Prove that

**Solution**

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

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# Tag: Integral

## Multiple logarithmic integral

## Limit of a multiple integral

## An extraordinary sin integral

## Integral with floor and ceiling function

## Multiple integral

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Prove that

**Solution**

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

Prove that

**Solution**

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.

So,

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 somethingĀ 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 .

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

**Solution**

We have successively

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