o t h e r w a y w e s e e t h a t J = \bar{I} \Rightarrow J = - \frac{i π}{2} e^{\frac{i π}{4}} .

2) w e h a v e I - J = \int_{0}^{\infty} {\frac{1}{x^{2} - i} - \frac{1}{x^{2} + i}} d x

= \int_{0}^{\infty} \frac{2 i}{x^{4} + 1} d x \Rightarrow \int_{0}^{\infty} \frac{d x}{x^{4} + 1} = \frac{1}{2 i} {I - J}

= \frac{1}{2 i} {\frac{i π}{2} e^{- \frac{i π}{4}} + \frac{i π}{2} e^{\frac{i π}{4}}} = \frac{π}{4} {e^{\frac{i π}{4}} + e^{- \frac{i π}{4}}} = \frac{π}{4} (2 c o s (\frac{π}{4})} = \frac{π}{2} . \frac{1}{\sqrt{2}} = \frac{π}{2 \sqrt{2}} .