Fitting parameter inside an integral using python (or another useful language)

Question

I have a set of data, basically with the information of f(x) as a function of x, and x itself. I know from the theory of the problem that I'm working on the format of f(x), which is given as the expression below:

Essentially, I want to use this set of data to find the parameters a and b. My problem is: How can I do that? What library should I use? I would like an answer using Python. But R or Julia would be ok as well.

From everything I had done so far, I've read about a functionallity called curve fit from the SciPy library but I'm having some trouble in which form I would do the code as long my x variable is located in one of the integration limit.

For better ways of working with the problem, I also have the following resources:

A sample set, for which I know the parameters I'm looking for. To this set I know that a = 2 and b = 1 (and c = 3). And before it rises some questions about how I know these parameters: I know they because I created this sample set using this parameters from the integration of the equation above just to use the sample to investigate how can I find them and have a reference.

I also have this set, for which the only information I have is that c = 4 and want to find a and b.

I would also like to point out that:

i) right now I have no code to post here because I don't have a clue how to write something to solve my problem. But I would be happy to edit and update the question after reading any answer or help that you guys could provide me.

ii) I'm looking first for a solution where I don't know a and b. But in case that it is too hard I would be happy to see some solution where I suppose that one either a or b is known.

EDIT 1: I would like to reference this question to anyone interested in this problem as it's a parallel but also important discussion to the problem faced here

Answer 1

They are three variables a,b,c which are not independent. One of them must be given if we want compute the two others thanks to regression. With given c, solving for a,b is simple :

The example of numerical calculus below is made with a small data (n=10) in order to make it easy to check.

Note that the regression is for the function t(y) wich is not exactly the same as for y(x) when the data is scattered (The result is the same if no scatter).

If it is absolutely necessary to have the regression for y(x) a non-linear regression is necessary. This involves an iterative process starting from good enought initial guess for a,b. The above calculus gives very good initial values.

IN ADDITION :

Meanwhile Andrea posted a pertinent answer. Of course the fitting with his method is better because this is a non-linear regression instead of linear as already pointed out in the above note.

Nevertheless, dispite the different values (a=1.881 ; b=1.617) compared to (a=2.346 , b=-0.361) the respective curves drawn below are not far one from the other :

Blue curve : from linear regression (above method)

Green curve : from non-linear regression ( Andrea's )

CASE OF THE SECOND SET OF DATA

https://mega.nz/#!echEjQyK!tUEx0gpFND7gucvsTONiB_wn-ewBq-5k-pZlfLxmfvw

The regression fails because the assumption c=3 is false.

In the case c=0 the analytic calculus of the integral is different from above :

Answer 2

I would use a pure numeric approach, which you can use even when you can not directly solve the integral. Here's a snipper for fitting only the a parameter:

import numpy as np
from scipy.optimize import curve_fit
import pandas as pd
import matplotlib.pyplot as plt

def integrand(x, a):
    b = 1
    c = 3
    return 1/(a*np.sqrt(b*(1+x)**3 + c*(1+x)**4))

def integral(x, a):
    dx = 0.001
    xx = np.arange(0, x, dx)
    arr = integrand(xx, a)
    return np.trapz(arr, dx=dx, axis=-1)

vec_integral = np.vectorize(integral)

df = pd.read_csv('data-with-known-coef-a2-b1-c3.csv')
x = df.domin.values
y = df.resultados2.values
out_mean, out_var = curve_fit(vec_integral, x, y, p0=[2])

plt.plot(x, y)
plt.plot(x, vec_integral(x, out_mean[0]))
plt.title(f'a = {out_mean[0]:.3f} +- {np.sqrt(out_var[0][0]):.3f}')
plt.show()

vec_integral = np.vectorize(integral)

Of course, you can lower the value of dx to get the desired precision. While for fitting just the a, when you try to fir b as well, the fit does not converge properly (in my opinion because a and b are strongly correlated). Here's what you get:

def integrand(x, a, b):
    c = 3
    return 1/(a*np.sqrt(np.abs(b*(1+x)**3 + c*(1+x)**4)))

def integral(x, a, b):
    dx = 0.001
    xx = np.arange(0, x, dx)
    arr = integrand(xx, a, b)
    return np.trapz(arr, dx=dx, axis=-1)

vec_integral = np.vectorize(integral)

out_mean, out_var = sp.optimize.curve_fit(vec_integral, x, y, p0=[2,3])
plt.title(f'a = {out_mean[0]:.3f} +- {np.sqrt(out_var[0][0]):.3f}\nb = {out_mean[1]:.3f} +- {np.sqrt(out_var[1][1]):.3f}')

plt.plot(x, y, alpha=0.4)
plt.plot(x, vec_integral(x, out_mean[0], out_mean[1]), color='green', label='fitted solution')
plt.plot(x, vec_integral(x, 2, 1),'--', color='red', label='theoretical solution')
plt.legend()
plt.show()

As you can see, even if the resulting a and b parameters form the fit are "not good", the plot is very similar.