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I wanted to compute the following integral $$I(\epsilon)=\int_{-\infty}^\infty \dfrac{d\lambda}{2\pi}e^{\epsilon+i\lambda}(\epsilon+i\lambda)^{\beta}\tag{1}$$

where $\epsilon>0$ is a regulator that I want to take to zero at the end. Doing (1) in Mathematica through

Integrate[1/(2 Pi)*Exp[I*x + \[Epsilon]]*(I*x + \[Epsilon])^\[Beta], {x, -\[Infinity], \[Infinity]}] //FullSimplify

I get the result $$\dfrac{i e^{-\frac{1}{2}i\pi\beta}((-i\epsilon)^{\beta}-e^{i\pi \beta}(i\epsilon)^\beta)\epsilon^{-\beta}\csc(\pi\beta)}{2\Gamma(-\beta)},\quad {\rm Re}(\beta)<0, \text{ && } {\rm Re}(\epsilon)>0\tag{2}.$$

Now it is clear that the $\epsilon$ dependence cancels since we can factor $\epsilon^\beta$ and cancel the $\epsilon^{-\beta}$, and the limit gets trivial. Still I find it interesting that if I ask Mathematica to evaluate this limit it says that it is indeterminate.

Why is that the case? Why Mathematica says that the limite is indeterminate when the $\epsilon$ dependence actually cancels rendering the limit trivial? I want to understand this because I'm probably doing something wrong and while this is a quite a trivial case here could be other cases in which the answer is far from obvious and having Mathematica do it would be very helpful.

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The simplification you identify is probably not valid for all values of β and ϵ. If we specify some assumptions, Mathematica makes the simplifications

integral = 
  Integrate[1/(2 Pi)*Exp[I*x + ϵ]*(I*x + ϵ)^β, {x, -∞, ∞}];

Assuming[β < 0 && ϵ > 0, 
 FullSimplify[integral]]
(* 1/Gamma[-β] *)
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