Symbolic Integration: Explicit integral could not be found

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

[David]

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?

Answer 1

[Andrei Bobrov]

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)

[geeks]

Thanks! But mu and sigma are parameters that I am going to estimate later on, so I can't really assign specific values to them. Is there any other way that you think might work?

[geeks]

Thanks a lot!! It works perfect now.

[Yingying]

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]

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.

Answer 2

[Walter Roberson]

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()