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Question Number 39787 by abdo mathsup 649 cc last updated on 10/Jul/18

$$calculste{I}_{\lambda }={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(\lambda x^n\right)}{1+x^2}dxwith$$$$\lambda fromRandnintegrnatural$$$$2)findthevslueof{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(3x^9\right)}{1+x^2}dx.$$

Commented by abdo mathsup 649 cc last updated on 11/Jul/18

$$1)wehave{I}_{\lambda }=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{i\lambda x^n}}{1+x^2}dx\right)let$$$$\phi \left(z\right)=\frac{e^{i\lambda z^n}}{1+z^2}$$$$\phi \left(z\right)=\frac{e^{i\lambda z^n}}{(z-i)(z+i)}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,i)$$$$Res(\phi ,i)={lim}_{z\to i}(z-i)\phi \left(z\right)$$$$={lim}_{z\to i}\frac{e^{i\lambda z^n}}{z+i}=\frac{e^{\lambda i^{n+1}}}{2i}=\frac{e^{\lambda \{cos(\frac{(n+1)\pi }{2}+isin\left(\frac{(n+1)\pi }{2}\right)\}}}{2i}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi e^{\lambda cos\left(\frac{(n+1)\pi }{2}\right)}\frac{cos)\left(\lambda sin\left(\frac{(n+1)\pi }{2}\right)\right)+isin\left(\lambda sin\left(\frac{(n+1)\pi }{2}\right)\right)}{2i}$$$$\Rightarrow I_{\lambda }=\pi e^{\lambda cos\left(\frac{(n+1)\pi }{2}\right)}cos\left\{\lambda sin\left(\frac{(n+1)\pi }{2}\right)\right\}.$$

Commented by math khazana by abdo last updated on 12/Jul/18

$$2)lettake\lambda =3andn=9weget$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(3x^9\right)}{1+x^2}dx=\pi e^{3cos\left(5\pi \right)}cos\left\{3sin\left(5\pi \right)\right\}$$$$=\pi e^{-3}=\frac{\pi }{e^3}.$$$$$$

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