Please help me. I want to integrate the following function from 0 to plus infinity.

syms r r0 sg g(r)=r*exp(-(log(r)-log(r0))^2/(2*sg^2));

[John D'Errico]

I would assume you REALLY want to integrate from r0 to infinity.

[El Houssain Sabour]

I'm supposed to integrate from 0 to infinity. I'm almost desperate. I have to calculate that.

[Torsten]

It's the lognormal distribution.

Look here which functions MATLAB supplies:

https://de.mathworks.com/help/stats/lognormal-distribution.html

[David Goodmanson]

Hi ES,

As Torsten has pointed out, you have a log-normal distribution, and that's all you need to start searching around. In your 18.55.57 jpeg expression for zeta, the integral is the log-normal's second moment about the origin. If you go to Wikipedia for example, you can find an algebraic expression for the nth moment..

[El Houssain Sabour]

It's also about the integral, which is in the equation. And that's my problem. I can't implement this with Matlab.

[Torsten]

As you can see under

https://www.statlect.com/probability-distributions/log-normal-distribution

integral_{r=0}^{r=Inf} r^n * 1/(r*sqrt(2*pi*sigma^2)) * exp(-1/2 * (log(r)-mu)^2 / sigma^2) dr

=

exp( n*mu + 1/2 * n^2*sigma^2)

Both of your two integrals in question follow from this relation for n=2 and n=3.

So your integral becomes

integral_{r=0}^{r=Inf} r^(n-1) * exp(-1/2 * (log(r)-log(r0))^2 / sg^2) dr =

sqrt(2*pi*sg^2) * exp( n*log(r0) + 1/2 * n^2*sg^2)

for n = 2: sqrt(2*pi*sg^2) * r0^2* exp( 2*sg^2)

for n = 3: sqrt(2*pi*sg^2) * r0^3* exp( 4.5*sg^2)

Answer 1

[Dr. JANAK TRIVEDI]

syms r r0 sg
g(r) = r * exp(-(log(r) - log(r0))^2 / (2 * sg^2));
int_g = int(g,r,0,inf)

int_g = 

Note that the inf keyword represents infinity. The integral of g(r) is calculated over the interval [0,inf). You can also specify other intervals of integration as needed.

[Torsten]

I don't see that the integral can be calculated by using "int".