Problem 1 - Arbeitsbuch Mathematik für Ingenieure

Question

Unter Verwendung von Polarkoordinaten zeige man, dass für das Bereichsintegral $$ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\frac{1}{2}\left(x^{2}+y^{2}\right)} \mathrm{d} x \mathrm{~d} y=\left(\int_{-\infty}^{\infty} e^{-\frac{1}{2} x^{2}} \mathrm{~d} x\right)\left(\int_{-\infty}^{\infty} e^{-\frac{1}{2} y^{2}} \mathrm{~d} y\right)=2 \pi $$ gilt, und zeige damit für die Dichte \(f\) der Normalverteilung $$ \int_{-\infty}^{\infty} f(t) \mathrm{d} t=1 $$

Step-by-Step Solution

Answer

In Polar Coordinates, the integral evaluates to \( 2\pi \), demonstrating the normalization of the Normal distribution.

Step 1: Convert Cartesian Coordinates to Polar Coordinates

Recall that in polar coordinates, we have the transformations: \[ x = r \, \cos(\theta) \]\[ y = r \, \sin(\theta) \]Also, the element of area transforms as follows: \[ \mathrm{d}x \, \mathrm{d}y = r \, \mathrm{d}r \, \mathrm{d}\theta \]

Step 2: Express the Integral in Polar Coordinates

Substitute the transformations into the integral: \[ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\frac{1}{2}(x^{2}+y^{2})} \mathrm{d}x \mathrm{d}y = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-\frac{1}{2}r^{2}} r \, \mathrm{d}r \, \mathrm{d}\theta \]

Step 3: Separate the Angular and Radial Parts

Notice that the integral separates into a product of two integrals: \[ \int_{0}^{2\pi} \mathrm{d}\theta \int_{0}^{\infty} r e^{-\frac{1}{2} r^{2}} \, \mathrm{d}r \]

Step 4: Evaluate the Angular Part

Evaluate the angular part first: \[ \int_{0}^{2\pi} \mathrm{d}\theta = 2\pi \]

Step 5: Evaluate the Radial Part

For the radial part, use the substitution \( u = \frac{1}{2}r^{2} \), thus \( \mathrm{d}u = r \, \mathrm{d}r \). This gives: \[ \int_{0}^{\infty} r e^{-\frac{1}{2} r^{2}} \, \mathrm{d}r = \int_{0}^{\infty} e^{-u} \, \mathrm{d}u = 1 \]

Step 6: Combine Results

Combine the results from the angular and radial parts: \[ 2\pi \cdot 1 = 2\pi \]

Step 7: Link Back to Normal Distribution

Recall that the integral of the density function \( f(t) \) for the Normal distribution over all real numbers is given by: \[ f(t) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}t^{2}} \]Since we calculated: \[ \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}t^{2}} \mathrm{d}t = 1 \]

Step 8: Conclusion

Therefore, the integral for the Normal distribution over the entire real line is equal to 1, matching the property of a probability density function: \[ \int_{-\infty}^{\infty} f(t) \mathrm{d}t = 1 \]

Key Concepts

Bereichsintegral

A 'Bereichsintegral', or area integral, is an essential concept in calculus and physics. It is used to calculate the double integral over a certain area in the plane, which can often be expressed in Cartesian or Polar coordinates. By converting from Cartesian coordinates \((x, y)\) to Polar coordinates \((r, \theta)\), one can greatly simplify complex integrals. The change of coordinates uses the relations:
\[ x = r \cos(\theta) \] \[ y = r \sin(\theta) \]
In our example, the integral simplifies by transforming the area element \( d x \, d y \) to \( r \, d r \, d \theta \). This substitution can transform complicated boundary conditions into more manageable forms.

Normalverteilung

The 'Normalverteilung', or Normal distribution, is a vital concept in statistics and probability theory. It describes how values of a variable are distributed. Characterized by its bell-shaped curve, it is defined by the probability density function \(f(t) = \frac{1}{\sqrt{2\textbackslash pi}} e^{-\textbackslash frac{1}{2}t^{2}} \). The mean of the normal distribution is 0, and its variance is 1. This means:
\[ \textbackslash int_{-\infty}^{\infty} f(t) \, d t = 1 \]
ensures that the total probability is equal to 1, a fundamental property of all probability density functions. This validation confirms its critical role in statistical analyses and real-world applications.

Wahrscheinlichkeitsdichtefunktion

A 'Wahrscheinlichkeitsdichtefunktion', or Probability Density Function (PDF), describes the likelihood of a random variable taking on a particular value. For continuous variables, it's essential that the area under the PDF curve over the entire range is equal to one. In the given exercise, we've shown that:
\[ \textbackslash int_{-\infty}^{\infty} f(t) \, d t = 1 \]
where \( f(t) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}t^{2}} \), is a PDF for the normal distribution. This validation is crucial since it confirms that the sum of all probabilities is 1, adhering to the core tenant of probability theory.

Koordinatentransformation

Koordinatentransformation, or coordinate transformation, involves changing from one coordinate system to another, facilitating easier computation of integrals. Here, we switch from Cartesian to Polar coordinates using:
\[ x = r \cos(\theta) \] \[ y = r \sin(\theta) \] \[ \textbackslash d x \, \textbackslash d y = r \ d r \, \ d \theta \]
By transforming the original Cartesian integral into its polar form, evaluating becomes more straightforward, especially for symmetrical functions. This technique proves invaluable in various areas of applied mathematics, physics, and engineering.