Consider the functions
$$ f_1(x) := \frac{1}{\sqrt{|x|}} $$
and
$$ f_2(x) := \frac{1}{|x|}. $$
Over the interval $I := [-1,1]$, the function $f_1$ integrates to 4, whereas the integral of $f_2$ diverges. We can still go ahead and apply the Monte Carlo method to estimate the integral of either function, i.e., we can evaluate
$$ \hat{I}_N := \frac{1}{N}\sum_{i=1}^N f(X_i), $$
where $X_1, \ldots, X_N$ are independent random values drawn from the uniform distribution on $[-1,1]$. If we do this for $f=f_1$, then the estimate converges as expected as $N \to \infty$; for $f=f_2$ it jumps all over the place, with no trend toward a definite value. This would seem to imply a positive answer to the following question:
Question: is there a statistically reliable way to numerically estimate whether or not a given function has a finite integral? E.g., a quantity that (with probability 1) tends toward zero if and only if the integral is well-defined.
(I have the sinking feeling I should just be recalling a textbook answer from my intro probability course…)
I think the Answer is Negative.
When the Integral is converging , we might use random Points on the Curve to "estimate" the value & make some way to quantify the "accuracy".
When the Integral is not converging , where we still blindly use random Points on the Curve to "estimate" the value , we might not get "converging" value or we might get a value co-incidentally converging to some spurious value.
Let us say we use a large value for $N$ in the given formula.
Based on that , we claim that the Integral is not converging , via some Statistical calculations & tests.
It is entirely Possible that beyond $N$ , eg $2N$ or $N^2$ , the Integral converges though our calculations have not even considered such large values.
Hence our Conclusion would be wrong :Integral will converge , except for a large "offset" from the early $N$ values.
Basically , we can never check "infinite" number of Points on the curve & we can never consider "$N \sim \infty$" , whereas the "finite" number of Points occurring early may give a wrong Conclusion.
[[ Alternately , we can cast this into the "Halting Problem" , which still gives the Negative Conclusion.
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