Problem 6 - Differential Equations and Linear Algebra

Question

(a) By making the change of variables \(t=x^{2}\) in the integral that defines the gamma function, show that $$ \Gamma(1 / 2)=2 \int_{0}^{\infty} e^{-x^{2}} d x $$ (b) Use your result from (a) to show that $$ [\Gamma(1 / 2)]^{2}=4 \int_{0}^{\infty} \int_{0}^{\infty} e^{-\left(x^{2}+y^{2}\right)} d x d y $$ (c) By changing to polar coordinates, evaluate the double integral in (b) and hence show that $$ \Gamma(1 / 2)=\sqrt{\pi} $$

Step-by-Step Solution

Answer

The solution can be summarized in three parts: 1. By making the change of variables \(t = x^2\), we showed $$ \Gamma(1/2) = 2\int_0^\infty e^{-x^2}dx $$ 2. By squaring the result, we found $$ [\Gamma(1/2)]^2 = 4\int_0^\infty\int_0^\infty e^{-(x^2+y^2)}dxdy $$ 3. Changing to polar coordinates and evaluating the double integral, we concluded $$ \Gamma(1/2) = \sqrt{\pi} $$

Step 1: Part a: Use Change of Variables for Gamma Function Integral

We need to remember the definition of the Gamma Function: $$ \Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt $$ We have to prove that: $$ \Gamma(1/2) = 2\int_0^\infty e^{-x^2}dx $$ We make a substitution, let \(t = x^2\). Then \(dt = 2xdx\). We also need to change the limits of integration; when \(x=0\), \(t=0\), and when \(x\to\infty\), \(t\to\infty\). Now the integral becomes: $$ \Gamma(1/2) = \int_0^\infty t^{(1/2)-1}e^{-t}dt = \int_0^\infty e^{-t}2xdx $$ With this substitution, we have shown that: $$ \Gamma(1/2) = 2\int_0^\infty e^{-x^2}dx $$

Step 2: Part b: Square the Gamma Function Integral

We start by squaring the result obtained in part a: $$ [\Gamma(1/2)]^2 = \left[2\int_0^\infty e^{-x^2}dx\right]^2 $$ Now, we rewrite the square of the integral as a double integral over the same domain: $$ [\Gamma(1/2)]^2 = 4\int_0^\infty\int_0^\infty e^{-x^2} e^{-y^2}dxdy = 4\int_0^\infty\int_0^\infty e^{-(x^2+y^2)}dxdy $$

Step 3: Part c: Change to Polar Coordinates and Evaluate Double Integral

First, we need to change the double integral to polar coordinates. We have that \(x = r\cos \theta\) and \(y = r\sin \theta\), so: $$ x^2 + y^2 = r^2\cos^2 \theta + r^2\sin^2 \theta = r^2(\cos^2 \theta + \sin^2 \theta) = r^2 $$ Also, the area element in polar coordinates is \(dA = rdrd\theta\). Now, the integral becomes: $$ [\Gamma(1/2)]^2 = 4\int_0^\infty\int_0^\infty e^{-r^2}(rdrd\theta) $$ Now, we change the limits of integration; when \(\theta=0\), \(x=0\), and when \(\theta=\pi/2\), \(y=\infty\), and when \(\theta=\pi\), \(x=\infty\), and when \(\theta=2\pi\), \(x=0\), and when \(r=0\), \(x=0\), and when \(r\to\infty\), \(x\to\infty\): $$ [\Gamma(1/2)]^2 = 4\int_0^{\pi/2}\int_0^\infty e^{-r^2}rdrd\theta $$ Now, we separate the integral into two parts, since they are independent of each other: $$ [\Gamma(1/2)]^2 = 4\left(\int_0^{\pi/2} d\theta\right)\left(\int_0^\infty e^{-r^2}rdr\right) $$ Now, evaluating the definite integrals: $$ [\Gamma(1/2)]^2 = 4\left[\frac{\pi}{2}\right]\left[-\frac{1}{2}e^{-r^2}\Bigg|_0^\infty\right] = \pi\left(0 - \left(-\frac{1}{2}\right)\right) = \pi $$ Now, we take the square root of both sides to find the value of \(\Gamma(1/2)\): $$ \Gamma(1/2) = \sqrt{\pi} $$ Thus, we have shown that \(\Gamma(1/2) = \sqrt{\pi}\).

Key Concepts

Integration Techniques

Integration techniques are vital tools in calculus used to find the integral of a function. One common technique is the change of variables, which simplifies integrals by transforming them into a form that is easier to evaluate. When we substitute variables in an integral, we also change the differential accordingly, ensuring the integral remains equivalent.
For example, in step 1 of our exercise, we used this method by letting \( t = x^2 \), transforming our exponential integral into an expression involving a simpler variable \( t \). This substitution allowed us to find a new integral representation of \( \Gamma(1/2) \), which is crucial for further analysis.
While employing the change of variables, it's essential to adjust the limits of integration according to the new variable, ensuring the new integral covers the same range as the original.

Polar Coordinates

Polar coordinates provide a powerful alternative to Cartesian coordinates, especially for dealing with problems involving circular symmetry. In polar coordinates, a point is identified by the distance from the origin \( r \), and the angle \( \theta \) from a reference direction.
This system simplifies our work significantly when integrating over circular regions. In the context of our exercise, transforming the double integral from Cartesian to polar coordinates allowed us to exploit the circular symmetry of the integrand \( e^{-(x^2 + y^2)} \). The expression \( x^2 + y^2 \) simplifies to \( r^2 \), streamlining the integration process.
Remember, converting to polar coordinates changes not only the variables but also the differential element: \( dx \, dy \) transforms to \( r \, dr \, d\theta \). This change is crucial for correctly evaluating integrals in this system.

Double Integrals

Double integrals extend the concept of integration to two-dimensional spaces, allowing us to determine the volume under a surface or compute areas and densities. They work similarly to single integrals but require evaluating an integral over two axes.
In our exercise, the double integral \( \int_0^\infty \int_0^\infty e^{-(x^2 + y^2)} \, dx \, dy \) represents the area under a surface defined by the function \( e^{-(x^2 + y^2)} \). By changing this to polar coordinates, we simplified its evaluation.
When solving double integrals, visualization of the region and an appropriate change of order or coordinates can make integration more manageable.

Change of Variables

Change of variables is a critical strategy when evaluating integrals, particularly in multiple dimensions. It involves substituting one or more variables to transform an integral into an easier form.
In part (a) and part (c) of our exercise, this technique streamlined solving the problem. First, we used it by substituting \( t = x^2 \) to match the format of the Gamma Function integral. Then, moving from Cartesian to polar coordinates by setting \( x = r\cos\theta \) and \( y = r\sin\theta \), further transformed the integrand into a form more straightforward to evaluate.
The key to using the change of variables effectively lies in identifying a transformation that simplifies the integral, re-calculating the differential element, and adjusting the limits of integration accordingly. This technique is indispensable in various fields, including physics and engineering, emphasizing its importance in problem-solving across disciplines.