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 \(I^{2}=\left(\int_{-\infty}^{\infty} e^{-x^{2}} d x\right)\left(\int_{-\infty}^{\infty} e^{-y^{2}} d y\right)=\) \(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^{2}-y^{2}} d x d y,\) 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 \(I=\sqrt{\pi}\) 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

Question: What is the value of the integral \(I = \int_{-\infty}^{\infty} e^{-x^2} dx\) and the following integrals: a) \(\int_{0}^{\infty} e^{-x^2} dx\) b) \(\int_{0}^{\infty} xe^{-x^2} dx\) c) \(\int_{0}^{\infty} x^2e^{-x^2} dx\) Answer: The values of the integrals are: \(I = \sqrt{\pi}\) a) \(\frac{1}{2}\sqrt{\pi}\) b) \(\frac{1}{2}\) c) \(\frac{1}{4}\sqrt{\pi}\)

Step-by-step solution

Set up the given integral by squaring I

According to the problem, 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 double integral to polar coordinates

To convert the given double integral to polar coordinates, we have to replace \(x^2+y^2\) with \(r^2\), and replace \(dxdy\) with \(r dr d\theta\). So, the integral becomes: \(I^2=\int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2}r dr d\theta\).

Evaluate the inner integral

Now, we need to evaluate the inner integral with respect to r: \(\int_{0}^{\infty} e^{-r^2}r dr\). Let's use substitution: \(u = r^2\), so \(du = 2r dr\). This changes the integral to: \(\frac{1}{2} \int_{0}^{\infty} e^{-u} du\). Now integrate: \(-\frac{1}{2}e^{-u} \Big|_{0}^{\infty} = \frac{1}{2}\). So, the inner integral becomes \(\frac{1}{2}\).

Evaluate the outer integral

Next, we need to evaluate the outer integral with respect to \(\theta\): \(\int_{0}^{2\pi} \frac{1}{2} d\theta = \frac{1}{2} (\theta \Big|_{0}^{2\pi}) = \pi\).

Find the value of I

Since \(I^2 = \pi\), we have: \(I=\sqrt{\pi}\).

Evaluate the other integrals

Now, we have to evaluate the other integrals using part (a) if needed: a) \(\int_{0}^{\infty} e^{-x^2} dx\) = \(\frac{1}{2}\sqrt{\pi}\), since it is half of the original integral \(I\) (integral from \(-\infty\) to \(\infty\)). b) \(\int_{0}^{\infty} xe^{-x^2} dx\): Using substitution, let \(u = x^2\) and \(du = 2x dx\). Then, the integral becomes: \(\frac{1}{2} \int_{0}^{\infty} e^{-u}du = -\frac{1}{2}e^{-u} \Big|_{0}^{\infty} = \frac{1}{2}\). c) \(\int_{0}^{\infty} x^2e^{-x^2}dx\): Integrate by parts with \(u=x\), \(dv=e^{-x^2}(2x dx)\), \(du=dx\), and \(v=-\frac{1}{2}e^{-x^2}\). The integral becomes: \(x\left(-\frac{1}{2}e^{-x^2}\right) \Big|_{0}^{\infty} - \int_{0}^{\infty}-\frac{1}{2}e^{-x^2} dx\). The first term evaluates to 0 and the second term to \(\frac{1}{2} \int_{0}^{\infty}e^{-x^2} dx\). So, \(\int_{0}^{\infty} x^2e^{-x^2}dx = \frac{1}{2} \cdot \frac{1}{2}\sqrt{\pi} = \frac{1}{4}\sqrt{\pi}\).

Key concepts

Polar Coordinates

Switching to polar coordinates is a handy method when we evaluate certain integrals, especially those involving expressions like \(x^2 + y^2\). In polar coordinates, these are simply replaced with \(r^2\), which can significantly simplify the integral.

Here's how it works:

• Transform \(x\) and \(y\) to polar coordinates: \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\).
• Replace \(x^2 + y^2\) with \(r^2\).
• The differential area element \(dxdy\) becomes \(rdrd\theta\).

These transformations make the integral easier to work with, especially for expressions symmetric around the origin.

Evaluation of Integrals

Evaluating integrals involves finding the area under the curve of a function. When working with a normal distribution, many integrals become crucial for statistical calculations. For example, the Gaussian integral \(\int_{-\infty}^{\infty} e^{-x^2} dx\) evaluates to \(\sqrt{\pi}\).

Evaluating integrals over infinite limits often requires:

• Rewriting the problem, like squaring the integral and converting it to a double integral.
• Leveraging known results, such as symmetry, to simplify the calculation.
• Using coordinate transformations, like switching to polar coordinates, for easier integration.

These techniques collectively simplify complex integrals, making them manageable.

Substitution Method

The substitution method is a fundamental technique for integration, similar to the reverse of the chain rule in differentiation. It involves changing the variable to simplify the integral.

For example, in the integral \(\int_{0}^{\infty} e^{-r^2}r dr\), we can use:

• Substitution: let \(u = r^2\), then \(du = 2rdr\).
• Change the bounds of the integral accordingly, but in our case only the function is transformed.

Using substitution, the integral becomes much easier to solve.

This method is especially useful when dealing with functions that are products of a polynomial and an exponential, trigonometric, or logarithmic function.

Integration by Parts

Integration by parts is a technique that simplifies the integration of products of functions. It's based on the product rule from differentiation.

Consider the integral \(\int_{0}^{\infty} x^2e^{-x^2} dx\). Here’s a quick rundown:

• Identify \(u\) and \(dv\). Set \(u = x\) and \(dv = e^{-x^2}(2x)dx\).
• Differentiate to get \(du = dx\) and integrate to find \(v = -\frac{1}{2}e^{-x^2}\).

The formula for integration by parts is \(\int u \, dv = uv - \int v \, du\).

This helps break down complicated integrals into simpler parts, sometimes making use of previously solved integrals.