# Why does the following integral not evaluate? [closed]

I am a complete beginner to Mathematica. I am trying to evaluate a Fourier integral (see below). Here $$\mu(\vec{r})$$ is a mass density with units $$[\frac{u}{mm^3}]$$, and its Fourier transform is defined as

$$\mu(\vec{k}) = \mu(\vec{k}_p, k_z) = \int_0^Rdr\int_0^{2\pi}d\phi\int_0^ddz e^{-i\left(k_zz + k_p rcos(\phi)\right)}r\mu(\vec{r})$$,

where the integral is evaluated in cylindrical coordinates and the $$r$$ is due to the Jacobi determinant.

FT[mu_, R_, d_, kz_, kp_] :=
Integrate[r*mu*e^(-i*kz*z)*e^(-i*kp*r*Cos[\[Phi]]),
{r, 0, R}, {\[Phi], 0, 2 Pi}, {z, 0, d}]
FT[\[Mu], R, d, kz, kp]


I get the following errors:

Integrate::units: Integrate was unable to determine the units of quantities that appear in the input.

Equal::nord2: Comparison of \!$$\*TemplateBox[{\"0.25\", \"\\\"mm\\\"\", \ \"millimeters\", \"\\\"Millimeters\\\"\"},\"Quantity\"]$$ and 1 is \
invalid"

Integrate::idiv: Integral of Sec[\[Phi]]^2 does not converge on {0,\[Pi]}"


• You need to use correct syntax for it to work. Try FT[mu_, R_, d_, kz_, kp_] := Integrate[ r*mu*Exp[-I*kz*z]*Exp[-I*kp*r*Cos[\[Phi]]], {r, 0, R}, {\[Phi], 0, 2 Pi}, {z, 0, d}] Commented Mar 19, 2022 at 14:54
• E is $e$, I`is $I$.... Commented Mar 19, 2022 at 14:54
• Ok, some error messages are now gone. However, the first one (unable to determine inputs) still prevails. But thanks so far. Commented Mar 19, 2022 at 14:59
• can you show screen shot of what you get? Here is mine !Mathematica graphics no errors. Make sure to start with clean kernel. Commented Mar 19, 2022 at 15:06
• @Nasser Added it to the question Commented Mar 19, 2022 at 15:10