# Evaluation of integral2. Error

조회 수: 5(최근 30일)
댓글: DIMITRIS GEORGIADIS 2021년 2월 23일
Using the following code I get a matrix dimension error. I suspect something goes wrong when I integrate out z from func_3. Could anyone help on that?
clear all; clc; close all;
% Define functions:
func_1 = @(x) (1/(0.1*sqrt(2*pi))).*exp(-0.5.*((x - 1)./0.1).^2);
func_2 = @(y) (1/(0.01*sqrt(2*pi))).*exp(-0.5.*((y - 0.1)./0.01).^2);
func_12 = @(x, y) func_1(x).*func_2(y);
% Visualization:
fcontour(func_12,[0.8 1.2 0.05 0.15])
% Define new function:
c = 1.05;
sigma_e = 0.05;
func_3 = @(x, y, z) (1./y).*exp(-0.5.*((z - x)./y).^2).*...
exp(-0.5.*((c - z)./sigma_e).^2);
% Intergate out z:
l_bound = -Inf;
u_bound = Inf;
func_4 = @(x, y) integral(@(z) func_3(x, y, z), l_bound, u_bound);
% Get the new function:
func_5 = @(x, y) func_4(x, y).*func_12(x, y);
% Compute the integral:
q = integral2(func_5, -Inf, Inf, 0, Inf);
w = 1/q;

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### 채택된 답변

Shashank Gupta 2021년 2월 23일
Hi Dimitris,
It does look like problem is in integration of func_3 and looking at the error I feel like some kind of matrix or vector is involved in the calculation. My first attempt to solve this issue is to enable "ArrayValued" flag in integral function. This will make sure the whenever the matrix or vector calculation involved will smoothly run. I am attaching the changes below. It should work.
func_4 = @(x, y) integral(@(z) func_3(x, y, z), l_bound, u_bound,"ArrayValued",1);
I hope this resolves the issue.
Cheers
##### 댓글 수: 1표시숨기기 없음
That works perfect indeed! Thank you very much.

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