## An extraordinary sin integral

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

where .

**Solution**

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 .

## On a nested sin sequence

Consider the sequence defined recursively as

Prove that .

**Solution**

**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

Therefore

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

converges.

## An interesting limit

Let . Prove that

**Solution**

We note that

Thus,

and the result follows from the sandwich theorem.

## Nested radical inequality

Let . Prove that

**Solution**

The LHS is equal to which by AM – GM is less or equal to

where . Since it follows from Bernoulli inequality that .

## No invertible matrices

Show that there do not exist invertible matrices such that and .

**Solution**

Suppose, on the contrary, that such matrices do exist. Then

and also

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.