Mathematica Asked on July 11, 2021
This simple integration:
Integrate[ E^(I q x)/(q + I [Kappa]), {q, -[Infinity], [Infinity]}, Assumptions -> { x [Element]
Reals, [Kappa] [Element] Reals, x != 0, [Kappa] != 0 }]
gives a complicated answer in terms of MeijerG function. Whereas the answer is simply
E^(- [Kappa] x) (2 [Pi] I ) UnitStep[[Kappa] x] Sign[-[Kappa]]
Up to George Pólya, two small steps are better than one big step, so
Integrate[Cos[(q x)]/(q + I*[Kappa]), {q, -[Infinity], [Infinity]}, Assumptions -> x > 0]
ConditionalExpression[ I*E^(x*[Kappa])*Pi, Re[[Kappa]] < 0]
and
Integrate[Sin[(q x)]/(q + I*[Kappa]), {q, -[Infinity], [Infinity]},Assumptions -> x > 0]
ConditionalExpression[E^(x [Kappa]) [Pi], Re[[Kappa]] < 0]
Answered by user64494 on July 11, 2021
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