find-f-x-0-pi-d-x-2-2xsin-2-1-x-real-

Question Number 74513 by mathmax by abdo last updated on 25/Nov/19

f i n d f (x) = \int_{0}^{π} \frac{d θ}{x^{2} - 2 x s i n (2 θ) + 1} (x r e a l)

Commented by mathmax by abdo last updated on 26/Nov/19

f (x) =_{2 θ = t} \int_{0}^{2 π} \frac{d t}{2 (x^{2} - 2 x s i n t + 1)} \Rightarrow 2 f (x) = \int_{0}^{2 π} \frac{d t}{x^{2} - 2 x s i n t + 1}

c h a n g e m e n t e^{i t} = z g i v e

2 f (x) = \int_{∣ z ∣= 1} \frac{1}{x^{2} - 2 x \frac{z - z^{- 1}}{2 i} + 1} \frac{d z}{i z}

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

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

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

l e t φ (z) = \frac{1}{{x z}^{2} - i (x^{2} + 1) z - x} p o l e o f φ ?

Δ = - {(x^{2} + 1)}^{2} + 4 x^{2} = 4 x^{2} - (x^{4} + 2 x^{2} + 1) = - x^{4} + 2 x^{2} - 1

= - (x^{4} - 2 x^{2} + 1) = - {(x^{2} - 1)}^{2} =

z_{1} = \frac{i (x^{2} + 1) + i ∣ x^{2} - 1 ∣}{2 x} a n d z_{2} = \frac{i (x^{2} + 1) - i ∣ x^{2} - 1 ∣}{2 x}

c a s e 1 ∣ x ∣< 1 \Rightarrow z_{1} = \frac{i (x^{2} + 1) + i (1 - x^{2})}{2 x} = \frac{i}{x}

∣ z_{1} ∣= \frac{1}{∣ x ∣} > 1 (o u t o f c i r c l e)

z_{2} = \frac{i (x^{2} + 1) - i (1 - x^{2})}{2 x} = \frac{2 {i x}^{2}}{2 x} = i x \Rightarrow∣ z_{2} ∣=∣ x ∣< 1

φ (z) = \frac{1}{x (z - z_{1}) (z - z_{2})} a n d \int_{∣ z ∣= 1} φ (z) d z = 2 i π R e s (φ, z_{2})

= 2 i π \times \frac{1}{x (z_{2} - z_{1})} = \frac{2 π i}{x (i x - \frac{i}{x})} = \frac{2 π}{(x^{2} - 1)} \Rightarrow 2 f (x) = \frac{- 2 π}{x^{2} - 1} \Rightarrow

f (x) = \frac{π}{1 - x^{2}} w i t h ∣ x ∣< 1

c a s e 2 ∣ x ∣> 1 \Rightarrow z_{1} = \frac{i (x^{2} + 1) + i (x^{2} - 1)}{2 x} = i x \Rightarrow∣ z_{1} ∣=∣ x ∣> 1 (o u t o f

c i r c l e) z_{2} = \frac{i (x^{2} + 1) - i (x^{2} - 1)}{2 x} = \frac{i}{x} \Rightarrow∣ z_{2} ∣= \frac{1}{∣ x ∣} < 1

\int_{∣ z ∣= 1} φ (z) d z = 2 i π R e s (φ, z_{2}) = 2 i π \times \frac{1}{x (z_{2} - z_{1})}

= \frac{2 i π}{x (\frac{i}{x} - i x)} = \frac{2 π}{1 - x^{2}} \Rightarrow 2 f (x) = \frac{- 2 π}{1 - x^{2}} \Rightarrow f (x) = \frac{π}{x^{2} - 1}

f i n a l l y f (x) = \frac{π}{1 - x^{2}} i f ∣ x ∣< 1 a n d f (x) = \frac{π}{x^{2} - 1} i f ∣ x ∣> 1

Answered by mind is power last updated on 25/Nov/19

f (x) = {\int_{0}}^{π} \frac{d θ}{{(x - s i n (2 θ))}^{2} + {c o s}^{2} (2 θ)}

x \neq 1

s i n c e 2 - 2 s i n (2 θ) = 0 f o r θ = \frac{π}{2}

f (x) = f (- x) w e s u p p o s e x ⩾ 0

= \int_{0}^{π} \frac{d θ}{(x - s i n (2 θ) - i c o s (θ)) (x - s i n (2 θ) + i c o s (2 θ))}

= \int_{0}^{π} \frac{d θ}{(x - {i e}^{- i 2 θ}) (x + {i e}^{i 2 θ})}

z = e^{2 i θ} \Rightarrow d z = 2 {i e}^{2 i θ} d θ

= \int_{C} \frac{d z}{2 i z (x - \frac{i}{z}) (x + i z)} = \int_{C} \frac{d z}{2 i (x z - i) (x + i z)}

p o l e s a r e \frac{i}{x}, - i x

i f x = 0

f (0) = π

f (- 1) = \int_{0}^{π} \frac{d θ}{2 + 2 s i n (2 θ)} = \frac{d θ}{2 (1 + 2 s i n (θ) c o s (θ))}

= \int_{0}^{\frac{π}{2}} \frac{1 + {t g}^{2} (θ)}{2 {(1 + t g (θ))}^{2}} d θ + \int_{\frac{π}{2}}^{π}

= {[- \frac{1}{2 (1 + t g (θ)}]}_{0}^{\frac{π}{2}} + {[\frac{- 1}{2 (1 + t g (θ))}]}_{\frac{π}{2}}^{π} = 0

i f ∣ x ∣> 1 \Rightarrow∣ \frac{i}{x} ∣< 1 \in C

f (x) = π . R e s (\frac{1}{(x z - i) (x + i z)}, z = \frac{i}{x})

= π . \frac{x}{x^{2} - 1} x > 1

∣ x ∣< 1 \Rightarrow∣ - i x ∣< 1 \Rightarrow f (x) = π R e s (\frac{1}{(x z - i) (x + i z)}, z = i x}

= \frac{π}{1 - x^{2}} ∣ x ∣< 1

Commented by mathmax by abdo last updated on 26/Nov/19

t h a n k x s i r .

Commented by mind is power last updated on 26/Nov/19

y^{'} r e w e l c o m