Question Number 150728 by ali1245 last updated on 14/Aug/21
Answered by tabata last updated on 15/Aug/21
![2I=_(z=x ) Re (∫_(−∞) ^( ∞) (z^( a−1) /((z^2 +c)(z^2 +b)))dz) Res(f,i(√c))=lim_(z→i(√c)) ((z^(a−1) /((z+i(√c))(z^2 +b))))=((((i(√c))^(a−1) )/((2i(√c))(b−c))))=((((i(√c))^a )/(2(c^2 −bc)))) Res(f,i(√b))=lim_(z→i(√b)) ((z^(a−1) /((z^2 +c)(z+i(√b)))))=((((i(√b))^(a−1) )/((c−b)(2i(√b)))))=((((i(√b))^a )/(2(b^2 −bc)))) 2I=Re(2πi[ Res(f ,i(√c))+Res(f,i(√b))]=2πi×(((i)^a [((√b))^a +((√c))^a ])/(2(c−b)^2 )) ) 2I=(((i)^(a+1) [((√b))^a +((√c))^a ]π)/((c−b)^2 ))⇒I=Re( (((e^(i(π/2)) )^(a+1) [((√b))^a +((√c))^a ]π)/(2(c−b)^2 ))) I=(([((√b))^a +((√c))^a ]π)/(2(c−b)^2 )) cos(((π(a+1))/2)) ⟨ M . T ⟩](https://www.tinkutara.com/question/Q150774.png)
$$\mathrm{2}{I}=_{{z}={x}\:} {Re}\:\left(\int_{−\infty} ^{\:\infty} \:\frac{{z}^{\:{a}−\mathrm{1}} }{\left({z}^{\mathrm{2}} +{c}\right)\left({z}^{\mathrm{2}} +{b}\right)}{dz}\right) \\ $$$$ \\ $$$${Res}\left({f},{i}\sqrt{{c}}\right)={lim}_{{z}\rightarrow{i}\sqrt{{c}}} \left(\frac{{z}^{{a}−\mathrm{1}} }{\left({z}+{i}\sqrt{{c}}\right)\left({z}^{\mathrm{2}} +{b}\right)}\right)=\left(\frac{\left({i}\sqrt{{c}}\right)^{{a}−\mathrm{1}} }{\left(\mathrm{2}{i}\sqrt{{c}}\right)\left({b}−{c}\right)}\right)=\left(\frac{\left({i}\sqrt{{c}}\right)^{{a}} }{\mathrm{2}\left({c}^{\mathrm{2}} −{bc}\right)}\right) \\ $$$$ \\ $$$${Res}\left({f},{i}\sqrt{{b}}\right)={lim}_{{z}\rightarrow{i}\sqrt{{b}}} \left(\frac{{z}^{{a}−\mathrm{1}} }{\left({z}^{\mathrm{2}} +{c}\right)\left({z}+{i}\sqrt{{b}}\right)}\right)=\left(\frac{\left({i}\sqrt{{b}}\right)^{{a}−\mathrm{1}} }{\left({c}−{b}\right)\left(\mathrm{2}{i}\sqrt{{b}}\right)}\right)=\left(\frac{\left({i}\sqrt{{b}}\right)^{{a}} }{\mathrm{2}\left({b}^{\mathrm{2}} −{bc}\right)}\right) \\ $$$$ \\ $$$$\mathrm{2}{I}={Re}\left(\mathrm{2}\pi{i}\left[\:{Res}\left({f}\:,{i}\sqrt{{c}}\right)+{Res}\left({f},{i}\sqrt{{b}}\right)\right]=\mathrm{2}\pi{i}×\frac{\left({i}\right)^{{a}} \:\left[\left(\sqrt{{b}}\right)^{{a}} +\left(\sqrt{{c}}\right)^{{a}} \right]}{\mathrm{2}\left({c}−{b}\right)^{\mathrm{2}} }\:\right) \\ $$$$ \\ $$$$\mathrm{2}{I}=\frac{\left({i}\right)^{{a}+\mathrm{1}} \left[\left(\sqrt{{b}}\right)^{{a}} +\left(\sqrt{{c}}\right)^{{a}} \right]\pi}{\left({c}−{b}\right)^{\mathrm{2}} }\Rightarrow{I}={Re}\left(\:\frac{\left({e}^{{i}\frac{\pi}{\mathrm{2}}} \right)^{{a}+\mathrm{1}} \left[\left(\sqrt{{b}}\right)^{{a}} +\left(\sqrt{{c}}\right)^{{a}} \right]\pi}{\mathrm{2}\left({c}−{b}\right)^{\mathrm{2}} }\right) \\ $$$$ \\ $$$${I}=\frac{\left[\left(\sqrt{{b}}\right)^{{a}} +\left(\sqrt{{c}}\right)^{{a}} \right]\pi}{\mathrm{2}\left({c}−{b}\right)^{\mathrm{2}} }\:{cos}\left(\frac{\pi\left({a}+\mathrm{1}\right)}{\mathrm{2}}\right) \\ $$$$ \\ $$$$\langle\:{M}\:.\:{T}\:\rangle \\ $$