After reading the Wikipedia page for Monte-Carlo integration, I have understood the basic idea but I am having trouble implementing it for a general case.
The integration that I am trying to do is $$ \int_{-\infty}^{\infty} \frac{\cos x}{x^2}- \frac{\sin x}{x^3} dx $$
This integral is equal to $-\frac{\pi}{2}$, but I do not see how to do it with the Monte-Carlo method. What is the algorithmic procedure in this case?
From Wikipedia:
The naive Monte Carlo approach is to sample points uniformly on Ω[...]
There is an implicit assumption here that a uniform distribution on $\Omega$ exists. It is well-known that such a distribution does not exist on any unbounded subinterval of $\mathbb{R}$. Therefore, using basic Monte Carlo will not work for this problem. One option is to do a $u$-substitution so that $u$ is integrated over a finite interval then perform uniform Monte Carlo sampling on that interval. Since the integrands are highly oscillatory, this will likely take a while to converge.
Note that the integrand is an even function, so we need to evaluate
$ I = 2 \int_0^{\infty} f(x)/x^2 dx, $
where $f(x) = \cos(x)-\sin(x)/x$
Let's write it as
$ I/2 = \int_0^{1} f(x)/x^2 dx + \int_1^{\infty} f(x)/x^2 dx $
The difficult part is the second half, and one way to deal with it is to draw the Monte-Carlo samples from the distribution $P(\xi)=1/\xi^2$ for $\xi \in [1,\infty]$. Then the improper integral is approximated by
$ \frac{1}{N} \sum_{k=1}^N f(x_k), $
where $N$ is the number of samples.
This approach may improve the accuracy, i.e., reduce the cost of the calculation.
