[208] [Short-term Intensive Series Part 4] Unraveling the Integral of the Normal Distribution: Utilizing Polar Coordinate Transformation and Circular Symmetry
Last time, we understood the core of Gauss's calculation strategy.
$${\displaystyle I^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2)} dx dy}$$
Since the integrand $${e^{-(x^2 + y^2)}}$$ is perfectly circularly symmetric, we found that using polar coordinate transformation is optimal.
This time, we will understand this polar coordinate transformation in detail and finally perform the integration to derive the beautiful result $${I^2 = \pi}$$.
What are polar coordinates: Development from high school mathematics
Two ways to represent the position of a point
Cartesian coordinates: Represent point P as $${(x, y)}$$
$${x}$$: Distance in the $${x}$$-axis direction
$${y}$$: Distance in the $${y}$$-axis direction
Polar coordinates: Represent the same point P as $${(r, \theta)}$$
$${r}$$: Distance from the origin
$${\theta}$$: Angle counterclockwise from the positive $${x}$$-axis
Coordinate transformation formulas
From polar to Cartesian coordinates
$${\begin{cases} x = r \cos \theta \\ y = r \sin \theta \end{cases}}$$
From Cartesian to polar coordinates
$${\displaystyle \begin{cases} r = \sqrt{x^2 + y^2} \\ \theta = \arctan\left(\frac{y}{x}\right) \end{cases}}$$ (Note: pay attention to the quadrant)
Verification with a simple example
Representing the point $${(1, \sqrt{3})}$$ in polar coordinates:
$${\displaystyle \begin{cases} r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2 \\ \theta = \arctan\left(\frac{\sqrt{3}}{1}\right) = \arctan(\sqrt{3}) = \frac{\pi}{3} \end{cases}}$$
This is the result. Checking it,
$${\displaystyle \begin{cases} x = 2 \cos \frac{\pi}{3} = 2 \times \frac{1}{2} = 1 \\ y = 2 \sin \frac{\pi}{3} = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \end{cases}}$$
it seems correct.
Why polar coordinates are effective
Simplification of the integrand
By polar coordinate transformation,
$${x^2 + y^2 = r^2}$$
so,
$${e^{-(x^2 + y^2)} = e^{-r^2}}$$
and...
Important observation : The transformed function does not depend on the angle $${\theta}$$!
This allows the complex 2D integral to be effectively separated into a 1D integral.
Transformation of the integration region
The region in Cartesian coordinates for the entire xy-plane is expressed as
$${-\infty < x < \infty}$$
$${-\infty < y < \infty}$$
but the same region in polar coordinates isthe same region in polar coordinates becomes
$${0 \leq r < \infty}$$ (distance is $${0}$$ or greater)
$${0 \leq \theta < 2\pi}$$ (angle for one full rotation)
With this transformation, the integration region becomes a "rectangle" in the world of polar coordinates.
Transformation of the area element: The most important part
Infinitesimal area in Cartesian coordinates
In Cartesian coordinates, the area of an infinitesimal rectangle is simple.
$${\text{Infinitesimal area} = dx \times dy}$$
Infinitesimal area in polar coordinates
In polar coordinates, the infinitesimal region becomes a "sector." Let's look at this in detail.
Shape of the infinitesimal sector
Inner radius: $${r}$$
Outer radius: $${r + dr}$$
Central angle: $${d\theta}$$
Area calculation of the infinitesimal sector
Since the area formula for a sector was $${\displaystyle S = \frac{1}{2}R^2\theta}$$,
Area of the outer sector: $${\displaystyle S_{\text{outer}} = \frac{1}{2}(r + dr)^2 d\theta}$$
Area of the inner sector: $${\displaystyle S_{\text{inner}} = \frac{1}{2}r^2 d\theta}$$
Therefore, the infinitesimal area is the difference:
$${\displaystyle dS = S_{\text{outer}} - S_{\text{inner}} }$$
$${\displaystyle = \frac{1}{2}(r + dr)^2 d\theta - \frac{1}{2}r^2 d\theta }$$
$${\displaystyle = \frac{1}{2}d\theta[(r + dr)^2 - r^2] }$$
$${\displaystyle = \frac{1}{2}d\theta[r^2 + 2r \cdot dr + (dr)^2 - r^2] }$$
$${\displaystyle = \frac{1}{2}d\theta[2r \cdot dr + (dr)^2]}$$
Since $${dr}$$ is an infinitesimal quantity, $${(dr)^2}$$ can be ignored, so
$${\displaystyle dS = \frac{1}{2}d\theta \cdot 2r \cdot dr = r \cdot dr \cdot d\theta}$$
becomes.
Conclusion
$${dx dy = r dr d\theta}$$
Another way to understand it: Rectangular approximation
You can also approximate the infinitesimal sector as a rectangle.
Vertical length : $${dr}$$ (change in radial direction)
Horizontal length : $${r d\theta}$$ (arc length)
Therefore, the area is $${dr \times r d\theta = r dr d\theta}$$
Both methods yield the same result, with the coefficient $${r}$$ appearing.
Studying mathematics properly: The Jacobian
To begin with, what is the $${r}$$ that appears here? It is the adjustment function for variable transformation that appeared in the case of $${1}$$ variable.
For example,
$${\displaystyle \int x(2x^2+1)^3 dx}$$
To calculate this, you could expand it, but if you let $${t=2x^2+1}$$, then since $${\displaystyle \frac{dt}{dx} = 2 \times 2x = 4x}$$, we have $${\displaystyle \frac14 dt = x dx}$$, so
$${\displaystyle \int x(2x^2+1)^3 dx = \int t^3 \times \frac14 dt}$$
and you can perform a variable transformation.
The $${2}$$-variable version of this $${\displaystyle \frac14dt = xdx}$$ is the Jacobian $${J}$$, which is the determinant of the Jacobian matrix.
When performing a variable transformation from $${(x, y)}$$ to $${(s, t)}$$,
$${J = \begin{vmatrix} \frac{\partial x}{\partial s} & \frac{\partial x}{\partial t} \\ \frac{\partial y}{\partial s} & \frac{\partial y}{\partial t} \end{vmatrix}}$$
This is how it works. $${\displaystyle \frac{\partial f}{\partial x}}$$ is the multivariable version of $${\displaystyle \frac{df}{dx}}$$, meaning to (partially) differentiate the function $${f}$$ with respect to the variable $${x}$$. Here, I used $${J}$$ to represent the Jacobian, but since it can also represent the Jacobian matrix, it is necessary to check which meaning is being used when looking at textbooks or papers.
Using this, we can write:
$${\displaystyle \int\int_D f(x, y) dxdy = \int\int_E f(x(s, t), y(x, t))|J|dsdt}$$
This allows us to perform the calculation.
In other words, in this case,
$${\begin{cases} x = r \cos \theta \\ y = r \sin \theta \end{cases}}$$
so,
$${J = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix}}$$
$${\displaystyle = \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix}}$$
$${= \cos\theta \times r \cos\theta - \sin\theta \times (-r \sin\theta)}$$
$${= r(\cos^2\theta+\sin^2\theta) = r}$$
Since this results in $${|J| = r}$$, we obtain the same result this way as well.
Performing the integration
Integral after transformation
By polar coordinate transformation,
$${\displaystyle I^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2)} dx dy = \int_0^{2\pi} \int_0^{\infty} e^{-r^2} \times r dr d\theta}$$
we get the following.
Separation of variables
The integrand $${e^{-r^2} \times r}$$ does not depend on the variable $${θ}$$, meaning it is a constant function with respect to $${θ}$$, so
$${\displaystyle I^2 = \int_0^{2\pi} d\theta \int_0^{\infty} e^{-r^2}r dr}$$
$${\displaystyle = \left(\int_0^{2\pi} d\theta\right) \times \left(\int_0^{\infty} e^{-r^2} r dr\right)}$$
This means we can calculate each integral separately and then take their product.
Integral with respect to θ (the simple part)
$${\displaystyle \int_0^{2\pi} d\theta = [\theta]_0^{2\pi} = 2\pi - 0 = 2\pi}$$
This is one full revolution of the circumference, which is the expected result.
Integral with respect to r (the essential part)
$${\displaystyle \int_0^{\infty} e^{-r^2} r dr}$$
This integral can be calculated using substitution.
If we let $${u = r^2}$$,
then $${du = 2r dr}$$ and so $${\displaystyle r dr = \frac{1}{2} du}$$.
Transformation of the integration range
$${\begin{array}{c|ccc} r & 0 & \rightarrow & \infty \\ \hline u & 0 & \rightarrow & \infty \end{array}}$$
This is the result.
Execution of the integral
$${\displaystyle \int_0^{\infty} e^{-r^2} r dr = \int_0^{\infty} e^{-u} \frac{1}{2} du}$$
$${\displaystyle = \frac{1}{2} \int_0^{\infty} e^{-u} du}$$
$${\displaystyle = \frac{1}{2} [-e^{-u}]_0^{\infty}}$$
$${\displaystyle = \frac{1}{2} [0 - (-e^0)]}$$
$${\displaystyle = \frac{1}{2} [0 + 1] = \frac{1}{2}}$$
This is the result.
Confirmation of Improper Integrals
Since $${e^{-u} \rightarrow 0}$$ converges rapidly as $${u \rightarrow \infty}$$, the integral converges.
Final Result
From the above,
$${\displaystyle I^2 = 2\pi \times \frac{1}{2} = \pi}$$
therefore,
$${\displaystyle I = \sqrt{\pi}}$$
In other words,
$${\displaystyle \int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}}$$
was obtained.
Confirmation of the Beautiful Result
Review of the Calculation Process
Problem Setup: Find $${I = \int_{-\infty}^{\infty} e^{-x^2} dx}$$
Change of Perspective: Calculate $${I^2}$$
Dimensional Expansion: Convert the 1D integral into a 2D integral
Coordinate Transformation: Utilize circular symmetry in polar coordinates
Separation of Variables: Separate into integrals of angle and distance
Substitution Integration: Reduce to a standard integral using $${u = r^2}$$
Significance of Each Step
Polar Coordinate Transformation: Essentially becomes a 1D problem due to circular symmetry
Area Element of $${r}$$: This coefficient makes the integration possible
Substitution Integration: Finally solved using high school mathematics techniques
Preparation for Application to the Normal Distribution
Conversion to the Standard Normal Distribution
What we found was
$${\displaystyle \int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}}$$
However, what is needed for the standard normal distribution is
$${\displaystyle \int_{-\infty}^{\infty} e^{-\frac{x^2}{2}} dx}$$
We will perform this conversion next time using substitution integration.
Preview of the Next Installment
Next time, we will finally derive the coefficient of the normal distribution, $${\displaystyle \frac1{\sqrt{2\pi}}}$$, in its entirety.
Integral transformation from $${e^{-x^2}}$$ to $${e^{-\frac{x^2}2}}$$
Determination of the coefficient
Extension to the general normal distribution ($${\mu \neq 0, \sigma \neq 1}$$)
The Beauty of Mathematics: The Appearance of Pi
Why does π appear?
Why does $${\pi}$$, a constant of a "circle," appear in the normal distribution, which should be a problem of a "straight line"?
The answer lies in polar coordinate transformation. When we integrate a circularly symmetric function, we are essentially using the geometry of a "circle."
$${2\pi}$$ is the circumference angle of a circle, and $${r dr d\theta}$$ is the area element of a circle.
Deep geometric structures were hidden behind the normal distribution.
The Genius of Gauss
The beauty of Gauss's idea is
Seeing the essence of the problem: Noticing circular symmetry
Choosing the right tools: Using polar coordinate transformation
Executing the calculation: Solving by combining techniques
This calculation is a perfect example of the fusion of mathematical creativity and technical precision.
Next time, we will complete this beautiful calculation. Finally, the true identity of the mysterious coefficient of the normal distribution, $$\displaystyle \frac1{\sqrt{2\pi}}$$, will be revealed.
To be continued