How to do Double numerical integration with a variable limits

Can someone please help me in doing numerical integration on the following function, I want to do double numerical integration with respect to y and x, such that the final answer will be a function of z only. Also the limit of integration is not a constant, it is basically something like (x-y, x+y), not sure if there exist a way to do this, can someone please advise me on how to proceed?

Thank you

[Torsten]

Please write out precisely the integral with limits of integration you want to compute.

[Abdulaziz Al-Amodi]

Thank you for responding;

I'll split the integral into two integrations such that

the integration limits of the first one is as follows:

dx > z-y to z+y ; dy > 0 to z

and the integration limit of the second one is:

dx > y-z to z+y ; dy > z to infinity

The final answer should be the summation of both integration.

Hope that is clear,

Thank you

[John D'Errico]

Are lamdas and lamdab known? If they are not, you can do nothing, since this would not be a numerical integration.

Regardless, since z lies internal to the kernel, you cannot solve for the integral in any form, unless you also know the value of z.

A numerical integration requires that all variables you will not be integrating out are known constants.

[Abdulaziz Al-Amodi]

Yes lamdab and lambdas are known, both are constant.

As for z, I'll integrate it as well after finishing the double integration, or should I make it a triple integration to include z?

Thank you,

[John D'Errico]

Again, you CANNOT do a numerical integration where z (or ANY variable) is an unknown parameter to the problem. If the end goal is to integrate over z also, then you do indeed need to perform a triple integration.

[Walter Roberson]

numeric integration can never be done on expressions that involve unresolved variables.

Consider for example that a numeric integration routine would not know whether z has a very small absolute value (such as 1e-200) that makes no difference to the integration... or if z will have a large enough absolute value that it overwhelms the x y contributions making the overall term very small.

[Abdulaziz Al-Amodi]

Thank you for clarifying, in case I wish to perform triple numerical integration is this correct?

fun = @(x,y,z) 4.*pi.*lambda1.*lambda2.*z.*exp(-pi.*(lambda2.*x.^2+lambda1.*y.^2))./(sqrt(1-((x.^2+y.^2-z.^2)/2.*x.*y).^2));
q = integral3(fun,d-y,d+y,0,d,-inf,inf);

[Torsten]

Your sqrt expression becomes negative if you integrate from -Inf to +Inf in the z-direction.

Is it a special geometrical object you try to integrate over ?

And

fun = @(x,y,z) 4.*pi.*lambda1.*lambda2.*z.*exp(-pi.*(lambda2.*x.^2+lambda1.*y.^2))./sqrt(1-((x.^2+y.^2-z.^2)/(2.*x.*y)).^2);

instead of

fun = @(x,y,z) 4.*pi.*lambda1.*lambda2.*z.*exp(-pi.*(lambda2.*x.^2+lambda1.*y.^2))./(sqrt(1-((x.^2+y.^2-z.^2)/2.*x.*y).^2));

[Torsten]

@John D'Errico @Walter Roberson

The integral (once it is correctly written) could be considered as a function of z. Then numerical integration can be carried out.

[Abdulaziz Al-Amodi]

z is always positive, I made a mistake when I wrote -inf, was just trying to understand the implementation of it.

I found 'vpaintegral' and it looks promising but it is very slow to compute.

Thank you for your help

Answer 1

[Matt J]

I'll demonstrate for a simpler function. The technique is the same regardless of what function we're integrating.

F=@(x,y,z) x.^2+y.^2+z.^2; %The input function to be integrated


Iyz=@(y,z)integral(   vect(@(q)F(q,y,z)),   z-y,z+y); %partial integral w.r.t. x

Iz=@(z) integral(  vect(@(q)Iyz(q,z))  ,   0,z);       %partial integral w.r.t. y


Iz(1)

ans = 2.6667

function fun=vect(fun)
%vectorize a non-vectorized function

   fun=@(x) arrayfun(fun,x);

end

[Matt J]

If F() is vectorized with respect to x, as it is in this case, then we don't actually need vect() for the innermost integral.

F=@(x,y,z) x.^2+y.^2+z.^2;


Iyz=@(y,z)integral(   @(q)F(q,y,z),   z-y,z+y);

Iz=@(z) integral(  vect(@(q)Iyz(q,z))  ,   0,z);

Iz(1)

ans = 2.6667

[Abdulaziz Al-Amodi]

Thank you for your help, I'll try it out

[Matt J]

You're welcome, but if you find it works for you, please Accept-click the answer.

[Abdulaziz Al-Amodi]

Thank you for your help,

I still couldn't do it with my function, I think it might require the use of Mathmatica.

[Torsten]

Yes, the square root in the denominator is quite ambitious ... When it gives real values and when it is different from 0 ... To take care of it in the integration limits for x,y and z won't be easy - be it with or without Mathematica.

[Abdulaziz Al-Amodi]

Right, I'll hopefully look more into it to figure it out.

Thanks again!