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

$${\int }_0^{\pi }{(\mathrm{a}-{\mathrm{e}}^{-\mathrm{ix}})}^{\mathrm{n}}{(\mathrm{a}-{\mathrm{e}}^{\mathrm{ix}})}^{\mathrm{n}}\cos \left(\mathrm{nx}\right)\mathrm{dx}$$

Answered by mindispower last updated on 14/Jul/21

$$={\int }_0^{2\pi }(a-e^{-ix}){(a-e^{ix})}^{n}cos\left(nx\right)dx$$$$={\int }_0^{\pi }{(a-e^{-ix})}^n{(a-e^{ix})}^{n}cos\left(nx\right)dx+$$$${\int }_{\pi }^{2\pi }{(a-e^{-ix})}^n{(a-e^{ix})}^{n}cos\left(nx\right)dx$$$$=A+B,B,x\to 2\pi -x$$$$B\iff {\int }_0^{\pi }{(a-e^{ix})}^n{(a-e^{-ix})}^{n}cos\left(nx\right)=A$$$$2A={\int }_0^{2\pi }{(a-\frac{1}{e^{ix}})}^n{(a-e^{ix})}^n.\frac{e^{inx}-e^{-inx}}{2}.$$$$e^{ix}=z$$$$\Rightarrow {\int }_C\frac{{(az-1)}^n{(a-z)}^n}{z^n}.\frac{z^{2n}-1}{2z^n}.\frac{1}{iz}dz=2i\pi Res(f,0)$$$$={\int }_C\frac{{(az-1)}^n{(a-z)}^n}{2iz}-\frac{1}{2{iz}^{2n+1}}(az-1)(a-z)dz$$$$=\pi {(-1)}^n{a}^n-\pi {(-1)}^n{a}^n=0$$$$$$$$$$$$$$$$$$

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

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

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