Question

An important integral in statistics associated with the normal distribution is \(I=\int_{-\infty}^{\infty} e^{-x^{2}} d x .\) It is evaluated in the following steps. a. Assume that $$\begin{aligned} I^{2} &=\left(\int_{-\infty}^{\infty} e^{-x^{2}} d x\right)\left(\int_{-\infty}^{\infty} e^{-y^{2}} d y\right) \\ &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^{2}-y^{2}} d x d y \end{aligned}$$ where we have chosen the variables of integration to be \(x\) and \(y\) and then written the product as an iterated integral. Evaluate this integral in polar coordinates and show that \(I=\sqrt{\pi} .\) Why is the solution \(I=-\sqrt{\pi}\) rejected? b. Evaluate \(\int_{0}^{\infty} e^{-x^{2}} d x, \int_{0}^{\infty} x e^{-x^{2}} d x,\) and \(\int_{0}^{\infty} x^{2} e^{-x^{2}} d x\) (using part (a) if needed).

Short answer

Answer: The significance of rejecting the negative solution \(I=-\sqrt{\pi}\) is because the normal distribution has positive values on both sides, and hence its integral represents the total area under the normal curve (probability) which should always be positive. The positive value for the integral (\(I=\sqrt{\pi}\)) ensures that we obtain valid results for the normal distribution and its related applications in statistics and probability.

Step-by-step solution

Write the integral as a double integral in cartesian coordinates

We are given that $$I^2 =\left(\int_{-\infty}^{\infty} e^{-x^{2}} dx\right)\left(\int_{-\infty}^{\infty} e^{-y^{2}} dy\right) =\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^{2}-y^{2}} dxdy$$

Convert the cartesian double integral to polar coordinates

We know that \(x^2+y^2=r^2\) and \(dx dy = r dr d\theta\). Therefore, the integral becomes $$I^2 = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} r drd\theta$$

Evaluate the polar integral

Now, we will evaluate the double integral: $$I^2 = \int_{0}^{2\pi} d\theta \int_{0}^{\infty} e^{-r^2}r dr = 2\pi \int_{0}^{\infty} e^{-r^2}r dr$$ Let's make the substitution \(u=r^2\). Then, \(du = 2r dr\). The integral becomes: $$I^2 = \pi \int_{0}^{\infty} e^{-u} du = \pi [-e^{-u}]_0^\infty = \pi(1-0)= \pi$$

Find the value of \(I\) and explain the rejection of \(I=-\sqrt{\pi}\)

Now, we can write \(I=\sqrt{I^2}=\sqrt{\pi}\). We reject the solution \(I=-\sqrt{\pi}\) because the normal distribution has positive values on both sides, and hence the integral should have a positive value. b) Using part (a), we can now evaluate the three given integrals: 1) \(\int_{0}^{\infty} e^{-x^2} dx = \frac{I}{2} = \frac{\sqrt{\pi}}{2}\) 2) For the integral \(\int_{0}^{\infty} xe^{-x^2} dx\), we can make use of the result \(\int_{0}^{\infty} e^{-ax^2} dx = \frac{\sqrt{\pi}}{2 \sqrt{a}}\). Here, \(a=1\). Therefore, the integral becomes \(\int_{0}^{\infty} xe^{-x^2} dx = \frac{\sqrt{\pi}}{2}\) 3) Similarly, for the integral \(\int_{0}^{\infty} x^2 e^{-x^2} dx\), we can use the result \(\int_{0}^{\infty} x^2 e^{-ax^2} dx = \frac{\sqrt{\pi}(2n-1)!!}{4a^n(2n)!!}\). Here, \(n=1\) and \(a=1\). Thus, the integral becomes \(\int_{0}^{\infty} x^2 e^{-x^2} dx = \frac{\sqrt{\pi}}{2} \cdot \frac{1}{2}\) = \(\frac{\sqrt{\pi}}{4}\).

Key concepts

Normal Distribution

The normal distribution, often referred to as the Gaussian distribution, is a fundamental concept in statistics. It represents a continuous probability distribution for a real-valued random variable. The curve is bell-shaped and symmetric around the mean. Understanding its properties is crucial for various statistical analyses.

• The mean, median, and mode of a normal distribution are all equal and located at the center of the distribution.
• The spread of the distribution is determined by its standard deviation.
• The total area under the curve is 1, which represents the entire probability space.

In the context of integral calculus, evaluating integrals related to the normal distribution, such as \(I = \int_{-\infty}^{\infty} e^{-x^2} \, dx\), is crucial. This particular integral is instrumental in determining probabilities and is often used in statistical mechanics and quantum physics. The integration stretches over the entire line and results in a value proportional to \(\sqrt{\pi}\), demonstrating the symmetry and specific decay properties of the normal distribution curve.

Polar Coordinates

Polar coordinates offer a different way of describing locations in a plane compared to the traditional Cartesian coordinates. In polar coordinates, each point is defined by a distance from a reference point and an angle from a reference direction.

• The position of a point is given by \((r, \theta)\), where \(r\) is the radius (distance from the origin) and \(\theta\) is the angle from the positive x-axis.
• Polar coordinates are particularly useful for dealing with problems involving circular symmetry or integrals like \(x^2 + y^2 = r^2\).

In terms of integration, converting a double integral from Cartesian coordinates to polar coordinates can simplify the calculation process significantly. The integral becomes easier to evaluate because of the circular symmetry. When we write double integrals in polar form, the term \(dx \, dy\) becomes \(r \, dr \, d\theta\), allowing smooth calculation over circular regions.

Double Integral

Double integrals extend the concept of integration to two-dimensional spaces, allowing calculation of areas and volumes under surfaces. In this context, they help evaluate probabilities and expectations within two-variable functions.The basic form of a double integral is:\[\int \int_R f(x, y) \, dx \, dy\]where \(R\) is the region of integration.

• Double integrals can be computed as iterated integrals, which simplifies the solving process by evaluating the integral layer by layer.
• Changing to polar coordinates is a common strategy to tackle double integrals involving circular or symmetrical regions.
• For example, \(\iint e^{-x^2-y^2} \, dx \, dy\) in rectangular coordinates translates to \(\int_0^{2\pi} \int_0^{\infty} e^{-r^2} \, r \, dr \, d\theta\) in polar coordinates.

This transformation facilitates the computation and reveals the symmetry of the problem. Understanding this process is crucial for solving complex integrals effectively.