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integration of l^c * exp(-l)

3 Ansichten (letzte 30 Tage)
Jonas Colmsjö
Jonas Colmsjö am 13 Okt. 2021
Why do the following code snippets give different results. The only difference is exp(-l*(m+1)) vs. exp(-l*3), l) where m=2. Any help is much appreciated!
>> syms l ct m cf
ncons_ = int(l^(c_1+c_2) * exp(-l*3), l)
ncons = eval(- subs(ncons_, l, 0))
eval(subs(ncons, [m, ct, cf, c_1, c_2], [2, sum([3,4]), prod(1./factorial([3,4])), 3, 4]))
ncons_ =
-((1/3)^(c_1 + c_2)*igamma(c_1 + c_2 + 1, 3*l))/3
ncons =
((1/3)^(c_1 + c_2)*gamma(c_1 + c_2 + 1))/3
ans =
0.7682
>> syms l ct m cf
ncons_ = int(l^(c_1+c_2) * exp(-l*(m+1)), l)
ncons = eval(- subs(ncons_, l, 0))
eval(subs(ncons, [m, ct, cf, c_1, c_2], [2, sum([3,4]), prod(1./factorial([3,4])), 3, 4]))
ncons_ =
-(l^(c_1 + c_2)*igamma(c_1 + c_2 + 1, l*(m + 1)))/((l*(m + 1))^(c_1 + c_2)*(m + 1))
ncons =
gamma(c_1 + c_2 + 1)/(m + 1)
ans =
1680
>>
``

Antworten (1)

Abhinav Aravindan
Abhinav Aravindan am 16 Feb. 2024
While computing the integral with symbolic variables using the “int” function, “int” by default, uses strict mathematical rules. For instance, these rules do not allow to rewrite “acos(cos(x))” as “x”. A potential solution is to use “int” with the argument “IgnoreAnalyticConstraints” set to “true”.
Note: This option can provide simpler results for expressions but can lead to results that do not always hold for all values of variables.
More information can be found in the documentation link below:
Assuming “c_1” and “c_2” are also symbolic variables, please find a code snippet for your reference below.
syms l ct m cf c_1 c_2
% Method 1
ncons_ = int(l^(c_1+c_2) * exp(-l*3), l)
ncons = eval(- subs(ncons_, l, 0))
method1_result = eval(subs(ncons, [m, ct, cf, c_1, c_2], [2, sum([3,4]), ...
prod(1./factorial([3,4])), 3, 4]))
% Method 2
ncons_ = int(l^(c_1+c_2) * exp(-l*(m+1)), l, "IgnoreAnalyticConstraints", true)
ncons = eval(- subs(ncons_, l, 0))
method2_result = eval(subs(ncons, [m, ct, cf, c_1, c_2], [2, sum([3,4]), ...
prod(1./factorial([3,4])), 3, 4]))

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R2021b

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