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

Question Number 56629 by maxmathsup by imad last updated on 19/Mar/19

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

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

Commented by maxmathsup by imad last updated on 19/Mar/19

J = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + i}

Commented by maxmathsup by imad last updated on 20/Mar/19

1) l e t φ (z) = \frac{1}{z^{2} - i} \Rightarrow φ (z) = \frac{1}{(z - \sqrt{i}) (z + \sqrt{i})} = \frac{1}{(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}} 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}})

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

= i π e^{- \frac{i π}{4}} \Rightarrow I = i π e^{- \frac{i π}{4}} l e t c a l c u l a t e J = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + i}

l e t W (z) = \frac{1}{z^{2} + i} \Rightarrow W (z) = \frac{1}{(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}}

r e s i d u s t h e o r e m g i v e \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}}) = {l i m}_{z \to - e^{- \frac{i π}{4}}} (z + e^{- \frac{i π}{4}}) W (z) = \frac{1}{- 2 e^{- \frac{i π}{4}}} = - \frac{1}{2} e^{\frac{i π}{4}} \Rightarrow

\int_{- \infty}^{+ \infty} W (z) d z = 2 i π (- \frac{1}{2} e^{\frac{i π}{4}}) = - i π e^{\frac{i π}{4}} = J a l s o w e c a n u s e t h a t

J = c o n j (I) = c o n j (i π e^{- \frac{i π}{4}}) = - i π e^{\frac{i π}{4}}

2) \int_{- \infty}^{+ \infty} \frac{d x}{x^{4} + 1} = \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} - i) (x^{2} + i)} = \frac{1}{2 i} \int_{- \infty}^{+ \infty} {\frac{1}{x^{2} - i} - \frac{1}{x^{2} + i}} d x

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