calculus and analysis - Bug in mathematica analytic integration?

I found Mathematica provides me a wrong answer for a relatively simple analytically solvable integral:

Integrate[Exp[-I θ]/(1 + b Cos[θ]), {θ, 0, 2 π}, 
     Assumptions -> b < 1 && b > -1]

Provides as a result (2π)/b, which is incorrect. The integral can be computed easily writing the integrand as a geometric series.

Additionally, you can compute it numerically

NIntegrate[Exp[-I θ]/(1 + b Cos[θ]), {θ, 0, 2 π}]

which produces the right result, different from (2π)/b.

Is this a very bad bug? or am I doing something wrong? What could be the reason for this error?

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plotting - Filling between two spheres in SphericalPlot3D

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I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Adding a thick curve to a regionplot

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