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I'm new to Mathematica so I'm not familiar with its potential use to solve tedious symbolically integrals. I'm trying to solve the following one:

Integrate[(U - 4 k (Sin[y] - 2 Cos[Sqrt[3] x/2] Sin[y/2]))/
  Sqrt[(U - 4 k (Sin[y] - 2 Cos[Sqrt[3] x/2] Sin[y/2]))^2 + 
  4*t^2 (3 + 2 Cos[y] + 4 Cos[y/2] Cos[Sqrt[3] x/2])], {y, 0, 
  1/Sqrt[3] (x + 2 Pi/Sqrt[3]) + 2 Pi/3}, {x, -2 Pi/Sqrt[3], 0}, 
  Assumptions -> U -> 0]

But it takes a long time without outputting a solution.

I'm not sure if by writing Integrate is all I can do to solve this integral or if there is something else I could do.

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  • 2
    $\begingroup$ Why not start with some simple integrals, perhaps using the help file for Integrate? Once you have solved a few simple ones, maybe you'll ee how to work the more complicated one. $\endgroup$ – bill s May 15 '16 at 21:23
  • $\begingroup$ "U -> 0" looks funny. Do you mean U equals zero? U is greater than or equal to zero? Is that just a typo? (just edit in a correction if you can) $\endgroup$ – Bill May 15 '16 at 21:24
  • $\begingroup$ sorry I thought I got the syntaxis right U->0 meant U<<1 (I swear to have seen it somewhere) $\endgroup$ – Sr Incerteza May 15 '16 at 21:35
  • $\begingroup$ If the limits of integration limits for y depend on x, then the limits for x are given first. In other words, {x, -2 Pi/Sqrt[3], 0} comes before {y, 0, 1/Sqrt[3] (x + 2 Pi/Sqrt[3]) + 2 Pi/3}. $\endgroup$ – JimB May 15 '16 at 21:45
  • $\begingroup$ I guess if there is no analytical expression for it, maybe solving perturbatively in U. How could I do that? $\endgroup$ – Sr Incerteza May 15 '16 at 22:11
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It seems unlikely that a symbolic solution can be obtained with Integrate, even for U == 0. So, I suggest solving the integral numerically. For U == 0, the result is,

f[t_] := NIntegrate[(-4  (Sin[y] - 2 Cos[Sqrt[3] x/2] Sin[y/2]))/
    Sqrt[(-4  (Sin[y] - 2 Cos[Sqrt[3] x/2] Sin[y/2]))^2 + 
    4*t^2 (3 + 2 Cos[y] + 4 Cos[y/2] Cos[Sqrt[3] x/2])], {x, -2 Pi/
    Sqrt[3], 0}, {y, 0, 1/Sqrt[3] (x + 2 Pi/Sqrt[3]) + 2 Pi/3}]
ListLinePlot[Quiet@Table[f[t], {t, 0, 10, .1}], DataRange -> {0, 10},  AxesLabel -> {t, f}]

enter image description here

Results for finite U can be obtained in a similar manner.

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