Symbolic Integration: Explicit integral could not be found

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geeks
geeks 2012-4-1
Hi all,
I got a problem when trying to run a symbolic integration. Since lognormal distribution is not defined in the symbolic tools, I just write it out in an explicit way. The code is as follows:
syms w mu sigma positive
t = (w-6.5)*(w*sigma)^(-1)*(1/sqrt(2*pi)*exp(-(log(w)-mu)^2/sigma^2/2));
c = int(t,w,6.5,inf);
And the error msg is:
Warning: Explicit integral could not be found.
Can anybody please give some hint how to get the integration?
Thanks very much!!
Sonia
  1 个评论
David
David 2012-10-22
Sonia, I had a similar problem and your post really help me, but I do not understand your formula:
(w-6.5).*(w*sigma).^(-1).*(1/sqrt(2*pi)*exp(-(log(w)-mu).^2/sigma^2/2));
could it be: (w-6.5).*(sigma).^(-1).*(1/sqrt(2*pi)*exp(-(log(w)-mu).^2/sigma^2/2));
Is this ok?

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采纳的回答

Andrei Bobrov
Andrei Bobrov 2012-4-1
try numeric solution
eg
sigma = .2;
mu = 2;
t = @(w)(w-6.5).*(w*sigma).^(-1).*(1/sqrt(2*pi)*exp(-(log(w)-mu).^2/sigma^2/2));
c = quadgk(t,6.5,inf)
ADD on Geeks comment
use function c
t = @(w,mu,sigma)(w-6.5).*(w*sigma).^(-1).*(1/sqrt(2*pi)*exp(-(log(w)-mu).^2/sigma^2/2));
c = @(mu,sigma)quadgk(@(w)t(w,mu,sigma),6.5,inf)
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Yingying
Yingying 2012-4-23
Can you explain the function C in more detail? I am having the similar problem now, The code is as follows:
>> syms a b y
>> a1=int('1/(a*y^3+b)',y,0,5)
And the msg is:
Warning: Explicit integral could not be found.
a1 =
piecewise([a <> 0 and b = 1 and abs(arg(a)) < pi, 5*hypergeom([1/3, 1], [4/3], (-125)*a)], [Otherwise, int(1/(a*y^3 + b), y = 0..5)])
Walter Roberson
Walter Roberson 2012-4-23
Yingying, for you the easiest solution might be to add assumptions. For example if you know that a > 0 then
syms a positive
syms b y
a1 = int(1/(a*y^3+b), y, 0, 5);
Notice here that I did not quote the expression to be integrated; otherwise the "positive" assumption on "a" would not have any effect.

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更多回答（1 个）

Walter Roberson
Walter Roberson 2012-4-1
limit(-(1/2)*exp(mu+(1/2)*sigma^2)*erf((1/2)*2^(1/2)*(sigma^2-U+mu)/sigma)+(13/4)*erf((1/2)*2^(1/2)*(-U+mu)/sigma)+(1/2)*exp(mu+(1/2)*sigma^2)*erf((1/2)*2^(1/2)*(-ln(13/2)+mu+sigma^2)/sigma)+(13/4)*erf((1/2)*2^(1/2)*(ln(13/2)-mu)/sigma), U = infinity)
Note, in this expression, U is an introduced variable for the purpose of the limit()

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