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 .
Consider the sequence defined recursively as
Prove that .
Lemma: If is a sequence for which then
Proof: In Stolz theorem we set and .
It is easy to see that is is monotonically decreasing to zero. Moreover, an application of L’Hospital’s rule gives
Now, due to the lemma we have and the result follows.
Remark : The asymptotic now follows to be .
Problem: Find what inequality should satisfy such that the series
Let . Prove that
We note that
and the result follows from the sandwich theorem.
Let . Prove that
The LHS is equal to which by AM – GM is less or equal to
where . Since it follows from Bernoulli inequality that .
Show that there do not exist invertible matrices such that and .
Suppose, on the contrary, that such matrices do exist. Then
Using the fact that we deduce that
The last means that which is impossible because both and are invertible ( and so must be the product ). Hence, the conclusion follows.