∫_0 ^( 2π) ln ( 1+ cos (x)).cos (nx )dx=?


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Question Number 165194 by mnjuly1970 last updated on 27/Jan/22

$\int_{0}^{2 π} l n (1 + c o s (x)) . c o s (n x) d x = ?$
	Answered by mindispower last updated on 27/Jan/22
	$=∫_0 ^(2π) ln(2cos^2 ((x/2)))cos(nx)dx x=2y =2∫_0 ^π ln(2cos^2 (y))cos(2ny)dy =2∫_0 ^(π/2) ln(2cos^2 (y))cos(2ny)dy+2∫_(π/2) ^π ln(2cos^2 (y))cos(2ny)dy =4∫_0 ^(π/2) cos(2ny)ln(2cos^2 (y))dy =4ln(2)∫_0 ^(π/2) cos(2ny)+8∫_0 ^(π/2) cos(2ny)ln(cos(y))dy =−8∫_0 ^(π/2) Σ_(k≥1) (((−1)^k )/k)cos(2ky)cos(2ny)dy−8ln(2)∫_0 ^(π/2) cos2ny)dy =−4Σ_(k≥1) (((−1)^k )/k)∫_0 ^(π/2) cos(2(k+n)x)+cos(2(k−n)x)dx n∈N,k+n≠0;∫_0 ^(π/2) cos(2(k+n)x)dx=0 =−4Σ_(k≥1) (((−1)^k )/k)∫_0 ^(π/2) cos(2(k−n)x)dx=−4(((−1)^n )/n)∫_0 ^(π/2) dx =−2π(((−1)^n )/n)=((2(−1)^(n−1) )/n)π,∀n≥1 n=0 ∫_0 ^(2π) ln(+cos(x))dx=∫_0 ^(2π) ln(2)+ln(cos^2 ((x/2)))dx =2πln(2)+∫_0 ^π ln(cos^2 (x))2dx =2πln(2)+8∫_0 ^(π/2) ln(sin(x))dx =−2πln(2) ∫_0 ^(2π) ln(1+cos(x))cos(x)dx= { ((−2πln(2) ,n=0)),((2π.(((−1)^(n−1) )/n);n≥1)) :}$
	$= \int_{0}^{2 π} l n (2 {c o s}^{2} (\frac{x}{2})) c o s (n x) d x$ $x = 2 y$ $= 2 \int_{0}^{π} l n (2 {c o s}^{2} (y)) c o s (2 n y) d y$ $= 2 \int_{0}^{\frac{π}{2}} l n (2 {c o s}^{2} (y)) c o s (2 n y) d y + 2 \int_{\frac{π}{2}}^{π} l n (2 {c o s}^{2} (y)) c o s (2 n y) d y$ $= 4 \int_{0}^{\frac{π}{2}} c o s (2 n y) l n (2 {c o s}^{2} (y)) d y$ $= 4 l n (2) \int_{0}^{\frac{π}{2}} c o s (2 n y) + 8 \int_{0}^{\frac{π}{2}} c o s (2 n y) l n (c o s (y)) d y$ $= - 8 \int_{0}^{\frac{π}{2}} \sum_{k ⩾ 1} \frac{{(- 1)}^{k}}{k} c o s (2 k y) c o s (2 n y) d y - 8 l n (2) \int_{0}^{\frac{π}{2}} c o s 2 n y) d y$ $= - 4 \sum_{k ⩾ 1} \frac{{(- 1)}^{k}}{k} \int_{0}^{\frac{π}{2}} c o s (2 (k + n) x) + c o s (2 (k - n) x) d x$ $n \in N, k + n \neq 0; \int_{0}^{\frac{π}{2}} c o s (2 (k + n) x) d x = 0$ $= - 4 \sum_{k ⩾ 1} \frac{{(- 1)}^{k}}{k} \int_{0}^{\frac{π}{2}} c o s (2 (k - n) x) d x = - 4 \frac{{(- 1)}^{n}}{n} \int_{0}^{\frac{π}{2}} d x$ $= - 2 π \frac{{(- 1)}^{n}}{n} = \frac{2 {(- 1)}^{n - 1}}{n} π, \forall n ⩾ 1$ $n = 0$ $\int_{0}^{2 π} l n (+ c o s (x)) d x = \int_{0}^{2 π} l n (2) + l n ({c o s}^{2} (\frac{x}{2})) d x$ $= 2 π l n (2) + \int_{0}^{π} l n ({c o s}^{2} (x)) 2 d x$ $= 2 π l n (2) + 8 \int_{0}^{\frac{π}{2}} l n (s i n (x)) d x$ $= - 2 π l n (2)$ $\int_{0}^{2 π} l n (1 + c o s (x)) c o s (x) d x = {\begin{cases} - 2 π l n (2), n = 0 \\ 2 π . \frac{{(- 1)}^{n - 1}}{n}; n ⩾ 1 \end{cases}$
	Commented by mnjuly1970 last updated on 27/Jan/22

	$v e r y n i c e . . r e a l l y v e r y n i c e s o l u t i o n$ $s i r p o w e r t h a n k s a l o t$
	Commented by mindispower last updated on 27/Jan/22

	$t h a n x S i r w i t h e P l e a s u r$ $H a v e a n i c e D a y$
	Answered by aleks041103 last updated on 27/Jan/22

	$I B P :$ $I_{n} = {[\frac{1}{n} l n (1 + c o s (x)) s i n (n x)]}_{0}^{2 π} + \frac{1}{n} \int_{0}^{2 π} \frac{s i n (x) s i n (n x)}{1 + c o s (x)} d x$ $\Rightarrow I_{n} = \frac{1}{n} \int_{0}^{2 π} \frac{s i n (x) s i n (n x)}{1 + c o s (x)} d x$ $l e t z = e^{i x}$ $\Rightarrow d z = {i e}^{i x} d x = i z d x \Rightarrow d x = \frac{d z}{i z}$ $s i n (x) = \frac{e^{i x} - e^{- i x}}{2 i} = \frac{z^{2} - 1}{2 i z}$ $s i n (n x) = \frac{z^{2 n} - 1}{2 {i z}^{n}}$ $c o s (x) = \frac{e^{i x} + e^{- i x}}{2} = \frac{z^{2} + 1}{2 z}$ $\Rightarrow \frac{s i n (x) s i n (n x)}{1 + c o s (x)} d x = \frac{(z^{2} - 1) (z^{2 n} - 1)}{- 4 {i z}^{n + 2} (1 + \frac{z^{2} + 1}{2 z})} d z =$ $= i \frac{(z - 1) (z + 1) (z^{2 n} - 1)}{2 z^{n + 1} {(z + 1)}^{2}} d z = \frac{i}{2} \frac{(z - 1) (z^{2 n} - 1)}{z^{n + 1} (z + 1)} d z$ $A l s o w h e n x \in [0, 2 π], z = e^{i x} \in Γ :∣ z ∣= 1 .$ $\Rightarrow I_{n} = \frac{i}{2 n} \oint_{Γ} \frac{(z - 1) (z^{2 n} - 1)}{z^{n + 1} (z + 1)} d z = \frac{i}{2 n} \oint_{Γ} f (z) d z$ $T h i s c a n b e e v a l u a t e d u s i n g t h e$ $R e s i d u e t h e o r e m .$ $P o l e s o f f (z) :$ $1) z = - 1$ $\underset{z \to - 1}{l i m} \frac{(z - 1) (z^{2 n} - 1)}{z^{n + 1} (z + 1)} = 2 {(- 1)}^{n} \underset{z \to - 1}{l i m} \frac{z^{2 n} - 1}{z + 1} =$ $\overset{l^{'} H}{=} 2 {(- 1)}^{n} \underset{z \to - 1}{l i m} \frac{2 {n z}^{2 n - 1}}{1} = 4 n {(- 1)}^{n + 1} ↛ \pm \infty$ $\Rightarrow z = - 1 i s n o t a p o l e!$ $2) z = 0 - o b v . a p o l e o f (n + 1) - t h o r d e r .$ $\Rightarrow \oint_{Γ} f (z) d z = 2 π i R e s (f, 0)$ $\Rightarrow I_{n} = - \frac{π}{n} R e s (f, 0)$ $R e s (f, 0) = \frac{1}{n!} \underset{z \to 0}{l i m} [\frac{d^{n}}{{d z}^{n}} (z^{n + 1} f (z))] =$ $= \frac{1}{n!} \underset{z \to 0}{l i m} [\frac{d^{n}}{{d z}^{n}} (\frac{(z - 1) (z^{2 n} - 1)}{(z + 1)})]$ $\frac{(z - 1) (z^{2 n} - 1)}{(z + 1)} = (z^{2 n} - 1) \frac{z - 1}{z + 1} =$ $= (z^{2 n} - 1) (1 - \frac{2}{z + 1}) =$ $= z^{2 n} + \frac{2}{z + 1} - 1 - \frac{2 z^{2 n}}{z + 1}$ $\Rightarrow \frac{d^{n}}{{d z}^{n}} (\frac{(z - 1) (z^{2 n} - 1)}{(z + 1)}) =$ $= \frac{(2 n)!}{n!} z^{n} + 2 {(- 1)}^{n} n! \frac{1}{{(z + 1)}^{n + 1}} - 2 \frac{d^{n}}{{d z}^{n}} (\frac{z^{2 n}}{z + 1})$ $\Rightarrow R e s (0, f) = 2 {(- 1)}^{n} - \frac{2}{n!} \frac{d^{n}}{{d z}^{n}} {(\frac{z^{2 n}}{z + 1})}_{z = 0}$ $S i n c e w e w a n t \frac{d^{n}}{{d z}^{n}} (\frac{z^{2 n}}{z + 1}) a t z = 0 (∣ z ∣= 0 < 1) w e c a n$ $u s e \frac{1}{1 + z} = \underset{i = 0}{\sum^{\infty}} {(- z)}^{i}$ $\Rightarrow \frac{d^{n}}{{d z}^{n}} {(\frac{z^{2 n}}{z + 1})}_{z = 0} = \frac{d^{n}}{{d z}^{n}} {(\underset{i = 0}{\sum^{\infty}} {(- z)}^{i + 2 n})}_{z = 0} =$ $= \underset{i = 0}{\sum^{\infty}} \frac{(2 n + i)!}{(n + i)!} {({(- z)}^{i + n})}_{z = 0}$ $s i n c e f o r i ⩾ 0 a n d n ⩾ 1, i + n ⩾ 1 t h e n$ ${({(- z)}^{i + n})}_{z = 0} = 0$ $\Rightarrow \frac{d^{n}}{{d z}^{n}} {(\frac{z^{2 n}}{z + 1})}_{z = 0} = 0$ $\Rightarrow R e s (0, f) = 2 {(- 1)}^{n}$ $\Rightarrow I_{n} = \frac{2 {(- 1)}^{n + 1} π}{n}$
	Commented by mindispower last updated on 28/Jan/22

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