Can an anonymous function be used as an input to another anonymous function, which will be integrated?

Hello, I've been having some issues with the output from a piece of my code, in that it gives infinities and sharp peaks at strange points. I'm not currently sure whether this is because I have done the maths leading me to the equations wrong, or the coding wrong, so I wanted to check the latter possibility.

What I am doing it trying to integrate an anonymous function of 3 variables, which is made up of a number of smaller functions multiplied together. The issue is that to get the correct dependence on the 3 variables, I must enter another function of them as one of the arguments in the smaller function. See the code below for what I mean. My question is, is this allowed in Matlab? If so, am I doing it wrong and how? The code is:

sigxyintegrand = @(k0,psi,psipr) vx(k0-(k00+k06*cos(6*phi+6.*psi)).*cos(psi)*tan(theta),psi).*vy(k0-(k00+k06*cos(6*phi+6.*psipr)).*cos(psipr)*tan(theta),psipr).*exp(psipr.*cos(theta)./wt0); %Defines the integrand
limit = @(k0,psi) psi*cos(theta);

Sxy = e^2*cos(theta)*m/(4*pi^3*hbar^2*omega_c)*integral3(sigxyintegrand,-pi/c,pi/c,0,(2*pi)/cos(theta),-Inf,limit);

The vx and vy functions are defined as simple functions of two variables involving sines and cosines, and e, theta, m, hbar, pi, phi, wt0 and omega_c are all constants. As you can see, changing variables from k0 to chi = the function used as an argument will not work as the two arguments are different - one depends on psi and the other on psipr. Any help in this would be hugely appreciated, thank you!

EDIT: I've now tried this via setting up another anonymous function, kz, as:

kz = @(k0,psi) k0-(k00+k06*cos(6*phi+6.*psi)).*cos(psi)*tan(theta);

sigxyintegrand = @(k0,psi,psipr) vx(kz(k0,psi),psi).*vy(kz(k0,psipr),psipr).*exp(psipr.*cos(theta)./wt0); %Defines the integrand

This clearly makes it easier to read, but doesn't seem to make it work better at all. Are there any other ways I could try and do this?