0-pi-a-e-ix-n-a-e-ix-n-cos-nx-dx-

Question Number 146669 by Ar Brandon last updated on 14/Jul/21

\int_{0}^{π} {(a - e^{- ix})}^{n} {(a - e^{ix})}^{n} \cos (nx) dx

Answered by mindispower last updated on 14/Jul/21

= \int_{0}^{2 π} (a - e^{- i x}) {(a - e^{i x})}^{n} c o s (n x) d x

= \int_{0}^{π} {(a - e^{- i x})}^{n} {(a - e^{i x})}^{n} c o s (n x) d x +

\int_{π}^{2 π} {(a - e^{- i x})}^{n} {(a - e^{i x})}^{n} c o s (n x) d x

= A + B, B, x \to 2 π - x

B \Leftrightarrow \int_{0}^{π} {(a - e^{i x})}^{n} {(a - e^{- i x})}^{n} c o s (n x) = A

2 A = \int_{0}^{2 π} {(a - \frac{1}{e^{i x}})}^{n} {(a - e^{i x})}^{n} . \frac{e^{i n x} - e^{- i n x}}{2} .

e^{i x} = z

\Rightarrow \int_{C} \frac{{(a z - 1)}^{n} {(a - z)}^{n}}{z^{n}} . \frac{z^{2 n} - 1}{2 z^{n}} . \frac{1}{i z} d z = 2 i π R e s (f, 0)

= \int_{C} \frac{{(a z - 1)}^{n} {(a - z)}^{n}}{2 i z} - \frac{1}{2 {i z}^{2 n + 1}} (a z - 1) (a - z) d z

= π {(- 1)}^{n} a^{n} - π {(- 1)}^{n} a^{n} = 0

Commented by Ar Brandon last updated on 15/Jul/21

Thank you Sir