Calculate the improper integral Monte Carlo method

Question

For exaple, I'm trying to transform the integral:

I need to transform it to an integral that goes from 0 to 1 (or from a to b) in order to apply the algorithm of Monte Carlo I implemented. I already have Monte Carlo function, which calculate definite integrals, but I need to transform this improper integral to definite integral and I really don't know how to do it.

Ideally, I want to transform this integral (from -inf to inf):

Also I tried to do transformation, which I found on Inthernet (integration by substitution: x=-ln(y), dx=-1/y), but it doesn't work:

So how can I get transformation of this integral?

Answer 1

I need to transform it to an integral that goes from 0 to 1 (or from a to b) in order to apply the algorithm of Monte Carlo I implemented.

No, you don't.

if you have integral

∫₀^∞ w(x) g(x) dx

you could sample from w(x) and compute mean value of g(x) at sampled points.

You integral

∫₀^∞ e^-x cos(x) dx

is pretty perfect for such approach - you sample from e^-x and compute E[cos(x)]

Along the lines (Python 3.9, Win10 x64)

import numpy as np

rng = np.random.default_rng()

N = 1000000
U = rng.random(N)

W = -np.log(1.0 - U) # sampling exp(-x)

G = np.cos(W)

ans = np.mean(G)
print(ans)

will print something like

0.5002769491719996

And concerning your second integral, see What is the issue in my array division step?