find ∫_0 ^π (dθ/(x^2 −2x cosθ +1)) with x real.


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Question Number 72908 by mathmax by abdo last updated on 04/Nov/19

$f i n d \int_{0}^{π} \frac{d θ}{x^{2} - 2 x c o s θ + 1} w i t h x r e a l .$
Commented by mind is power last updated on 05/Nov/19

$this is True for ∣ x ∣< 1$
Commented by Tanmay chaudhury last updated on 05/Nov/19

Commented by Tanmay chaudhury last updated on 05/Nov/19

$\frac{1}{1 - x^{2}} \int_{0}^{π} \frac{1 - x^{2}}{x^{2} - 2 x c o s θ + 1} d θ [w h e n x < 1]$ $\frac{1}{1 - x^{2}} \int_{0}^{π} (1 + 2 x c o s θ + 2 x^{2} c o s 2 θ + 2 x^{3} c o s 3 θ + . . .) d θ$ $\frac{1}{1 - x^{2}} [∣ θ ∣_{0}^{π} + o t h e r s t e r m s = 0$ $= \frac{π}{1 - x^{2}} a n s w e r$ $c a s e - I I w h e n x > 1$ $l e t k = \frac{1}{x} w h e n x > 1 s o k < 1$ $\int_{0}^{π} \frac{d θ}{x^{2} - 2 x c o s θ + 1} d θ$ $\int_{0}^{π} \frac{d θ}{\frac{1}{k^{2}} - \frac{2}{k} c o s θ + 1} d θ$ $\int_{0}^{π} \frac{k^{2} d θ}{1 - 2 k c o s θ + k^{2}} d θ [n o t e k < 1]$ $k^{2} \times \frac{π}{1 - k^{2}}$ $= \frac{π}{\frac{1}{k^{2}} - 1} = s o a n s w e r i s = \frac{π}{x^{2} - 1}$
Commented by mathmax by abdo last updated on 05/Nov/19
$let f(x)=∫_0 ^π (dθ/(x^2 −2xcosθ +1)) changement tan((θ/2))=tgive f(x)=∫_0 ^∞ ((2dt)/((1+t^2 )(x^2 −2x((1−t^2 )/(1+t^2 )) +1))) =∫_0 ^∞ ((2dt)/(x^2 (1+t^2 )−2x(1−t^2 )+1+t^2 )) =∫_0 ^∞ ((2dt)/(x^2 +x^2 t^2 −2x+2xt^2 +1+t^2 )) =∫_0 ^∞ ((2dt)/((x^2 +2x+1)t^2 +x^2 −2x+1)) =∫_0 ^∞ ((2dt)/((x+1)^2 t^2 +(x−1)^2 )) =(1/((x+1)^2 ))∫_0 ^∞ ((2dt)/(t^2 +(((x−1)/(x+1)))^2 ))( if x≠−1 and x≠1) ⇒f(x)=_(t=∣((x−1)/(x+1))∣u) (1/((x+1)^2 ))∫_0 ^∞ (2/((((x−1)/(x+1)))^2 {1+u^2 }))∣((x−1)/(x+1))∣du =(1/((x−1)^2 ))×((∣x−1∣)/(∣x+1∣))∫_0 ^∞ ((2du)/(1+u^2 )) =(1/(∣x^2 −1∣))(2×(π/2))=(π/(∣x^2 −1∣)) ⇒ f(x)=(π/(∣x^2 −1∣))$
$l e t f (x) = \int_{0}^{π} \frac{d θ}{x^{2} - 2 x c o s θ + 1} c h a n g e m e n t t a n (\frac{θ}{2}) = t g i v e$ $f (x) = \int_{0}^{\infty} \frac{2 d t}{(1 + t^{2}) (x^{2} - 2 x \frac{1 - t^{2}}{1 + t^{2}} + 1)} = \int_{0}^{\infty} \frac{2 d t}{x^{2} (1 + t^{2}) - 2 x (1 - t^{2}) + 1 + t^{2}}$ $= \int_{0}^{\infty} \frac{2 d t}{x^{2} + x^{2} t^{2} - 2 x + 2 {x t}^{2} + 1 + t^{2}} = \int_{0}^{\infty} \frac{2 d t}{(x^{2} + 2 x + 1) t^{2} + x^{2} - 2 x + 1}$ $= \int_{0}^{\infty} \frac{2 d t}{{(x + 1)}^{2} t^{2} + {(x - 1)}^{2}} = \frac{1}{{(x + 1)}^{2}} \int_{0}^{\infty} \frac{2 d t}{t^{2} + {(\frac{x - 1}{x + 1})}^{2}} (i f x \neq - 1 a n d$ $x \neq 1) \Rightarrow f (x) =_{t =∣ \frac{x - 1}{x + 1} ∣ u} \frac{1}{{(x + 1)}^{2}} \int_{0}^{\infty} \frac{2}{{(\frac{x - 1}{x + 1})}^{2} {1 + u^{2}}} ∣ \frac{x - 1}{x + 1} ∣ d u$ $= \frac{1}{{(x - 1)}^{2}} \times \frac{∣ x - 1 ∣}{∣ x + 1 ∣} \int_{0}^{\infty} \frac{2 d u}{1 + u^{2}} = \frac{1}{∣ x^{2} - 1 ∣} (2 \times \frac{π}{2}) = \frac{π}{∣ x^{2} - 1 ∣} \Rightarrow$ $f (x) = \frac{π}{∣ x^{2} - 1 ∣}$
Commented by Tanmay chaudhury last updated on 05/Nov/19

$w h o m y o u t a r g e t b y t h i s r e m a r k s . . . r e p l y n o w$
Commented by ajfour last updated on 05/Nov/19

$G r e a t m e t h o d, T a n m a y S i r . .$
Commented by Tanmay chaudhury last updated on 05/Nov/19

$T h a n k y o u s i r . . .$
Commented by ajfour last updated on 05/Nov/19

$s o r r y s i r, i c o n c l u d e d t h i s o n a$ $f l e e t i n g g l a n c e, f o r g i v e m e . .$
Commented by Tanmay chaudhury last updated on 05/Nov/19

Commented by Tanmay chaudhury last updated on 05/Nov/19

$t a k e n f r o m S L l o n e y$
Commented by Tanmay chaudhury last updated on 05/Nov/19

$t a k e n f r o m M L k h a n n a$
Commented by Tanmay chaudhury last updated on 05/Nov/19

Commented by Tanmay chaudhury last updated on 05/Nov/19

$s i r p l s c h e c k i h a v e c o n s i d e r d b o t h c a s e$ $x < 1 a n d x > 1 p l s$
Commented by mind is power last updated on 05/Nov/19

$nice solution$
Commented by mind is power last updated on 05/Nov/19

$nice sir$
	Answered by mind is power last updated on 05/Nov/19
	$x^2 −2xcos(θ)+1 =(x−e^(iθ) )(x−e^(−iθ) ) ∫_0 ^π (dθ/((x−e^(iθ) )(x−e^(−iθ) )))=∫_0 (dθ/((x−e^(iθ) )(xe^(iθ) −1))) f(x)=∫_0 ^π (dθ/(x^2 −2xcos(θ)+1))=f(−x) juste find f in x∈R_+ z=e^(iθ) ⇒e^(iθ) =−idz ⇒∫_1 ^(−1) ((−idz)/((x−z)(xz−1)))=i∫_1 ^(−1) (dz/((z−x)(xz−1))) =i∫_1 ^(−1) (1/((x^2 −1)(z−x)))dz+((xdz)/((1−x^2 )(xz−1))) =(i/(x^2 −1)){∫_1 ^(−1) (dz/(z−x))+∫_(−1) ^1 ((xdz)/(xz−1))} =(i/(x^2 −1))(ln(−1−x)−ln(1−x)+ln(x−1)−ln(−x−1) =(i/(x^2 −1))(ln(x−1)−ln(1−x)) ln(z_1 .z_2 )=ln(z_1 )+ln(z_2 )+2inπ if x>1⇒ln(1−x)=ln(−1(x−1)=iπ+ln(x−1) if x<1 ln(x−1)=iπ+ln(1−x) ⇒ ln(x−1)−ln(1−x)=−iπ x>1 ∫_0 ^π (dθ/(x^2 −2xcos(θ)+1))= { (((π/(x^2 −1)) x>1)),((((−π)/(x^2 −1)) 0≤ x<1)) :}$
	$x^{2} - 2 xcos (θ) + 1$ $= (x - e^{i θ}) (x - e^{- i θ})$ $\int_{0}^{π} \frac{d θ}{(x - e^{i θ}) (x - e^{- i θ})} = \int_{0} \frac{d θ}{(x - e^{i θ}) ({xe}^{i θ} - 1)}$ $f (x) = \int_{0}^{π} \frac{d θ}{x^{2} - 2 xcos (θ) + 1} = f (- x)$ $juste find f in x \in R_{+}$ $z = e^{i θ} \Rightarrow e^{i θ} = - idz$ $\Rightarrow \int_{1}^{- 1} \frac{- idz}{(x - z) (xz - 1)} = i \int_{1}^{- 1} \frac{dz}{(z - x) (xz - 1)}$ $= i \int_{1}^{- 1} \frac{1}{(x^{2} - 1) (z - x)} dz + \frac{xdz}{(1 - x^{2}) (xz - 1)}$ $= \frac{i}{x^{2} - 1} {\int_{1}^{- 1} \frac{dz}{z - x} + \int_{- 1}^{1} \frac{xdz}{xz - 1}}$ $= \frac{i}{x^{2} - 1} (\ln (- 1 - x) - \ln (1 - x) + \ln (x - 1) - \ln (- x - 1)$ $= \frac{i}{x^{2} - 1} (\ln (x - 1) - \ln (1 - x))$ $\ln (z_{1} . z_{2}) = \ln (z_{1}) + \ln (z_{2}) + 2 in π$ $if x > 1 \Rightarrow \ln (1 - x) = \ln (- 1 (x - 1) = i π + \ln (x - 1)$ $if x < 1 \ln (x - 1) = i π + \ln (1 - x)$ $\Rightarrow$ $\ln (x - 1) - \ln (1 - x) = - i π x > 1$ $\int_{0}^{π} \frac{d θ}{x^{2} - 2 xcos (θ) + 1} = {\begin{cases} \frac{π}{x^{2} - 1} x > 1 \\ \frac{- π}{x^{2} - 1} 0 ⩽ x < 1 \end{cases}$
	Answered by ajfour last updated on 05/Nov/19

	$I = \int_{0}^{π} \frac{d θ}{{(x - 1)}^{2} + 4 x \sin^{2} \frac{θ}{2}}$ $d i v i d i n g b y \cos^{2} \frac{θ}{2} a n d$ $l e t \tan \frac{θ}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{θ}{2} d θ = d t$ $\Rightarrow (\sec^{2} \frac{θ}{2}) d θ = 2 d t$ $I = \int^{} \frac{2 d t}{{(x - 1)}^{2} (1 + t^{2}) + 4 {x t}^{2}}$ $= \frac{1}{{(x + 1)}^{2}} \int \frac{2 d t}{t^{2} + {(\frac{x - 1}{x + 1})}^{2}} = \frac{2}{x^{2} - 1} \tan^{- 1} [\frac{t (x + 1)}{x - 1}] + c$ $= \frac{2}{x^{2} - 1} (\frac{π}{2}) = \frac{π}{x^{2} - 1} .$ $I = \frac{π}{x^{2} - 1} \cdot$
	Commented by mind is power last updated on 05/Nov/19

	$sir if \frac{x + 1}{x - 1} < 0$ $\lim t \to \infty \tan^{-} ((\frac{x + 1} t}{x - 1}) = - \frac{π}{2}$
	Commented by ajfour last updated on 05/Nov/19

	$t r u e, t h a n k s f o r t h e l i g h t .$
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