function of a variable obtained from integrating a two variable function

2014 9 16
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更新时间：2014 9 16
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I have a function that depends on two variables, say x and y; f(x,y). I want to integrate with respect to x only and get a new function that depends only on y; i.e., g(y) = @(y) integral(@(y)f(x,y),lim1,lim2), where lim1 and lim2 are the limits for for x. I want to use g(y) later to carry out other operations. I cannot find a way around this problem.

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Matt J
Matt J 2014-9-16
编辑：Matt J 2014-9-16

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I think you've answered your own question. Why can't you create an anonymous function for g() just as you did in your post
g = @(y) integral(@(x)f(x,y),lim1,lim2)
and carry that around for reuse?

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Mike Hosea
Mike Hosea 2014-9-16
You can make it more general (work with array inputs) like so
g = @(y1)arrayfun(@(y)integral(@(x)f(x,y),lim1,lim2),y1);
For example:
f = @(x,y)exp(-hypot(x,y));
lim1 = -inf;
lim2 = inf;
g = @(y1)arrayfun(@(y)integral(@(x)f(x,y),lim1,lim2),y1);
y = linspace(-10,10);
plot(y,g(y));

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Mauricio
Mauricio 2014-9-16
编辑：Matt J 2014-9-16

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Well you are right but I did not expressed myself correctly. The problem is the following: In the code below I have the function "Base(z)" which depends on "z", which I need to integrate with respect to z. This function "Base" depends in turn on another function called RR which I have to integrate with respect to one variable a to obtain "fun", the output of "Base(z)". Unfortunately, I keep on getting the error message you see below. Note: I can evaluate "Base(z)" for any z, but I cannot integrate "Base(z)" with respect to z. Thanks for your help
function out = Princip
clc;
function fun=Base(z)
RR = @(a,b)(z.*3.^a)./(b+1);
fun = integral(@(a)RR(a,5),0,10);
end
out = integral(@Base,1,10);
end
Error Message:
Error using integralCalc/finalInputChecks (line 515)
Output of the function must be the same size as the input. If FUN is an array-valued integrand, set the 'ArrayValued' option to
true.
Error in integralCalc/iterateScalarValued (line 315)
finalInputChecks(x,fx);
Error in integralCalc/vadapt (line 132)
[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);
Error in integralCalc (line 75)
[q,errbnd] = vadapt(@AtoBInvTransform,interval);
Error in integral (line 88)
Q = integralCalc(fun,a,b,opstruct);
Error in Princip (line 18)
out = integral(@Base,1,10);

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Matt J
Matt J 2014-9-16
Set the 'ArrayValued' option to true, as the error message instructs (in all your calls to integral()).
Mike Hosea
Mike Hosea 2014-9-16
Or make the Base function satisfy the requirements of INTEGRAL without the ArrayValued flag, i.e. to accept an array input and return an array of the same size.
function fun=Base(z)
fun = zeros(size(z));
for k = 1:numel(fun)
RR = @(a,b)(z(k).*3.^a)./(b+1);
fun(k) = integral(@(a)RR(a,5),0,10);
end
end

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