prove-that-0-pi-4-cos-nx-cos-n-x-dx-2-n-pi-8-k-1-n-1-sin-kpi-4-2k-2-k-n-N-

Question Number 83108 by M±th+et£s last updated on 28/Feb/20

p r o v e t h a t

\int_{0}^{\frac{π}{4}} \frac{c o s (n x)}{{c o s}^{n} (x)} d x = 2^{n} [\frac{π}{8} - \underset{k = 1}{\sum^{n - 1}} \frac{s i n (\frac{k π}{4})}{2 k {(\sqrt{2})}^{k}}] n \in N^{*}

Answered by mind is power last updated on 28/Feb/20

= R e \int_{0}^{\frac{π}{4}} \frac{e^{i n x}}{{(\frac{e^{i x} + e^{- i x}}{2})}^{n}}

= R e {\int_{0}^{\frac{π}{4}} \frac{2^{n} e^{i 2 n x}}{{(e^{2 i x} + 1)}^{n}} d x}

u = e^{2 i x}

d u = 2 {i e}^{2 i x} d x

= R e {2^{n} \int_{C} \frac{u^{n - 1}}{{(1 + u)}^{n}} . \frac{d u}{2 i}} C q u a r t u n i t e c i r c l e

\int_{C^{^{'}}} \frac{z^{n - 1}}{{(1 + z)}^{n}} d z = 0 w i t h e C^{'} = [0, 1] \cup [e^{i t}, 0 < t < \frac{π}{2}] \cup [i, 0]

s i c e f (z) = \frac{z^{n - 1}}{{(1 + z)}^{n}} \frac{1}{2 i} i s H o l o m o r p h i c

\int_{C^{'}} f (z) d z = 0 \Rightarrow \int_{C} f (z) = - \int_{0}^{1} f (z) d z - \int_{i}^{0} f (z) d z

R e {2^{n} \int f (z) d z} = R e {- \frac{2^{n}}{2 i} \int_{0}^{1} \frac{z^{n - 1}}{{(1 + z)}^{n}} d z - \frac{2^{n}}{2 i} \int_{i}^{0} \frac{z^{n - 1}}{{(1 + z)}^{n}} d z}

= R e {\frac{2^{n}}{2 i} \int_{0}^{i} \frac{z^{n - 1}}{{(1 + z)}^{n}} d z}

\frac{x^{n - 1}}{{(1 + x)}^{n}} = \frac{{(x + 1 - 1)}^{n - 1}}{{(1 + x)}^{n}}

= \underset{k = 0}{\sum^{n - 1}} C_{n - 1}^{k} {(x + 1)}^{n - 1 - k} {(- 1)}^{k}

= \underset{k = 0}{\sum^{n - 1}} \frac{C_{n - 1}^{k} {(- 1)}^{k}}{{(x + 1)}^{1 + k}} == \underset{j = 1}{\sum^{n}} \frac{C_{n - 1}^{j - 1} {(- 1)}^{j - 1}}{{(x + 1)}^{j}}

\int_{0}^{i} \underset{j = 1}{\sum^{n}} \frac{C_{n - 1}^{j - 1} {(- 1)}^{j - 1}}{{(x + 1)}^{j}} d x

= \int_{0}^{i} \frac{d x}{(x + 1)} + \underset{j = 2}{\sum^{n}} \int_{0}^{i} \frac{C_{n - 1}^{j - 1} {(- 1)}^{j - 1}}{{(x + 1)}^{j}}

= l n (1 + i) + \underset{j = 2}{\sum^{n}} C_{n - 1}^{j - 1} {(- 1)}^{j - 1} \int_{0}^{i} \frac{d x}{{(x + 1)}^{j}}

= l n (\sqrt{2}) + i \frac{π}{4} + \underset{j = 2}{\sum^{n}} C_{n - 1}^{j - 1} {(- 1)}^{j - 1} . {[\frac{{(x + 1)}^{1 - j}}{1 - j}]}_{0}^{i}

= l n (\sqrt{2}) + i \frac{π}{4} + \underset{j = 2}{\sum^{n}} C_{n - 1}^{j - 1} . \frac{{(1 + i)}^{1 - j} - 1}{1 - j}

= l n (\sqrt{2}) + \frac{i π}{4} + \underset{j = 2}{\sum^{n}} C_{n - 1}^{j - 1} \frac{{(\sqrt{2})}^{1 - j} e^{i \frac{(1 - j) π}{4}}}{1 - j} - \underset{j = 2}{\sum^{n}} \frac{C_{n - 1}^{j - 1}}{1 - j}

= \frac{i π}{4} + \underset{j = 1}{\sum^{n - 1}} C_{n - 1}^{j} \frac{{(\sqrt{2})}^{- j} e^{i (\frac{j π}{4})}}{- j} - \underset{j = 1}{\sum^{n - 1}} \frac{C_{n - 1}^{- j}}{- j}

R e {\frac{2^{n}}{2 i} . [\frac{i π}{4} - \underset{j = 1}{\sum^{n - 1}} C_{n - 1}^{j} \frac{e^{i \frac{j π}{4}}}{j {(\sqrt{2})}^{j}} + \underset{j = 1}{\sum^{n - 1}} \frac{C_{n - 1}^{j}}{j}]}

= \frac{2^{n} π}{8} - \frac{1}{2} \underset{j = 1}{\sum^{n - 1}} C_{n - 1}^{j} \frac{2^{n} s i n (\frac{j π}{4})}{j {(\sqrt{2})}^{j}}

= 2^{n} (\frac{π}{8} - \underset{k = 1}{\sum^{n - 1}} \frac{C_{n - 1}^{k} s i n (\frac{k π}{4})}{2 k {(\sqrt{2})}^{k}}), c h e c k i t S i r