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Question Number 90040 by abdomathmax last updated on 21/Apr/20

$$find{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{ch(acosx+bsinx)}{x^2-x+1}dx$$$$aandbrealsgiven$$

Commented by mathmax by abdo last updated on 22/Apr/20

$$A={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{ch(acosx+bsinx)}{x^2-x+1}dxlet\phi \left(z\right)=\frac{ch(acosz+bsinz)}{z^2-z+1}$$$$polesof\phi ?z^2-z+1=0\to \mathrm{\Delta }=-3\Rightarrow z_1=\frac{1+i\sqrt{3}}{2}=e^{\frac{i\pi }{3}}$$$$z_2=\frac{1-i\sqrt{3}}{2}=e^{-\frac{i\pi }{3}}\Rightarrow \phi \left(z\right)=\frac{ch(acosx+bsinx)}{(z-e^{\frac{i\pi }{3}})(z-e^{-\frac{i\pi }{3}})}residustheoremgive$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,e^{\frac{i\pi }{3}})=2i\pi \times \frac{ch(acos\left(e^{\frac{i\pi }{3}}\right)+bsin\left(e^{\frac{i\pi }{3}}\right))}{2isin\left(\frac{\pi }{3}\right)}$$$$=\pi \times \frac{2}{\sqrt{3}}ch(acos\left(e^{\frac{i\pi }{3}}\right)+bsin\left(e^{\frac{i\pi }{3}}\right))$$$$cos\left(e^{\frac{i\pi }{3}}\right)=cos(\frac{1}{2}+i\frac{\sqrt{3}}{2})=\frac{e^{i(\frac{1}{2}+i\frac{\sqrt{3}}{2})}+e^{-i(\frac{1}{2}+\frac{i\sqrt{3}}{2})}}{2}$$$$=\frac{1}{2}\left\{\frac{e^{-\frac{\sqrt{3}}{2}}(cos\left(\frac{1}{2}\right)+isin\left(\frac{1}{2}\right))+e^{\frac{\sqrt{3}}{2}}(cos\left(\frac{1}{2}\right)-isin\left(\frac{1}{2}\right))}{1}\right\}$$$$=\frac{1}{2}\{cos\left(\frac{1}{2}\right)(e^{\frac{\sqrt{3}}{2}}+e^{-\frac{\sqrt{3}}{2}})-isin\left(\frac{1}{2}\right)(e^{\frac{\sqrt{3}}{2}}-e^{-\frac{\sqrt{3}}{2}})\}$$$$=cos\left(\frac{1}{2}\right)ch\left(\frac{\sqrt{3}}{2}\right)-sin\left(\frac{1}{2}\right)sh\left(\frac{\sqrt{3}}{2}\right)...becontinued...$$

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