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

∫_0 ^π (a−e^(−ix) )^n (a−e^(ix) )^n cos(nx)dx

$$\int_{\mathrm{0}} ^{\pi} \left(\mathrm{a}−\mathrm{e}^{−\mathrm{ix}} \right)^{\mathrm{n}} \left(\mathrm{a}−\mathrm{e}^{\mathrm{ix}} \right)^{\mathrm{n}} \mathrm{cos}\left(\mathrm{nx}\right)\mathrm{dx} \\ $$

Answered by mindispower last updated on 14/Jul/21

=∫_0 ^(2π) (a−e^(−ix) )(a−e^(ix) )^n cos(nx)dx  =∫_0 ^π (a−e^(−ix) )^n (a−e^(ix) )^n cos(nx)dx+  ∫_π ^(2π) (a−e^(−ix) )^n (a−e^(ix) )^n cos(nx)dx  =A+B,B,x→2π−x  B⇔∫_0 ^π (a−e^(ix) )^n (a−e^(−ix) )^n cos(nx)=A  2A=∫_0 ^(2π) (a−(1/e^(ix) ))^n (a−e^(ix) )^n .((e^(inx) −e^(−inx) )/2).  e^(ix) =z  ⇒∫_C (((az−1)^n (a−z)^n )/z^n ).((z^(2n) −1)/(2z^n )).(1/(iz))dz=2iπ Res(f,0)  =∫_C (((az−1)^n (a−z)^n )/(2iz))−(1/(2iz^(2n+1) ))(az−1)(a−z)dz  =π(−1)^n a^n −π(−1)^n a^n =0

$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \left({a}−{e}^{−{ix}} \right)\left({a}−{e}^{{ix}} \right)^{{n}} {cos}\left({nx}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \left({a}−{e}^{−{ix}} \right)^{{n}} \left({a}−{e}^{{ix}} \right)^{{n}} {cos}\left({nx}\right){dx}+ \\ $$$$\int_{\pi} ^{\mathrm{2}\pi} \left({a}−{e}^{−{ix}} \right)^{{n}} \left({a}−{e}^{{ix}} \right)^{{n}} {cos}\left({nx}\right){dx} \\ $$$$={A}+{B},{B},{x}\rightarrow\mathrm{2}\pi−{x} \\ $$$${B}\Leftrightarrow\int_{\mathrm{0}} ^{\pi} \left({a}−{e}^{{ix}} \right)^{{n}} \left({a}−{e}^{−{ix}} \right)^{{n}} {cos}\left({nx}\right)={A} \\ $$$$\mathrm{2}{A}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \left({a}−\frac{\mathrm{1}}{{e}^{{ix}} }\right)^{{n}} \left({a}−{e}^{{ix}} \right)^{{n}} .\frac{{e}^{{inx}} −{e}^{−{inx}} }{\mathrm{2}}. \\ $$$${e}^{{ix}} ={z} \\ $$$$\Rightarrow\int_{{C}} \frac{\left({az}−\mathrm{1}\right)^{{n}} \left({a}−{z}\right)^{{n}} }{{z}^{{n}} }.\frac{{z}^{\mathrm{2}{n}} −\mathrm{1}}{\mathrm{2}{z}^{{n}} }.\frac{\mathrm{1}}{{iz}}{dz}=\mathrm{2}{i}\pi\:{Res}\left({f},\mathrm{0}\right) \\ $$$$=\int_{{C}} \frac{\left({az}−\mathrm{1}\right)^{{n}} \left({a}−{z}\right)^{{n}} }{\mathrm{2}{iz}}−\frac{\mathrm{1}}{\mathrm{2}{iz}^{\mathrm{2}{n}+\mathrm{1}} }\left({az}−\mathrm{1}\right)\left({a}−{z}\right){dz} \\ $$$$=\pi\left(−\mathrm{1}\right)^{{n}} {a}^{{n}} −\pi\left(−\mathrm{1}\right)^{{n}} {a}^{{n}} =\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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

Thank you Sir

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir} \\ $$

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