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Here's how to make Mathematica integrate this, but a lot is done by hand. Cos[β] Exp[I z Cos[β - α]] == Cos[β] Cos[z Cos[α - β]] + I Cos[β] Sin[z Cos[α - β]] The integral of the first summand is zero. Make the substitution γ == α - β and expand the Cos[β] to Cos[α] Cos[γ] + Sin[α] Sin[γ]. Also note that the function is Pi-periodic. The integral ...


This question is being automatically bumped as unanswered. However, we have an authoritative answer in comments: Investigating as a regression. You can put a "bugs" tag on it if you like. --Daniel Lichtblau


I believe this equation are quiet unstable for the initial values, so there is a two solutions. You can either specify AccuracyGoal: ListPlot@NDSolveValue[{-w''[x] + 2/x w'[x] + w[x] == 0, w[1/10^6] == 10^-2, w[5] == 1}, w, {x, 1/10^6, 5}, AccuracyGoal -> 10] Or use the DSolveValue, the equation are solvable analytically: ...


Trace[Integrate[ Integrate[ Integrate[ E^(-v1 - v2 - v3) (v1 + v2 + v3)/3, {v3, 2 v1 - 5/2 v2, v2}], {v2, 4 v1/7, 5 v1/7}], {v1, 0, Infinity}]] Check one line before the result. I think this is what you want.


Well, n[1, 1, 1] (* {1.38996, 1.85383, 1.37325} *) n[5, 5, 5] (* {1.38996, 1.85383, 1.37325} *) When you first define your variables and then assign n[...] to this expression, the expression will evaluate and then be stored as that value. Regardless of what values you pass to n it will always return the same thing. You can read more about how Mathematica ...

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