Question and Answers Forum
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 ⟩](Q150774.png)
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Question and Answers Forum | ||
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Question Number 150728 by ali1245 last updated on 14/Aug/21 | ||
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Answered by tabata last updated on 15/Aug/21 | ||
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