0-2pi-ln-1-cos-x-cos-nx-dx-

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

= \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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