integral3 requires the function handle you pass in as the first input to accept three input arrays and return an output array of the same size. Your f function assumes that x is a vector with 3 elements which is incompatible with that requirement. One way to resolve this incompatibility is to use arrayfun. But that uncovers a different problem:
t = 15;
f = @(x) ( x(1)*(t - x(2))^x(3) ).*(x(2) <= t);
newF = @(x, y, z) arrayfun(@(x, y, z) f([x, y, z]), x, y, z);
f_int = integral3(newF, 0, Inf, 0, Inf, 0, Inf)
Rather than including (x(2) <= t) in the integrand, why not change the limits of integration? This doesn't avoid the NaN issue but does make the integrand simpler to debug.
t = 15;
g = @(x) ( x(1)*(t - x(2))^x(3) );
newG = @(x, y, z) arrayfun(@(x, y, z) g([x, y, z]), x, y, z);
f_int = integral3(newG, 0, Inf, 0, t, 0, Inf)
At some point your integrand boils down to 0*Inf I believe. You will need to determine how to handle that.
