Integrate[((Sqrt[3] d (d + (\[Pi]^(-d/2)
r^-d ((2.309401076758503` - 1.1547005383792515` d) \[Pi]^(d/2) r^
d + (-25.719005343255322` + 25.719005343255322` d) r^3 Gamma[
1/2 (-1 + d)]))/((-2 + d) Sqrt[
1 + (0.06 r^2)/(2 - 3 d + d^2) - (
8 \[Pi]^(3/2 - d/2) r^(3 - d) Gamma[1/2 (-1 + d)])/(-2 +
d)])))/(8 r^2 \[Omega])),{r,a,b}]
I need the solution to be in d,\omega form but it is not giving any solution. Can anyone help?
If you experiment with parameter ranges (as the answer showed in your previous question, link), the integral can be done. For example, your integrand f[d,w,r] is singular at d=2, so we avoid this point; as a test, I then plotted the integrand for d=2.5, and found that it is real for r > 50 (roughly). Then,
In[1]:= Integrate[ f[2.5, 1, r], {r, a, b}, Assumptions -> {a > 50, b > a}, GenerateConditions -> False]
Out[1]= (1.35316 - 0.625 Sqrt[1. - 26.103 a^0.5 + 0.08 a^2])/a + (-1.35316 + 0.625 Sqrt[1. - 26.103 b^0.5 + 0.08 b^2])/b
So my comment was wrong, Mathematica can evaluate the integral, as long as the integrand is real.
EDIT: 1) The dependence on $\omega$ is trivial, 1/$\omega$, 2) for a fixed value of $d$, the valid range of $r$ is determined by the range where the square root in the denominator is real - Mathematica cannot solve for this explicitly (at least in my attempts) because of the $\Gamma$ function. You will have to get around this somehow by working out each case.
