Solve Integral Problem: L.H.S w/ a,b,x0

[princepolo]

Can you you solve this Integral problem

Dear Friends
Can youn help me to integrate the follownig formula
L.H.S = Integrate[1-6*Lambda/x*Coth(x/2*Lambda)+12*(Lambda^2/x^2))*x^3*g(x)dx] / Integrate[x^3*g(x)dx]
Where g(x) = a*exp[-0.5/b^2*ln(x/x0)^2] is the distribution function
where a= 51.5801
b=0.9585
x0=5.4073
I like to solve for Lambda, where the L.H.S is determined experimentally
Thank you very much for your cooperation.

 

[saltydog]

Dear Friends
Can youn help me to integrate the follownig formula
L.H.S = Integrate[1-6*Lambda/x*Coth(x/2*Lambda)+12*(Lambda^2/x^2))*x^3*g(x)dx] / Integrate[x^3*g(x)dx]
Where g(x) = a*exp[-0.5/b^2*ln(x/x0)^2] is the distribution function
where a= 51.5801
b=0.9585
x0=5.4073
I like to solve for Lambda, where the L.H.S is determined experimentally
Thank you very much for your cooperation.

Hey Prince. That's hard to follow. Welcome to PF. We use LaTex in here and you may wish to learn to use it if you post frequently. Check out the thread in the Physics Forum about using LaTex. This is what I think it is:

$$\frac{\int_u^v \left(1-\frac{6\lambda}{x}Coth(\frac{\lambda x}{2})+12(\frac{\lambda^2}{x^2})x^3 g(x)\right)dx}{\int_u^v x^3 g(x)dx}$$

With u and v as the limits of integration

Maybe that's close. Correct it if necessary. Really, I'd just program it into Mathematica as a function and solve it numerically:

f($\lambda$)=NIntegrate[g(x,$\lambda$),{x,u,v}]

Plot it for starters and see where the LHS meets your value. Then can use NSolve:

NSolve[f($\lambda$)==my value]

(or whatever else it takes in Mathematica to get a numerical answer)

 

[quicknote]

Dear Friend,

Thank you for reaching out to us with your question. I am always happy to help with solving problems.

To solve this integral problem, we will need to use some techniques from calculus. First, we can simplify the integrand by expanding the Coth function using its definition: Coth(x) = (e^x + e^(-x)) / (e^x - e^(-x)). This will give us a polynomial expression in terms of x and Lambda.

Next, we can use integration by parts to integrate the first term in the numerator, and then use substitution to evaluate the remaining integrals. This will give us a final expression in terms of Lambda.

To solve for Lambda, we can set the expression equal to the experimental value of the L.H.S and use numerical methods, such as Newton's method, to find the root of the equation.

I hope this helps with your problem. Let me know if you need any further assistance. Best of luck with your calculations!