let-A-0-dx-x-2-3-x-4-e-i-with-0-lt-lt-pi-2-1-calculate-A-interms-of-2-determine-also-0-dx-x-2-3-x-4-e-i-2-

Question Number 67932 by mathmax by abdo last updated on 02/Sep/19

l e t A (θ) = \int_{0}^{\infty} \frac{d x}{(x^{2} + 3) (x^{4} - e^{i θ})} w i t h 0 < θ < \frac{π}{2}

1) c a l c u l a t e A (θ) i n t e r m s o f θ

2) d e t e r m i n e a l s o \int_{0}^{\infty} \frac{d x}{(x^{2} + 3) {(x^{4} - e^{i θ})}^{2}}

Commented by MJS last updated on 02/Sep/19

I get (1) A (θ) = - \frac{π (2 \sqrt{3} e^{i \frac{3 θ}{4}} - {3 e}^{i \frac{θ}{2}} - 9)}{{12 e}^{i \frac{3 θ}{4}} (e^{i θ} - 9)}

can this be right ?

Commented by mathmax by abdo last updated on 02/Sep/19

1) A (θ) = \int_{0}^{\infty} \frac{d x}{(x^{2} + 3) (x^{4} - e^{i θ})} \Rightarrow 2 A (θ) = \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} + 3) (x^{4} - e^{i θ})}

l e t W (z) = \frac{1}{(z^{2} + 3) (z^{4} - e^{i θ})} p o l e s o f W ?

z^{4} - e^{i θ} = 0 \Rightarrow z^{4} = e^{i θ} l e t z = r e^{i α} \Rightarrow r = 1 a n d 4 α = θ + 2 k π \Rightarrow

α_{k} = \frac{θ + 2 k π}{4} \Rightarrow t h e r o o t s a r e Z_{k} = e^{i \frac{θ + 2 k π}{4}} a n d k \in [[0, 3]]

Z_{0} = e^{\frac{i θ}{4}}, Z_{1} = e^{i (\frac{θ + 2 π}{4})} = e^{\frac{i θ}{4} + \frac{i π}{2}}, Z_{2} = e^{i (\frac{θ + 4 π}{4})} = e^{\frac{i θ}{4} + i π}

Z_{3} = e^{i (\frac{θ + 6 π}{4})} = e^{\frac{i θ}{4} + i \frac{3 π}{2}} = e^{\frac{i θ}{4} + i (2 π - \frac{π}{2})} = e^{i (\frac{θ}{4} - \frac{π}{2})}

Z_{1} = i e^{\frac{i π}{4}} = i {c o s (\frac{π}{4}) + i s i n (\frac{π}{4})} = - s i n \frac{π}{4} + i c o s (\frac{π}{4}) \Rightarrow i m (z_{1}) > 0

i m (Z_{0}) > 0 i m (Z_{2}) < 0 i m (z_{3}) < 0

z^{2} + 3 = z^{2} - {(i \sqrt{3})}^{2} = (z - i \sqrt{3}) (z + i \sqrt{3}) s o t h e r o o t s a r e \bar{+} i \sqrt{3}

r e s i d u t h e o r e m \Rightarrow \int_{- \infty}^{+ \infty} W (z) d z = 2 i π {R e s (W, i \sqrt{3})

+ R e s (W, Z_{0}) + R e s (W, Z_{1})} w e h a v e

W (z) = \frac{1}{(z - i \sqrt{3}) (z + i \sqrt{3}) (z - Z_{0}) (z - Z_{1}) (z - Z_{2}) (z - Z_{3})} \Rightarrow

R e s (W, i \sqrt{3}) = \frac{1}{(2 i \sqrt{3}) ({(i \sqrt{3})}^{4} - e^{i θ})} = \frac{1}{(2 i \sqrt{3}) (9 - e^{i θ})}

R e s (W, Z_{0}) = \frac{1}{(Z_{0}^{2} + 3) (Z_{0} - Z_{1}) (Z_{0} - Z_{2}) (Z_{0} - Z_{3})}

R e s (W, Z_{1}) = \frac{1}{(Z_{1}^{2} + 3) (Z_{1} - Z_{0}) (Z_{1} - Z_{2}) (Z_{1} - Z_{3})}

r e s t t o f i n i c h t h e c a l c u l u s \dots .

Commented by mathmax by abdo last updated on 02/Sep/19

p e r h a p s b e c a u s e i d o n t c o m p l e t e t h e c a l c u l u s . .

Commented by mathmax by abdo last updated on 02/Sep/19

2) w e h a v e A^{^{'}} (θ) = \int_{0}^{\infty} \frac{e^{i θ}}{(x^{2} + 3) {(x^{4} - e^{i θ})}^{2}} d x \Rightarrow

\int_{0}^{\infty} \frac{d x}{(x^{2} + 3) {(x^{4} - e^{i θ})}^{2}} = e^{- i θ} A^{^{'}} (θ) r e s t t o c a l c u l a t e A^{^{'}} (θ) .

Answered by mind is power last updated on 02/Sep/19

l e t f (z) = \frac{d z}{(z^{2} + 3) (x^{4} - e^{i θ})}

p o l e o f f a r e \underline{+} i \sqrt{3}, e^{i \frac{θ + 2 k π}{4}} k ⩽ 4

w e s h o s e u p e r [h a l f i m (z) > 0 p l a n a n d u s e r e s i d u s t h e o r e m

R e s (f . i \sqrt{3}) = \frac{1}{2 i \sqrt{3} (9 - e^{i θ})}

r e s (f, e^{i θ}) = \frac{1}{4 e^{i 3 θ} (e^{2 i θ} + 3)}

r e s (f, e^{i (θ + \frac{π}{2})}) = \frac{1}{4 e^{i (3 θ + \frac{3 π}{2})} (- e^{i 2 θ} + 3)}

\int_{0}^{+ \infty} \frac{d z}{(z^{2} + 3) (z^{4} + e^{i θ})} = \frac{1}{2} \int_{- \infty}^{+ \infty} f (z) d z = i π \sum_{r e s > 0} f (z)

= i π [\frac{1}{2 i \sqrt{3} (9 - e^{i θ})} + \frac{1}{4 e^{3 i θ} (e^{2 i θ} + 3)} + \frac{1}{- 4 {i e}^{3 i θ} (3 - e^{2 i θ})}]

= \frac{i π}{} [\frac{- i (9 - e^{- i θ})}{2 \sqrt{3} (82 - 18 c o s (θ))} + \frac{e^{- 3 i θ} (3 + e^{- 2 i θ})}{4 (10 + 6 c o s (2 θ))} + \frac{{i e}^{- i 3 θ} (3 - e^{- 2 i θ})}{4 (10 - 6 c o s (2 θ))})

2) \frac{d f}{d θ} = \int \frac{e^{i θ}}{(z^{2} + 3) {(z^{4} - e^{i θ})}^{2}} d z = e^{i θ} \int \frac{1}{(z^{2} + 3) (z^{4} - e^{i θ})} d z

\Rightarrow \int \frac{d x}{(x^{2} + 3) (x^{4} - e^{i θ})} = f^{'} (θ) e^{- i θ}