let f(x)=(1/(x^2 −2(cosθ)x +1)) 1) calculate f^((n)) (x) 2) find ∫_0 ^∞ f(x)dx 3) find ∫


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Question Number 81165 by abdomathmax last updated on 09/Feb/20

$l e t f (x) = \frac{1}{x^{2} - 2 (c o s θ) x + 1}$ $1) c a l c u l a t e f^{(n)} (x)$ $2) f i n d \int_{0}^{\infty} f (x) d x$ $3) f i n d \int_{0}^{\infty} f^{(n)} (x) d x$
Commented by mathmax by abdo last updated on 10/Feb/20
$1)let solve x^2 −2cosθ)x +1=0 →Δ^′ =cos^2 x−1=−sin^2 x ⇒ x_1 =cosθ +isinθ =e^(iθ) and x_2 =cosθ−isinθ =e^(−iθ) f(x)=(1/((x−e^(iθ) )(x−e^(−iθ) ))) =(1/(2isinθ))((1/(x−e^(iθ) ))−(1/(x−e^(−iθ) ))) (if θ ≠kπ) ⇒f^((n)) (x)=(1/(2isinθ)){ (((−1)^n n!)/((x−e^(iθ) )^(n+1) ))−(((−1)^n n!)/((x−e^(−iθ) )^(n+1) ))} =(((−1)^n n!)/(2isinθ)){(((x−e^(−iθ) )^(n+1) −(x−e^(iθ) )^(n+1) )/((x^2 −2cosθ +1)^(n+1) ))} =(((−1)^(n+1) n!)/(2isinθ)){((2iIm((x−e^(iθ) )^(n+1) ))/((x^2 −2cosθ +1)^(n+1) ))} ⇒ f^((n)) (x)=(((−1)^(n+1) n! ×Im((x−e^(iθ) )^(n+1) ))/(sinθ(x^2 −2cosθ +1)^(n+1) ))$
$1) l e t s o l v e x^{2} - 2 c o s θ) x + 1 = 0 \to Δ^{^{'}} = {c o s}^{2} x - 1 = - {s i n}^{2} x \Rightarrow$ $x_{1} = c o s θ + i s i n θ = e^{i θ} a n d x_{2} = c o s θ - i s i n θ = e^{- i θ}$ $f (x) = \frac{1}{(x - e^{i θ}) (x - e^{- i θ})} = \frac{1}{2 i s i n θ} (\frac{1}{x - e^{i θ}} - \frac{1}{x - e^{- i θ}}) (i f θ \neq k π)$ $\Rightarrow f^{(n)} (x) = \frac{1}{2 i s i n θ} {\frac{{(- 1)}^{n} n!}{{(x - e^{i θ})}^{n + 1}} - \frac{{(- 1)}^{n} n!}{{(x - e^{- i θ})}^{n + 1}}}$ $= \frac{{(- 1)}^{n} n!}{2 i s i n θ} {\frac{{(x - e^{- i θ})}^{n + 1} - {(x - e^{i θ})}^{n + 1}}{{(x^{2} - 2 c o s θ + 1)}^{n + 1}}}$ $= \frac{{(- 1)}^{n + 1} n!}{2 i s i n θ} {\frac{2 i I m ({(x - e^{i θ})}^{n + 1})}{{(x^{2} - 2 c o s θ + 1)}^{n + 1}}} \Rightarrow$ $f^{(n)} (x) = \frac{{(- 1)}^{n + 1} n! \times I m ({(x - e^{i θ})}^{n + 1})}{s i n θ {(x^{2} - 2 c o s θ + 1)}^{n + 1}}$
Commented by mathmax by abdo last updated on 10/Feb/20

$2) i f θ = k π \Rightarrow f (x) = \frac{1}{{(x \bar{+} 1)}^{2}} \Rightarrow \int f (x) d x = - \frac{1}{x \bar{+} 1} + c$ $i f θ \neq k π \Rightarrow \int_{0}^{\infty} f (x) d x = \int_{0}^{\infty} \frac{d x}{x^{2} - 2 c o s θ x + {c o s}^{2} θ + {s i n}^{2} θ}$
Commented by mathmax by abdo last updated on 10/Feb/20
$∫_0 ^∞ f(x)dx=∫_0 ^∞ (dx/((x−cosθ)^2 +sin^2 θ)) =_(x−cosθ =∣sinθ∣u) ∫_(−((cosθ)/(∣sinθ∣))) ^(+∞) (1/(sin^2 θ(1+u^2 )))∣sinθ∣ du =(1/(∣sinθ∣)) [arctanu]_(−((cosθ)/(∣sinθ∣))) ^(+∞) =(1/(∣sinθ∣)){(π/2) +arctan(((cosθ)/(∣sinθ∣)))} ⇒ if sinθ>0 ⇒∫_0 ^∞ f(x)dx=(1/(sinθ)){(π/2) +arctan((1/(tanθ)))} =(1/(sinθ)){π−θ} if sinθ<0 ⇒∫_0 ^∞ f(xdx=−(1/(sinθ)){(π/2)−arctan((1/(tanθ)))} =−(1/(sinθ)){(π/2)−(−(π/2) −θ)} =−(1/(sinθ)){π +θ}$
$\int_{0}^{\infty} f (x) d x = \int_{0}^{\infty} \frac{d x}{{(x - c o s θ)}^{2} + {s i n}^{2} θ}$ $=_{x - c o s θ =∣ s i n θ ∣ u} \int_{- \frac{c o s θ}{∣ s i n θ ∣}}^{+ \infty} \frac{1}{{s i n}^{2} θ (1 + u^{2})} ∣ s i n θ ∣ d u$ $= \frac{1}{∣ s i n θ ∣} {[a r c t a n u]}_{- \frac{c o s θ}{∣ s i n θ ∣}}^{+ \infty} = \frac{1}{∣ s i n θ ∣} {\frac{π}{2} + a r c t a n (\frac{c o s θ}{∣ s i n θ ∣})} \Rightarrow$ $i f s i n θ > 0 \Rightarrow \int_{0}^{\infty} f (x) d x = \frac{1}{s i n θ} {\frac{π}{2} + a r c t a n (\frac{1}{t a n θ})}$ $= \frac{1}{s i n θ} {π - θ}$ $i f s i n θ < 0 \Rightarrow \int_{0}^{\infty} f (x d x = - \frac{1}{s i n θ} {\frac{π}{2} - a r c t a n (\frac{1}{t a n θ})}$ $= - \frac{1}{s i n θ} {\frac{π}{2} - (- \frac{π}{2} - θ)} = - \frac{1}{s i n θ} {π + θ}$
	Answered by mind is power last updated on 09/Feb/20
	$x^2 −2cos(θ)x+1=(x−e^(iθ) )(x−e^(−iθ) ) f(x)=(1/(2isin(θ))).(1/(x−e^(iθ) ))−(1/(2isin(θ))).(1/(x−e^(−iθ) )) f(x)=2Re((1/(2i.sin(θ)(x−e^(iθ) )))) g(x)=(1/(x−e^(iθ) )).(1/(2isin(θ))) f^n =2Re(g^n (x)) g^n (x)=(((−1)^n n!.)/((x−e^(iθ) )^(n+1) )).(1/(2isin(θ))) f(x)=2Re((((−1)^n n!)/((x−e^(iθ) )^(n+1) )).(1/(2isin(θ)))) =(−1)^n n! Re{(((x−e^(−iθ) )^(n+1) )/((x^2 −2cos(θ)x+1)^(n+1) )).(1/(2isin(θ)))} x−e^(−iθ) =x−cos(θ)+isin(θ) =(√(x^2 −2cos(θ)x+1)).(((x−cos(θ))/(√(x^2 −2xcos(θ)+1)))+((isin(θ))/(√(x^2 −2xcos(θ)+1)))) =(√(x^2 −2cos(θ)x+1)).e^(i(kπ+arctan(((sin(θ))/(x−cos(θ)))))) ,k∈{−1,0,1} =(−1)^n n!((sin((n+1){kπ+arctan(((sin(θ))/(x−sin(θ))))}))/((x^2 −2xcos(θ)+1)^((n+1)/2) sin(θ))) may bee error of calculus 2)(x^2 −2cos(θ)x+1)=(x−cos(θ))^2 +sin^2 (θ) =sin^2 (θ)(((x/(sin(θ)))−cot(θ))^2 +1) ∫(dx/((x^2 −2cos(θ)x+1)))=∫(dx/(sin^2 (θ)(((x/(sin(θ)))−cot(θ))^2 +1))) ∀θ∈]0,2π[ ∣sin(θ)≠0 if sin(θ)=0⇒cos(θ)∈{−1,1} since f(−x)=x^2 −2xcos(θ)+1⇒cos(θ)=1 is ennough to find f(x), ⇒f(x)=(1/((x−1)^2 )) not integrable over IR_+ ⇒θ≠0 θ=π⇒f(x)=(1/((x+1)^2 )) ⇒∫_0 ^(+∞) (dx/((1+x)^2 ))=1 ∀θ∈[0,2π[−{0,π} f(x)=(1/(sin^2 (θ){((x/(sin(θ)))−cot(θ))^2 +1})) ∫_0 ^(+∞) f(x)dx=(1/(sin(θ)))[arctan((x/(sin(θ)))−cot(θ))]_0 ^(+∞) if θ∈]0,π[ f(x)=(1/(sin(θ))){(π/2)+arctan(cot(θ))} =(1/(sin(θ))){π−θ} if θ∈]π,2π[ f(x)=(1/(sin(θ))){−(π/2)+arctan(cot(θ))} =(1/(sin(θ))){−(π/2)+arctan(cot(π+(θ−π))} =(1/(sin(θ))){−(π/2)+arctan(cot(θ−π))}=(1/(sin(θ))){−(π/2)+arctan(tan(((3π)/2)−θ)) =(1/(sin(θ))){−(π/2)+((3π)/2)−θ}=(1/(sin(θ)))(π−θ) ⇒∫_0 ^(+∞) f(x)dx=((π−θ)/(sin(θ))),θ∈[0,2π]−{0,π} θ=π⇒f(x)=1 we can see that lim_(x→π) ((π−θ)/(sin(θ)))=−(1/(cos(π)))=1 3 ∀n≥2 =lim_(x→∞) f^(n−1) (x)−f^(n−1) (0)$
	$x^{2} - 2 c o s (θ) x + 1 = (x - e^{i θ}) (x - e^{- i θ})$ $f (x) = \frac{1}{2 i s i n (θ)} . \frac{1}{x - e^{i θ}} - \frac{1}{2 i s i n (θ)} . \frac{1}{x - e^{- i θ}}$ $f (x) = 2 R e (\frac{1}{2 i . s i n (θ) (x - e^{i θ})})$ $g (x) = \frac{1}{x - e^{i θ}} . \frac{1}{2 i s i n (θ)}$ $f^{n} = 2 R e (g^{n} (x))$ $g^{n} (x) = \frac{{(- 1)}^{n} n! .}{{(x - e^{i θ})}^{n + 1}} . \frac{1}{2 i s i n (θ)}$ $f (x) = 2 R e (\frac{{(- 1)}^{n} n!}{{(x - e^{i θ})}^{n + 1}} . \frac{1}{2 i s i n (θ)})$ $= {(- 1)}^{n} n! R e {\frac{{(x - e^{- i θ})}^{n + 1}}{{(x^{2} - 2 c o s (θ) x + 1)}^{n + 1}} . \frac{1}{2 i s i n (θ)}}$ $x - e^{- i θ}$ $= x - c o s (θ) + i s i n (θ)$ $= \sqrt{x^{2} - 2 c o s (θ) x + 1} . (\frac{x - c o s (θ)}{\sqrt{x^{2} - 2 x c o s (θ) + 1}} + \frac{i s i n (θ)}{\sqrt{x^{2} - 2 x c o s (θ) + 1}})$ $= \sqrt{x^{2} - 2 c o s (θ) x + 1} . e^{i (k π + a r c t a n (\frac{s i n (θ)}{x - c o s (θ)}))}, k \in {- 1, 0, 1}$ $= {(- 1)}^{n} n! \frac{s i n ((n + 1) {k π + a r c t a n (\frac{s i n (θ)}{x - s i n (θ)})})}{{(x^{2} - 2 x c o s (θ) + 1)}^{\frac{n + 1}{2}} s i n (θ)}$ $m a y b e e e r r o r o f c a l c u l u s$ $2) (x^{2} - 2 c o s (θ) x + 1) = {(x - c o s (θ))}^{2} + {s i n}^{2} (θ)$ $= {s i n}^{2} (θ) ({(\frac{x}{s i n (θ)} - c o t (θ))}^{2} + 1)$ $\int \frac{d x}{(x^{2} - 2 c o s (θ) x + 1)} = \int \frac{d x}{{s i n}^{2} (θ) ({(\frac{x}{s i n (θ)} - c o t (θ))}^{2} + 1)}$ $\forall θ \in] 0, 2 π [∣ s i n (θ) \neq 0$ $i f s i n (θ) = 0 \Rightarrow c o s (θ) \in {- 1, 1}$ $s i n c e f (- x) = x^{2} - 2 x c o s (θ) + 1 \Rightarrow c o s (θ) = 1 i s e n n o u g h$ $t o f i n d f (x),$ $\Rightarrow f (x) = \frac{1}{{(x - 1)}^{2}} n o t i n t e g r a b l e o v e r {I R}_{+}$ $\Rightarrow θ \neq 0$ $θ = π \Rightarrow f (x) = \frac{1}{{(x + 1)}^{2}}$ $\Rightarrow \int_{0}^{+ \infty} \frac{d x}{{(1 + x)}^{2}} = 1$ $\forall θ \in [0, 2 π [- {0, π}$ $f (x) = \frac{1}{{s i n}^{2} (θ) {{(\frac{x}{s i n (θ)} - c o t (θ))}^{2} + 1}}$ $\int_{0}^{+ \infty} f (x) d x = \frac{1}{s i n (θ)} {[a r c t a n (\frac{x}{s i n (θ)} - c o t (θ))]}_{0}^{+ \infty}$ $i f θ \in] 0, π [$ $f (x) = \frac{1}{s i n (θ)} {\frac{π}{2} + a r c t a n (c o t (θ))}$ $= \frac{1}{s i n (θ)} {π - θ}$ $i f θ \in] π, 2 π [$ $f (x) = \frac{1}{s i n (θ)} {- \frac{π}{2} + a r c t a n (c o t (θ))}$ $= \frac{1}{s i n (θ)} {- \frac{π}{2} + a r c t a n (c o t (π + (θ - π))}$ $= \frac{1}{s i n (θ)} {- \frac{π}{2} + a r c t a n (c o t (θ - π))} = \frac{1}{s i n (θ)} {- \frac{π}{2} + a r c t a n (t a n (\frac{3 π}{2} - θ))$ $= \frac{1}{s i n (θ)} {- \frac{π}{2} + \frac{3 π}{2} - θ} = \frac{1}{s i n (θ)} (π - θ)$ $\Rightarrow \int_{0}^{+ \infty} f (x) d x = \frac{π - θ}{s i n (θ)}, θ \in [0, 2 π] - {0, π}$ $θ = π \Rightarrow f (x) = 1 w e c a n s e e t h a t$ $\lim_{x \to π} \frac{π - θ}{s i n (θ)} = - \frac{1}{c o s (π)} = 1$ $3 \forall n ⩾ 2 = \lim_{x \to \infty} f^{n - 1} (x) - f^{n - 1} (0)$
	Commented by mathmax by abdo last updated on 10/Feb/20

	$t h a n k y o u s i r .$
	Commented by mind is power last updated on 10/Feb/20

	$w i t h e p l e a s u r s i r$
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