# Integrate is not working [closed]

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?

• Mathematica (version 11.2.0) returns the indefinite integral unevaluated - it is not able to perform the integration, so not likely to with arbitrary limits $a$ and $b$. Mar 17, 2022 at 21:19
• Can we integrate it using a>2 and b>a? I am trying it but still not getting the solution.
– AAA
Mar 17, 2022 at 21:22

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.

• yes. that we can do but I wanted the final result in terms of d and \omega. I am not getting that.
– AAA
Mar 18, 2022 at 7:55