Marche quantique

April 21, 2026

Dirichlet Integral

1. Setup: The Square Wave and its Transform

Consider a square wave function $\psi(x)$ defined as:

$\psi(x) = \begin{cases} \frac{1}{\sqrt{2}}, & |x| \leq 1 \\ 0, & |x| > 1 \end{cases}$

The Fourier transform $\phi(k)$ is calculated as follows:

$\phi (k)=\frac{1}{\sqrt{2\pi}} \int_{-1}^{1} \frac{1}{\sqrt{2}} e^{-ikx} dx = \frac{1}{\sqrt{\pi}} \frac{\sin k}{k}$

2. Solving $\int_{-\infty}^{\infty} \frac{\sin^2 k}{k^2} dk$

By Parseval’s Theorem (Total Probability), the integral of the square of the wave function in position space must equal the integral of the square of its Fourier transform:

$\int_{-\infty}^{\infty}|\psi (x)|^2 dx = \int_{-\infty}^{\infty}|\phi (k)|^2 dk = 1$

Substituting $\phi(k) = \frac{1}{\sqrt{\pi}} \frac{\sin k}{k}$ into the probability identity:

$\frac{1}{\pi} \int_{-\infty}^{\infty} \frac{\sin^2 k}{k^2} dk = 1 \implies \int_{-\infty}^{\infty} \frac{\sin^2 k}{k^2} dk = \pi$

3. Solving $\int_{-\infty}^{\infty} \frac{\sin k}{k} dk$

We use the inverse Fourier transform to find the value of the wave function at the origin $x = 0$ :

$\psi(0) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \phi(k) dk$

Given $\psi(0) = \frac{1}{\sqrt{2}}$ and substituting our expression for $\phi(k)$ :

$\frac{1}{\sqrt{2}} = \frac{1}{\pi\sqrt{2}} \int_{-\infty}^{\infty} \frac{\sin k}{k} dk \implies \int_{-\infty}^{\infty} \frac{\sin k}{k} dk = \pi$

Xu Dong Liu

Applied Mathematics

Dirichlet Integral

1. Setup: The Square Wave and its Transform

2. Solving

3. Solving

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2. Solving $\int_{-\infty}^{\infty} \frac{\sin^2 k}{k^2} dk$

3. Solving $\int_{-\infty}^{\infty} \frac{\sin k}{k} dk$