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We begin by stating a lemma:
Lemma: Let be an analytic function on some closed disk which has center and radius . Let denote the the boundary of the disk. It holds that
Proof: By the Cauchy integral formula we have that
The equation of a circle of radius and centre is given by . Hence,
and the proof is complete.
Something quickie: Given the assumptions in Gauss’ MVT, we have
The proof of the result is pretty straight forward by using the fact that
Back to the problem the result now follows by the lemma.
Evaluate the integral
using Riemann sums.
Partition the interval into subintervals of length . This produces,
On the other hand, assuming is even:
since it holds that
Euler’s Gamma reflection formula was used at line . Letting we get that
If is odd we work similarly.
First of all we observe that the integral as well as the integral diverge whereas the proposed integral converges which is an interesting fact. Now,
since is determined by the inequalities
Let be independent and uniform random variables. By the law of large numbers we have
in probability as . Therefore ,
in probability as .
The ratio random variables are bounded below by zero and above by one. This guarantees convergence of the expectations, as well.
which is the required result.
Remark: In general it holds that
because the distribution of is the Lebesgue measure on hence for every measurable function ,
I was surfing the net today and I fell on this cute integral
I have seen integrals of such kind before like for instance this .In fact something more general holds
The original integral does not fall into this category which is a real shame. Yet it does have a closed form and it does not contain an in its final answer. Strange, huh? So similar but so different at the same time these two integrals. A sign changes everything.
We begin by exploring the integral
Manipulating the integral ( substitutions and known Gaussian results) reveals that
where . Taking the imaginary part of the last expression we get that
and this is the final answer. See, no !. Of course we can also extract the real part and calculate the corresponding integral involving .