How to solve an integral with symbolic values as borders in another integral?

In my problem I have to evaluate the integral of a function with respect to x. However within that function is an integral with respect to variable h with borders from 0 to 56-x. To model this I used the following code.

if true
  % code
end
      scale = 0.00095
      shape = 8.907
      lambda = 0.00174
      syms x h
      func1 = @(x) (wblpdf(x,1/scale,shape))/(1-wblcdf(0,1/scale,shape));
      func2 = @(x) expcdf(56-x,1/lambda);
      f3 = int((1-wblcdf(56-x-h,1/scale,shape)),h,0,56-x);
      func3 = matlabFunction(f3);
      func4 = @(x) func1(x).*func2(x).*func3(x);
      forecast = forecast + integral(func4,0,28);
end

If I try this I get the error message: Error using symengine. Unable to prove '56 - x - h < 0' literally. Use 'isAlways' to test the statement mathematically. I know that this has something to do with the variable of the weibullcdf having to be bigger than 0. How do I use isAlways to solve my error?

Answer 1

[Torsten]

Use integral2:

scale = 0.00095;
shape = 8.907;
lambda = 0.00174;
func1 = @(x) (wblpdf(k_1t-tr+x,1/scale,shape))/(1-wblcdf(k_1t-tr,1/scale,shape));
func2 = @(x) expcdf(56-x,1/lambda);
func3 = @(x,y) 1-wblcdf(56-x-y,1/scale,shape);

fun = @(x,y)func1(x).*func2(x).*func3(x,y);
ymax = @(x) 56 - x;
q = integral2(fun,0,28,0,ymax)

Best wishes

Torsten.

[Jeroen Vlak]

Thanks a lot!!!!!