Function within an integral

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Hi,
How can I solve an Integral which looks like:
f(z) = e^A(z)*ln(z), where as A(z) = -z^2
I tried with this code:
fun = @(x) exp(A).*log(x).^2;
A = test_test(x);
q = integral(fun,0,Inf);
function A = test_test(x)
A= -x^2;
end
what am I doing wrong???? PLEASE HELP!!!!!
thanks a lot!!!

采纳的回答

Star Strider
Star Strider 2016-3-9
This works for me:
test_test = @(x) -x.^2;
fun = @(x) exp(test_test(x)).*log(x).^2;
q = integral(fun,0,Inf)
q =
1.9475
You have to call your function as ‘test_test(x)’ inside ‘fun’, not as ‘A’. (I created an anonymous function for of ‘test_test(x)’ for convenience. If you are using it as a function file, it would work the same way in ‘fun’.)
  2 个评论
beginner
beginner 2016-3-9
Thank you soooo much!!!! you are a genius!!!! <3
Star Strider
Star Strider 2016-3-9
My pleasure!
Thank you!
If my Answer solved your problem, please Accept it!

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