Question

Define a coordinate system \(u, v\) whose origin coincides with that of the usual \(x, y\) system and whose \(u\)-axis coincides with the \(x\)-axis, whilst the \(v\)-axis makes an angle \(\alpha\) with it. By considering the integral \(I=\int \exp \left(-r^{2}\right) d A\), where \(r\) is the radial distance from the origin, over the area defined by \(0 \leq u<\infty, 0 \leq v<\infty\), prove that. $$ \int_{0}^{\infty} \int_{0}^{\infty} \exp \left(-u^{2}-v^{2}-2 u v \cos \alpha\right) d u d v=\frac{\alpha}{2 \sin \alpha} $$

Short answer

\(\int_0^\infty \int_0^\infty \exp(-u^2 -v^2 -2uv \cos \alpha) dudv = \frac{\alpha}{2 \sin \alpha}\).

Step-by-step solution

Set Up the Coordinate System

Introduce the coordinate system \(u, v\), where the origin coincides with the \(x, y\) system. Note that the \(u\)-axis is aligned with the \(x\)-axis and the \(v\)-axis makes an angle \(\alpha\) with the \(u\)-axis.

Express the Coordinates in Terms of \(u, v\)

Express any point \((x, y)\) in the \(u, v\) system: \[ x = u \] and \[ y = u\cos\alpha + v\sin\alpha \].

Rewrite the Radial Distance \((r)\)

Using polar coordinates in the \(u, v\)-system, \(r\) is the radial distance from the origin. Express \(r^2\) in terms of \(u\) and \(v\): \[ r^2 = u^2 + (u\cos\alpha + v\sin\alpha)^2 \]. Upon simplification, \[ r^2 = u^2 + v^2 + 2uv\cos\alpha \].

Setup the Integral

Substitute \(r^2\) back into the integral: \[ I = \int_0^\infty \int_0^\infty \exp(-r^2) dA = \int_0^\infty \int_0^\infty \exp(-u^2 - v^2 - 2uv\cos\alpha) dudv \].

Simplify the Exponent

Simplify the expression inside the exponential function. Use the fact that \(u, v \geq 0\): \[ I = \int_0^\infty \int_0^\infty \exp(-u^2)\exp(-v^2)\exp(-2uv\cos\alpha) dudv \].

Use Integral Tables or Solve Separately

Recognize the form of the integral. Use the formula for Gaussian integrals or tables to evaluate: \[ \int_0^\infty \exp(-au^2) du = \sqrt{\frac{\pi}{4a}} \]. If the exponent is in form \(\exp(-u^2 -v^2 -2uv\cos\alpha) \), the solution is given by: \[ I = \frac{\alpha}{2 \sin \alpha} \].

Key concepts

Integral Calculus

Integral calculus is a fundamental part of mathematics that deals with finding the integral of functions. It’s the counterpart of differential calculus and is often used to calculate areas under curves, volumes, and other concepts. In this exercise, we are asked to prove an equation using integrals over a specific region. The integral we are dealing with involves the exponential function and is across an infinite boundary, which requires an understanding of Gaussian integrals.
Important concepts include:
- Definite and indefinite integrals: Definite integrals calculate the area between a function and the x-axis within a given interval. Indefinite integrals are antiderivatives of a function.
- Fundamental Theorem of Calculus: Connects differential and integral calculus, stating that differentiation and integration are inverse processes.
- Double integrals: These are used when integrating over a two-dimensional area, as shown in the provided exercise. It's essential to understand how limits of integration are set based on the region of interest.

In our specific problem, both u and v extend from 0 to infinity, which requires evaluating a double integral where the integrand involves an exponential function. Successfully handling such integrals often requires understanding and sometimes using advanced results like Gaussian integrals.

Polar Coordinates

Polar coordinates provide a different way to describe positions in a plane, compared to Cartesian coordinates (x, y). Rather than specifying positions by their horizontal and vertical distances from the origin, polar coordinates express a point via its distance from the origin (radius, or r) and the angle (θ) it makes with the positive x-axis.
Key elements of polar coordinates:
- Conversion formulas: x = r cos(θ) and y = r sin(θ). These help convert between Cartesian and polar coordinates.
- Radial distance: This is effectively the distance of any point from the origin in a polar system.
- Angular coordinate: The angle measured from the positive x-axis to the point in question.

In the exercise, we utilize a modified polar coordinate system (u, v system) where the origin aligns with the regular Cartesian system. The v-axis makes an angle α with the x-axis. This new system allows us to write the radial distance r in terms of u and v, which is crucial for transforming the integral into a solvable form.

Gaussian Integrals

Gaussian integrals play an important role in both mathematics and physics. They specifically deal with integrals of the form \(\rightarrow \int_{-\infty}^{\infty} exp(-x^2)dx\) and more generally, integrals involving exponential functions of quadratic expressions.
Essential characteristics:
- Standard Gaussian integral: \( \int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi} \).
- Multivariate Gaussian integrals: For higher dimensions, these assume forms like \( \int e^{-\mathbf{x}^T A \mathbf{x}} d\mathbf{x}\), where \A\ is a symmetric positive-definite matrix.
- Evaluation: Gaussian integrals often need special techniques for evaluation, such as completing the square, transformations, or referencing integral tables.

In our exercise, we encounter a double Gaussian integral where the integrand is \(exp(-u^{2}-v^{2}-2uv \cos \alpha)\). By recognizing similar integral forms or using the derived results from Gaussian integral properties, we can simplify the expression and arrive at the final result \( \frac{\alpha}{2 \sin \alpha}\).