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(\int_{-\infty}^{+\infty} e^{-x^2} dx)^{2 \cdot \frac{1}{2}}

@mathineer

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Why it worked

The video effectively uses visual mathematics to explain a complex concept, the Gauss integral. The step-by-step transformation of the integral, combined with clear on-screen text, makes the derivation accessible and engaging for viewers interested in mathematics and calculus.

Summary

The video visually demonstrates the calculation of the Gauss integral. It starts with the integral of e^(-x^2) dx from negative infinity to positive infinity, raised to the power of 1/2. The calculation proceeds through several steps, transforming the integral into polar coordinates and eventually simplifying to pi^(1/2).

Structure

1Initial integral setup
2Transformation to double integral
3Conversion to polar coordinates
4Jacobian determinant calculation
5Simplification and integration
6Final result

On-screen text

(\int_{-\infty}^{+\infty} e^{-x^2} dx)^{2 \cdot \frac{1}{2}} (\int_{-\infty}^{+\infty} e^{-x^2} dx \int_{-\infty}^{+\infty} e^{-y^2} dy)^{\frac{1}{2}} ((\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{-x^2} e^{-y^2} dx dy)^{\frac{1}{2}}) ((\int_{\mathbb{R}^2} e^{-(x^2+y^2)} d(x,y))^{\frac{1}{2}}) ((\int_{\mathbb{R}+ \times [0,2\pi]} e^{-(\rho \cos \theta)^2 + (\rho \sin \theta)^2} |\frac{d(x,y)}{d(\rho, \theta)}| d(\rho, \theta))^{\frac{1}{2}}) ((\int_{\mathbb{R}+ \times [0,2\pi]} e^{-(\rho^2 \cos^2 \theta + (\rho \sin \theta)^2)} \cdot |\begin{pmatrix} \frac{\partial}{\partial \rho} \rho \cos \theta & \frac{\partial}{\partial \theta} x(\rho, \theta) \\ \frac{\partial}{\partial \rho} y(\rho, \theta) & \frac{\partial}{\partial \theta} y(\rho, \theta) \end{pmatrix}| d \rho \theta)^{\frac{1}{2}}) ((\int\int e^{-\rho^2} (\cos^2 \theta + \sin^2 \theta) \cdot |\begin{pmatrix} \frac{\partial}{\partial \rho} \rho \cos \theta & \frac{\partial}{\partial \theta} \rho \cos \theta \\ \frac{\partial}{\partial \rho} \rho \sin \theta & \frac{\partial}{\partial \theta} \rho \sin \theta \end{pmatrix}| d\rho d\theta)^{\frac{1}{2}}) ((\int_{0}^{\infty} d\rho \int_{0}^{2\pi} d\theta \cdot |\begin{pmatrix} \cos \theta & -\rho \sin \theta \\ \sin \theta & \rho \cos \theta \end{pmatrix}|)^{\frac{1}{2}}) ((\int_{0}^{\infty} d\rho e^{-\rho^2} \int_{0}^{2\pi} d\theta \cdot (\cos \theta \rho \cos \theta + \rho \sin \theta \sin \theta))^{\frac{1}{2}}) ((\int_{0}^{\infty} d\rho \rho e^{-\rho^2} \int_{0}^{2\pi} d\theta)^{\frac{1}{2}}) ((-\pi \int_{0}^{\infty} d\rho \frac{\partial}{\partial \rho} (- \rho^2) e^{-\rho^2})^{\frac{1}{2}}) ((-\pi [ e^{-\rho^2} ]_{0}^{\infty})^{\frac{1}{2}}) ((-\pi [0-1])^{\frac{1}{2}}) \pi^{\frac{1}{2}}