How to get the integral expression?

I am trying to solve following expression in Matlab.

And my script is

clc;close all;clear all;
syms s phis r phir
A=1;m=1;n=1;L=100;alpha=1;
k=(2*pi)/(550*10^-9);
us=A*((-1j*sqrt(2*k*alpha)*s*exp(-1j*phis))^m)*exp(-k*alpha*s*s)*laguerreL(n,m,2*k*alpha*s*s);
C=((-1j*k*exp(1j*k*L))/(2*pi*L))*exp((1j*k*r*r)/(2*L));
f = @(s,phis,r,phir) C*s*us*exp(((1j*k)/(2*L))*(-2*r*s*cos(phir-phis)+(s^2)));
g = @(s,phis,r,phir) integral2(@(s,phis) f(s,phis,r,phir),0,Inf,0,2*pi)

g = function_handle with value:
    @(s,phis,r,phir)integral2(@(s,phis)f(s,phis,r,phir),0,Inf,0,2*pi)

I need to get the final result as

How could I get the result like this?

[Torsten]

Symbolic variables can only be used together with "int", not "integral2".

[Athira T Das]

Ok.

Then by doing integration in two steps like

clc;close all;clear all;
syms s phis r phir
A=1;m=1;n=1;L=100;alpha=1;
k=(2*pi)/(550*10^-9);
us=A*((-1j*sqrt(2*k*alpha)*s*exp(-1j*phis))^m)*exp(-k*alpha*s*s)*laguerreL(n,m,2*k*alpha*s*s);
C=((-1j*k*exp(1j*k*L))/(2*pi*L))*exp((1j*k*r*r)/(2*L));
f = @(s,phis,r,phir) C*s*us*exp(((1j*k)/(2*L))*(-2*r*s*cos(phir-phis)+(s^2)));
g1 = @(s,phis,r,phir) int(@(phis) f(s,phis,r,phir),0,2*pi)

g1 = function_handle with value:
    @(s,phis,r,phir)int(@(phis)f(s,phis,r,phir),0,2*pi)

g2 = @(s,phis,r,phir) int(@(s) g1(s,r,phir),0,Inf)

g2 = function_handle with value:
    @(s,phis,r,phir)int(@(s)g1(s,r,phir),0,Inf)

Still my problem is not solved. How to get the expression?

[Walter Roberson]

syms s phis r phir
Pi = sym(pi);
Q = @(v) sym(v);
A = Q(1); m = Q(1); n = Q(1); L = Q(100); alpha = Q(1);
k = (2*Pi)/(Q(550)*Q(10)^-9);
us = A*((-1j*sqrt(2*k*alpha)*s*exp(-1j*phis))^m)*exp(-k*alpha*s*s)*laguerreL(n,m,2*k*alpha*s*s);
C = ((-1j*k*exp(1j*k*L))/(2*Pi*L))*exp((1j*k*r*r)/(2*L));
f(s,phis,r,phir) = C*s*us*exp(((1j*k)/(2*L))*(-2*r*s*cos(phir-phis)+(s^2)));
g1(s,r,phir) = int(f(s,phis,r,phir), phis, 0,2*Pi, 'hold', true)
g2(r,phir) = int(g1(s,r,phir), s, 0, Inf, 'hold', true)

%decent speed up to this point. But the below is somewhat slow.
%and all it ends up doing is stripping the 'hold' status from the int()

G2 = release(g2)

If you are planning to go to a numeric function handle using integral(), then you can matlabFunction(g2) without having to release() . At least in the current version you can; my recollection is that in the previous release you could not do that.

The function generated by matlabFunction() looks like

@(r,phir)integral(@(s)integral(@(phis)sqrt(1.1e+1).*8.944271909999159e+3.*s.^2.*sqrt(pi).*exp(pi.*6.363636363636364e-1i).*exp(phis.*-1i).*exp(pi.*(s.^2-r.*s.*cos(phir-phis).*2.0).*1.818181818181818e+4i).*exp(r.^2.*pi.*1.818181818181818e+4i).*exp(s.^2.*pi.*(-3.636363636363636e+6)).*(s.^2.*pi.*7.272727272727273e+6-2.0).*(-1.652892561983471e+3),0.0,pi.*2.0),0.0,Inf)