Question Number 150728 by ali1245 last updated on 14/Aug/21
Answered by tabata last updated on 15/Aug/21
$$\mathrm{2}{I}=_{{z}={x}\:} {Re}\:\left(\int_{−\infty} ^{\:\infty} \:\frac{{z}^{\:{a}−\mathrm{1}} }{\left({z}^{\mathrm{2}} +{c}\right)\left({z}^{\mathrm{2}} +{b}\right)}{dz}\right) \\ $$$$ \\ $$$${Res}\left({f},{i}\sqrt{{c}}\right)={lim}_{{z}\rightarrow{i}\sqrt{{c}}} \left(\frac{{z}^{{a}−\mathrm{1}} }{\left({z}+{i}\sqrt{{c}}\right)\left({z}^{\mathrm{2}} +{b}\right)}\right)=\left(\frac{\left({i}\sqrt{{c}}\right)^{{a}−\mathrm{1}} }{\left(\mathrm{2}{i}\sqrt{{c}}\right)\left({b}−{c}\right)}\right)=\left(\frac{\left({i}\sqrt{{c}}\right)^{{a}} }{\mathrm{2}\left({c}^{\mathrm{2}} −{bc}\right)}\right) \\ $$$$ \\ $$$${Res}\left({f},{i}\sqrt{{b}}\right)={lim}_{{z}\rightarrow{i}\sqrt{{b}}} \left(\frac{{z}^{{a}−\mathrm{1}} }{\left({z}^{\mathrm{2}} +{c}\right)\left({z}+{i}\sqrt{{b}}\right)}\right)=\left(\frac{\left({i}\sqrt{{b}}\right)^{{a}−\mathrm{1}} }{\left({c}−{b}\right)\left(\mathrm{2}{i}\sqrt{{b}}\right)}\right)=\left(\frac{\left({i}\sqrt{{b}}\right)^{{a}} }{\mathrm{2}\left({b}^{\mathrm{2}} −{bc}\right)}\right) \\ $$$$ \\ $$$$\mathrm{2}{I}={Re}\left(\mathrm{2}\pi{i}\left[\:{Res}\left({f}\:,{i}\sqrt{{c}}\right)+{Res}\left({f},{i}\sqrt{{b}}\right)\right]=\mathrm{2}\pi{i}×\frac{\left({i}\right)^{{a}} \:\left[\left(\sqrt{{b}}\right)^{{a}} +\left(\sqrt{{c}}\right)^{{a}} \right]}{\mathrm{2}\left({c}−{b}\right)^{\mathrm{2}} }\:\right) \\ $$$$ \\ $$$$\mathrm{2}{I}=\frac{\left({i}\right)^{{a}+\mathrm{1}} \left[\left(\sqrt{{b}}\right)^{{a}} +\left(\sqrt{{c}}\right)^{{a}} \right]\pi}{\left({c}−{b}\right)^{\mathrm{2}} }\Rightarrow{I}={Re}\left(\:\frac{\left({e}^{{i}\frac{\pi}{\mathrm{2}}} \right)^{{a}+\mathrm{1}} \left[\left(\sqrt{{b}}\right)^{{a}} +\left(\sqrt{{c}}\right)^{{a}} \right]\pi}{\mathrm{2}\left({c}−{b}\right)^{\mathrm{2}} }\right) \\ $$$$ \\ $$$${I}=\frac{\left[\left(\sqrt{{b}}\right)^{{a}} +\left(\sqrt{{c}}\right)^{{a}} \right]\pi}{\mathrm{2}\left({c}−{b}\right)^{\mathrm{2}} }\:{cos}\left(\frac{\pi\left({a}+\mathrm{1}\right)}{\mathrm{2}}\right) \\ $$$$ \\ $$$$\langle\:{M}\:.\:{T}\:\rangle \\ $$