Triple integral with absolute value limits

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MOHAMMED MEHDI SALEH 2022-12-4

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1870972-triple-integral-with-absolute-value-limits

评论： MOHAMMED MEHDI SALEH 2023-9-9

采纳的回答： Walter Roberson

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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).

采纳的回答

Walter Roberson 2022-12-4

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1870972-triple-integral-with-absolute-value-limits#answer_1119927

在 MATLAB Online 中打开

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

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Walter Roberson 2022-12-4

编辑：Walter Roberson 2022-12-4

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 MEHDI SALEH 2023-9-9

I want l1<(l0+l2) The problem is that it is a triple integral, and as the exponent of l1 increases, the integration becomes slow, for example, l1^3 or l1^4. I will explain to you what I want to do.

The distribution BS* and BS+ in 2D space are given as follows: The BSs* are randomly distributed in a 2D area. The user is located in the center of the area and will connect with the nearest BS*. The pdf of l0 is f(l0) = 2 * (λBS*) * π * l0 * exp(-(λBS*) * π * l0^2), where l0 is the distance between the user and the nearest BS*. The BSs+ are randomly distributed in the same area. The user will connect with the nearest BSs+. The pdf of l2 is f(l2) = 2 * (λBSs+) * π * l2 * exp(-(λBSs+) * π * l2^2), where l2 is the distance between the user and the nearest (BS+). I want to find the pdf of the distance l1 between the nearest BS* and nearest BS+ to the user, where l1=sqrt(l0^2 + l2^2-l0*l2* cos(φ)) as shown in figure below The possible positioning of the nearest BS+ within the captured annular of the inner circle with a radius of l2, centred at the origin.

l1<l0+l2

l0 0 to infinity

l2 0 to infinity

How can I simplify the integration, or is there a simpler derivation than that, as I need it to find the expectation of l1^-3 and l1^-4? so I need pdf of l1

thank you so much

MOHAMMED MEHDI SALEH 2023-9-9

编辑：MOHAMMED MEHDI SALEH 2023-9-9

SORRY fun=fun2

its mistake

fun =@(l1,l2,l0) ((l1.^(1-4)).*(exp(-pi*(lambda_2*l2.^2+lambda_1*l0.^2))./(sqrt(1-((l0.^2+l2.^2-l1.^2)./(2*l0.*l2)).^2))));

const=4*pi*lambda_2*lambda_1;

z=const*integral3(fun,0,inf,0,inf,@(l0,l2) 1+eps*(l0-l2),@(l0,l2) 1-eps*(l0+l2));

I got it, but there are big differences with the simulation result.

if iw ant to change like this

z=@(l2) const*integral2(@(l0,l1) fun,0,inf,@(l0,l2) 1+eps*(l0-l2),@(l0,l2) 1-eps*(l0+l2)); its possible ?

i want z as functions of l2.

MOHAMMED MEHDI SALEH 2023-9-9

i did like that but get error Not enough input arguments.

fun =@(l0,l2,l1) ((l1.^(1-4)).*(exp(-pi*(lambda_2*l2.^2+lambda_1*l0.^2))./(sqrt(1-((l0.^2+l2.^2-l1.^2)./(2*l0.*l2)).^2))));

z = @(l2) integral2(@(l0, l1) fun(l0,l2,l1), 0,inf, @(l0,l2) 1+eps*(l0-l2),@(l0,l2) 1-eps*(l0+l2));

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