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I want to integrate this integral $$\int_0^{\pi}d \theta\sin ^{d-2}(\theta ) \left(-1+e^{i k x \cos (\theta )}\right)$$ So first I tried

Integrate[(E^(I k2 x Cos[\[Theta]]) - 1) Sin[\[Theta]]^(  d - 2),
   {\[Theta], 0, \[Pi]},  Assumptions -> x > 0 && k2 > 0 ]

which came back nothing. Then I tried

Integrate[(E^(I k2 x Cos[\[Theta]]) - 1) Sin[\[Theta]]^(  d),
   {\[Theta], 0, \[Pi]},  Assumptions -> x > 0 && k2 > 0 ]

the result was

ConditionalExpression[(Sqrt[\[Pi]] Gamma[(1+d)/2] (-1
         +Hypergeometric0F1[1+d/2,-(1/4) k2^2 x^2]))/Gamma[(2+d)/2],Re[d]>-1]

Why didn't the first one work?

If I add another assumption

Integrate[(E^(I k2 x Cos[\[Theta]]) - 1) Sin[\[Theta]]^(  d - 2),
   {\[Theta], 0, \[Pi]},  Assumptions -> x > 0 && k2 > 0 && d>1]

the result looks normal but why can't mma guess the condition $d>1$ in the first integration?

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  • $\begingroup$ With Version 8.0 MMA does integrate your first integral and find the condition Re[d]>1. Integrate[(E^(I k2 x Cos[\[Theta]]) - 1) Sin[\[Theta]]^(d - 2), {\[Theta], 0, \[Pi]}, Assumptions -> x > 0 && k2 > 0] Result: ConditionalExpression[(1/Gamma[d/2]) Sqrt[\[Pi]] Gamma[1/2 (-1 + d)] (-1 + Hypergeometric0F1[d/2, -(1/4) k2^2 x^2]), Re[d] > 1] $\endgroup$ – Akku14 Jan 19 '18 at 15:21
  • $\begingroup$ @Akku14 That's weird. I'm using MMA 11.2 so maybe Wolfram decided to make it easier for competitors xd. $\endgroup$ – Turgon Jan 21 '18 at 17:27
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With MMA ver 11.2, I have tried to simplify a bit the integral with a variable change and taking into account that $k2>0$ and $x>0$. Thus $t=\cos(x)$ and the integral now reads

Assuming[a > 0, Integrate[(E^(I a t) - 1) (1 - t^2)^((d - 3)/2), {t, -1, 1}, 
Assumptions -> True]]

You get your expected normal (?) result:

ConditionalExpression[(Sqrt[\[Pi]]
Gamma[1/2 (-1 + d)] (-1 + Hypergeometric0F1[d/2, -(a^2/4)]))/
Gamma[d/2], Re[d] > 1]
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