Triple integral with absolute value limits

I would like to solve the following integral numerically:

I wrote this code, but the result is not coming. Can anyone tell me what my mistake is?

lambda1=25/(1000)^2;
lambda1=100/(1000)^2

lambda1 = 1.0000e-04

fun =integral3(@(l0,l2,l1) (l1.^-3).*exp(-pi*(lambda2.*l2.^(2)+lambda1.*l0.^(2)))./(sqrt(1-((l0.^(2)+l2.^(2)-l1.^(2))/(2*l0.*l2)).^2)),0,Inf,0,@(l0,l2) abs(l0-l2),@(l0,l2) (l0+l2))

Error using integral3
Invalid argument list. Function requires 1 more input(s).

Answer 1

[Walter Roberson]

lambda1=25/(1000)^2;
lambda2=100/(1000)^2

lambda2 = 1.0000e-04

fun =integral3(@(l0,l2,l1) (l1.^-3).*exp(-pi*(lambda2.*l2.^(2)+lambda1.*l0.^(2)))./(sqrt(1-((l0.^(2)+l2.^(2)-l1.^(2))/(2*l0.*l2)).^2)),0,Inf,0,Inf,@(l0,l2) abs(l0-l2),@(l0,l2) (l0+l2))

Warning: Inf or NaN value encountered.

Warning: The integration was unsuccessful.

fun = NaN

[Walter Roberson]

With l0 and l2 both ranging from 0 to +inf, then when they are both 0, 4*l0^2*l2^2 would be 0 and you would have a division by 0. The numerator in that situation would be (-l1^2)^2 which would be l1^4

Is it possible that in the limit case, the overall numerator compensates? Well, in this situation l0^2*lambda1 + l2^2*lambda2 would be 0, so exp(-pi*0) which would be 1, so the overall numerator would be l1^-3. What is l1 ? Well, l1 ranges from abs(l0-l2) to l0+l2 and both of those are 0 so l1 is 0

Stepping back a second... that means that in the other part, the l1^4 numerator is going to 0^4 and the denominator is going to plain 0, so in the limit I guess that would be an overall 0, leading to sqrt(1-0) and so in the limit case the denominator might be defined after all... in the limit case.

But then when we apply that logic to the overall numerator and see that we dealing with 0^(-3) that is a "faster" division by 0 than problems in the denominator, so we get problems all over again.

But l0 == l2 an infinite number of times with those limits, so l1 would have a lower bound of 0 an infinite number of times, and that gives you the problems noted above even when l0 and l2 are not 0.

[mohammed saleh]

Thank you for your response.

In fact, the value of l2 should not be greater than 10, so l2 < l0.

Since we are dealing with an area, the terms of integration must be positive |l0-l2| .

Values of l0 can be any value from 0 to inf.

Here I am trying to calculate pdf of l1 Where the value of l1 depends on the values of l0&l2

Should the code be in this way with changing limits depending on the above data?

fun =integral3(@(l0,l2,l1) (l1.^-3).*exp(-pi*(lambda2.*l2.^(2)+lambda1.*l0.^(2)))./(sqrt(1-((l0.^(2)+l2.^(2)-l1.^(2))/(2*l0.*l2)).^2)),0,Inf,0,@(l0) l0,@(l0,l2) (l0-l2),@(l0,l2) (l0+l2))

[Walter Roberson]

@(l0) l0 would not prevent l2 from being exactly equal to l0 . You would need something like @(l0) lo*(1-eps)