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Question Number 130387 by Bird last updated on 25/Jan/21

$$calculate{\int }_0^{\mathrm{\infty }}\frac{cos\left(abx\right)}{(x^2+ax+1)(x^2+bx+1)}$$$$withaandbrealand\mid a\mid <2,\mid b\mid <2$$

Commented by mathmax by abdo last updated on 25/Jan/21

$$\mathrm{the}\mathrm{q}.\mathrm{is}{\int }_0^{\mathrm{\infty }}\frac{\cos \left(\mathrm{abx}\right)}{({\mathrm{x}}^2+\mathrm{ax}+1)({\mathrm{x}}^2+\mathrm{bx}+1)}\mathrm{dx}$$

Commented by mathmax by abdo last updated on 25/Jan/21

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\cos \left(\mathrm{abx}\right)}{({\mathrm{x}}^2+\mathrm{ax}+1)({\mathrm{x}}^2+\mathrm{bx}+1)}\mathrm{dx}\Rightarrow$$$$2\mathrm{I}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\cos \left(\mathrm{abx}\right)}{({\mathrm{x}}^2+\mathrm{ax}+1)({\mathrm{x}}^2+\mathrm{bx}+1)}\mathrm{dx}=\mathrm{Re}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{iabx}}}{({\mathrm{x}}^2+\mathrm{ax}+1)({\mathrm{x}}^2+\mathrm{bx}+1)}\mathrm{dx}\right)$$$$\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{iabz}}}{({\mathrm{z}}^2+\mathrm{az}+1)({\mathrm{z}}^2+\mathrm{bz}+1)}\mathrm{poles}\mathrm{of}\phi ?$$$${\mathrm{z}}^2+\mathrm{az}+1=0\to \mathrm{\Delta }={\mathrm{a}}^2-4<0\Rightarrow {\mathrm{z}}_1=\frac{-\mathrm{a}+\mathrm{i}\sqrt{4-{\mathrm{a}}^2}}{2}\mathrm{and}$$$${\mathrm{z}}_2=\frac{-\mathrm{a}-\mathrm{i}\sqrt{4-{\mathrm{a}}^2}}{2}\mathrm{also}\mathrm{the}\mathrm{roots}\mathrm{of}{\mathrm{x}}^2+\mathrm{bx}+1\mathrm{are}$$$${\mathrm{t}}_1=\frac{-\mathrm{b}+\mathrm{i}\sqrt{4-{\mathrm{b}}^2}}{2}\mathrm{and}{\mathrm{t}}_2=\frac{-\mathrm{b}-\mathrm{i}\sqrt{4-{\mathrm{b}}^2}}{2}\Rightarrow$$$$\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{iabz}}}{(\mathrm{z}-{\mathrm{z}}_1)(\mathrm{z}-{\mathrm{z}}_2)(\mathrm{z}-{\mathrm{t}}_1)(\mathrm{z}-{\mathrm{t}}_2)}\mathrm{residus}\mathrm{theorem}\mathrm{give}$$$${\int }_{\mathrm{R}}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\mathrm{Res}(\phi ,{\mathrm{z}}_1)+\mathrm{Res}(\phi ,{\mathrm{t}}_1)\}$$$$\mathrm{Res}(\phi ,{\mathrm{z}}_1)=\frac{{\mathrm{e}}^{{\mathrm{iabz}}_1}}{({\mathrm{z}}_1-{\mathrm{z}}_2)({\mathrm{z}}_1-{\mathrm{t}}_1)({\mathrm{z}}_1-{\mathrm{t}}_2)}$$$$=\frac{{\mathrm{e}}^{{\mathrm{iabz}}_1}}{-\mathrm{i}\sqrt{4-{\mathrm{a}}^2}({\mathrm{z}}_1^2+{\mathrm{bz}}_1+1)}=....$$$$\mathrm{Res}(\phi ,{\mathrm{z}}_2)=\frac{{\mathrm{e}}^{{\mathrm{iabz}}_2}}{({\mathrm{z}}_2-{\mathrm{z}}_1)({\mathrm{z}}_2^2+{\mathrm{bz}}_2+1)}=....$$$$\mathrm{rest}\mathrm{to}\mathrm{collect}\mathrm{the}\mathrm{results}....$$

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