## 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 .

## Integral with floor and ceiling function

Let denote the floor function and denote the ceiling function. Evaluate the integral

Solution

We have successively

## Multiple integral

Let denote the usual inner product of . Evaluate the integral

where is a positive symmetric matrix and .

Solution

Since is a positive symmetric matrix , so is . For a positive symmetric matrix there exists an positive symmetric matrix such that . Applying this to our integral becomes

where is the Euclidean norm. Applying a change of variables we have that

Since then by converting to polar coordinates we have that

Here denotes the surface area measure of the unit sphere and it is known to be

hence

where denotes the Gamma Euler function for which it holds that

## A curious logarithmic and trigonometric integral

Let denote the Euler – Mascheroni constant. Prove that

Solution

Let us consider the following

It is known that . Also, the Mellin transform of is known to be

Hence

On the other hand

What we are seeking is .

and the result follows.

## On a classic logarithmic computation

Let . Find a closed form for the integral

Solution

We are invoking a powerful theorem stating that:

Theorem: Let be a second degree polynomial (this implies ) . Then it holds that

where .

Making use of the above theorem we have that