Let \(f(x)\) and \(g(x)\) be a pair of Fourier transforms. Show that \(d f / d x\) and \(\operatorname{ixg}(x)\) are a pair of Fourier transforms. Hint. Differentiate the first integral in (4.2) under the integral sign with respect to \(x\). Use ( \(5.16\) ) to show that $$ \int_{-x}^{x} \alpha|g(x)|^{2} d x=\frac{1}{2 \pi} \int_{-x}^{x} \bar{f}(x) \frac{d}{d x} f(x) d x $$ Comment : This result is of interest in quantum mechanies where it would read, in the notation of Problem \(5.24\) $$ \int_{-x}^{x} p|\phi(p)|^{2} d p=\int_{-x}^{x} \psi^{*}(x)\left(\frac{-i h}{2 \pi} \frac{\partial}{\partial x}\right) \psi(x) d x $$

Differentiation under the integral sign is a powerful technique used to evaluate integrals where the integrand contains a parameter. It involves differentiating the integral with respect to this parameter while keeping the integral bounds constant. This method is invaluable in simplifying complex integrals and finding relationships between functions. In the context of Fourier transforms, it assists in deriving important properties and solving differential equations involving transformed functions.

For example, when we differentiate the Fourier transform integral with respect to a variable (e.g., \(x\)), we make use of this technique to manage the interplay between differentiation and integration. This principle can also be extended to more complex applications in physics, such as quantum mechanics.

Here’s a simplified example:
Suppose \( g(x, a) \) is a function where \(a\) is a parameter. If we need to differentiate \( \int_a^b \ g(x, a) \ dx \) with respect to \(a\), the differentiation under the integral sign approach lets us interchange the differentiation and integration operations.

The product rule in calculus allows us to differentiate a product of two functions. If we have two functions, \( u(x) \) and \( v(x) \), the product rule states:
\[ \frac{d}{dx} \ [u(x) v(x)] = u'(x)v(x) + u(x)v'(x). \]
This rule is essential when dealing with Fourier transforms, as seen in the exercise. When differentiating \(f(x) e^{-ikx}\) with respect to \(x\), we apply the product rule:

\[\frac{d}{dx} [f(x) e^{-ikx}] = f'(x) e^{-ikx} + f(x) (-ik)e^{-ikx}.\]
By using the product rule we break the differentiation into manageable parts, which then helps us in identifying the Fourier transform pairs. Essentially, it helps in revealing how derivatives interact inside the integral, and showcases the linearity and the symmetry properties of the Fourier transforms.

Quantum mechanics is a fundamental branch of physics that explains the behavior of particles at the atomic and subatomic levels.

In the context of Fourier transforms, quantum mechanics employs these mathematical tools to analyze wave functions and probability amplitudes. For instance, the exercise references integrals often seen in quantum mechanics, with the Fourier pairs representing physical quantities like position and momentum.

In our earlier example, the integral involving \(p|\phi(p)||\) and the wave function expressions relate to how particle momentum and positions are treated in the quantum domain. Here, Fourier transforms provide a bridge between the wave and particle descriptions of quantum states, offering a clearer understanding of phenomena like wave-particle duality.

So, when solving or understanding problems in quantum mechanics, familiarity with Fourier transforms and operations such as differentiation under the integral sign becomes crucial.

Fourier transforms have specific properties that make them a powerful tool in both mathematics and physics. Some critical properties include:

• Linearity: The transform of a sum of functions is the sum of their transforms.
• Scaling: Scaling the function’s argument scales the transform’s argument inversely.
• Duality: The Fourier transform of the Fourier transform of a function is the original function reversed.
• Differentiation: Differentiating a function in time (or space) domain corresponds to multiplying its transform by a polynomial.

In the provided problem, using differentiation confirms these properties. When we show the Fourier transform pair for the derivative \ (f'(x)\) and the functional transform becomes \(-ikF(k)).\ This highlights the property that differentiation in the time domain translates to multiplication by \(-ik\) in the frequency domain.

Understanding these properties allows us to exploit Fourier transforms to solve differential equations, analyze signals, and understand complex wave phenomena, all of which are fundamental in science and engineering.