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Question Number 48171 by Abdo msup. last updated on 20/Nov/18

$$calculate{\int }_0^{\mathrm{\infty }}\frac{sin\left(cosx\right)}{x^2+3}dx$$

Commented by Abdo msup. last updated on 21/Nov/18

$$letI={\int }_0^{\mathrm{\infty }}\frac{sin\left(cosx\right)}{x^2+3}dx\Rightarrow 2I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{sin\left(cosx\right)}{x^2+3}dx$$$$=Im\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{icosx}}{x^2+3}dx\right)letconsiderthecomplexfunction$$$$\phi \left(z\right)=\frac{e^{icosz}}{z^2+3}wehave\phi \left(z\right)=\frac{e^{icosz}}{(z-i\sqrt{3})(z+i\sqrt{3})}?sothepoles$$$$of\phi are\stackrel{-}{+}i\sqrt{3}residustheorem?give$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,i\sqrt{3})$$$$Res(\phi ,i\sqrt{3})={lim}_{z\to i\sqrt{3}}(z-i\sqrt{3})\phi \left(z\right)$$$$=\frac{e^{icos\left(i\sqrt{3}\right)}}{2i\sqrt{3}}butcos\left(i\sqrt{3}\right)=\frac{e^{i\left(i\sqrt{3}\right)}+e^{-i\left(i\sqrt{3}\right)}}{2}=\frac{e^{-\sqrt{3}}+e^{\sqrt{3}}}{2}$$$$=ch\left(\sqrt{3}\right)\Rightarrow Res(\phi ,i\sqrt{3})=\frac{e^{ich\left(\sqrt{3}\right)}}{2i\sqrt{3}}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi .\frac{e^{ich\left(\sqrt{3}\right)}}{2i\sqrt{3}}=\frac{\pi }{\sqrt{3}}\{cos(ch\left(\sqrt{3}\right)+isin\left(ch\left(\sqrt{3}\right)\right)\}\Rightarrow$$$$2I=\frac{\pi }{\sqrt{3}}sin\left(ch\left(\sqrt{3}\right)\right)\Rightarrow$$$$I=\frac{\pi }{2\sqrt{3}}sin\left(\frac{e^{\sqrt{3}}+e^{-\sqrt{3}}}{2}\right).$$

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