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Question Number 42680 by prof Abdo imad last updated on 31/Aug/18

$$calculale{A}_n\left(\alpha \right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(\alpha x^n\right)}{1+x^2}dxwith$$$$nintegrnatural.$$$$$$

Commented by prof Abdo imad last updated on 31/Aug/18

$$\alpha real.$$

Commented by maxmathsup by imad last updated on 01/Sep/18

$$wehave{A}_n\left(\alpha \right)=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{i\alpha x^n}}{1+x^2}dx\right)letconsiderthecomlexfunction$$$$\phi \left(z\right)=\frac{e^{i\alpha z^n}}{1+z^2}\Rightarrow \phi \left(z\right)=\frac{e^{i\alpha z^n}}{(z-i)(z+i)}thepolesof\phi are\stackrel{-}{+}i$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,i)butRes(\phi ,i)=\frac{e^{i\alpha i^n}}{2i}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \frac{e^{i\alpha i^n}}{2i}=\pi e^{i\alpha {(-1)}^{\frac{n}{2}}}=\pi \{cos\left({(-1)}^{\frac{n}{2}}\alpha \right)+isin\left({(-1)}^{\frac{n}{2}}\alpha \right)\}\Rightarrow$$$$A_n=\pi cos\left\{{(-1)}^{\frac{n}{2}}\alpha \right\}\Rightarrow A_{2n}=\pi cos\left\{{(-1)}^n\alpha \right\}and$$$$A_{2n+1}=\pi cos\left\{i{(-1)}^n\alpha \right\}butcosz=\frac{e^{iz}+e^{-iz}}{2}\Rightarrow cos\left(ix\right)=\frac{e^{-x}+e^x}{2}=ch\left(x\right)$$$$(x\in R)\Rightarrow A_{2n+1}=\pi ch\left\{{(-1)}^n\alpha \right\}.$$

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