calculate-0-dx-x-2-e-ia-x-2-e-ib-with-a-gt-0-andb-gt-0-

Question Number 67799 by mathmax by abdo last updated on 31/Aug/19

c a l c u l a t e \int_{0}^{\infty} \frac{d x}{(x^{2} - e^{i a}) (x^{2} - e^{i b})} w i t h a > 0 a n d b > 0

Commented by MJS last updated on 31/Aug/19

can we calculate \underset{0}{\int^{\infty}} \frac{d x}{(x^{2} - α) (x^{2} - β)} and then

insert α = e^{i a}; β = e^{i b} ?

Commented by mathmax by abdo last updated on 31/Aug/19

y e s s i r b u t y o u m u s t d e t e r m i n e \int \frac{d x}{x^{2} - z} w i t h z f r o m C .

Commented by mathmax by abdo last updated on 01/Sep/19

l e t I = \int_{0}^{\infty} \frac{d x}{(x^{2} - e^{i a}) (x^{2} - e^{i b})} \Rightarrow 2 I = \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} - e^{i a}) (x^{2} - e^{i b})}

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

t h e p o l e s o f φ a r e \bar{+} e^{\frac{i a}{2}} a n d \bar{+} e^{\frac{i b}{2}} 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 a}{2}}) + R e s (φ, e^{\frac{i b}{2}})}

R e s (φ, e^{\frac{i a}{2}}) = {l i m}_{z \to e^{\frac{i a}{2}}} (z - e^{\frac{i a}{2}}) φ (z) = \frac{1}{2 e^{\frac{i a}{2}} (e^{i a} - e^{i b})} = \frac{e^{- i \frac{a}{2}}}{2 (e^{i a} - e^{i b})}

R e s (φ, e^{\frac{i b}{2}}) = {l i m}_{z \to e^{\frac{i b}{2}}} (z - e^{\frac{i b}{2}}) φ (z) = \frac{e^{- \frac{i b}{2}}}{2 (e^{i b} - e^{i a})} \Rightarrow

\int_{- \infty}^{+ \infty} φ (2) d z = 2 i π {\frac{e^{- \frac{i a}{2}}}{2 (e^{i a} - e^{i b})} + \frac{e^{- \frac{i b}{2}}}{2 (e^{i b} - e^{i a})}}

= \frac{i π}{e^{i a} - e^{i b}} {e^{- \frac{i a}{2}} - e^{- \frac{i b}{2}}} = 2 I \Rightarrow

I = \frac{i π (e^{- \frac{i a}{2}} - e^{- \frac{i b}{2}})}{2 (e^{i a} - e^{i b})} .

Answered by mind is power last updated on 01/Sep/19

l e t γ_{r} = {z \in C s u c h t h a t ∣ z ∣⩽ r ∣ i m (z) > 0}

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

l e t f (z) = \frac{1}{(z^{2} - e^{i a}) (z^{2} - e^{i b})}

f (z) = \frac{1}{(r^{2} e^{2 i \emptyset} - e^{i a}) (r^{2} e^{2 i \emptyset} - e^{i b})} o v e r γ_{r}

∣ f (z) ∣⩽ \frac{1}{∣ r^{2} - 1 ∣^{2}}

w e u s e d ∣ z + z^{'} ∣⩾∣∣ z ∣ - ∣ z^{'} ∣∣ t w i c e

\Rightarrow \int_{γ r} f (z) d z ⩽ \int_{γ_{r}} \frac{1}{{(r^{2} - 1)}^{2}} d z = \frac{π r}{{(r^{2} - 1)}^{2}} \to 0 = a s r \to + \infty

r e s i d u t h e o r e m

p o l e s o f f i n s i d e γ_{r} \cup] - \infty . + \infty [a r e^{i \frac{a}{2}} . e^{\frac{i b}{2}}

l i m (z - e^{i \frac{a}{2}}) . f (z) = \frac{1}{2 e^{i \frac{a}{2}} . (e^{i a} - e^{i b})}

l i m (z - e^{i \frac{b}{2}}) f (z) = \frac{1}{2 e^{i \frac{b}{2}} (e^{i b} - e^{i a})}

\int_{- \infty}^{+ \infty} f (z) d z = 2 i π \sum_{r e s i d u} f (z) = 2 i π (\frac{e^{- i \frac{b}{2}}}{(e^{i b} - e^{i a}) 2} - \frac{e^{- \frac{i a}{2}}}{2 (e^{i b} - e^{i a})})

= i π (\frac{e^{- \frac{i b}{2}} - e^{\frac{- i a}{2}}}{(e^{i \frac{a + b}{2}} (e^{i \frac{b - a}{2}} - e^{i - \frac{b - a}{2}})}) = i π \frac{e^{- i b - i \frac{a}{2}} - e^{- i a - i \frac{b}{2}}}{2 i s i n (\frac{a - b}{2})}

Commented by mathmax by abdo last updated on 01/Sep/19

t h a n k y o u s i r .

Commented by mind is power last updated on 01/Sep/19

y : r e w e l c o m