In practical applications, especially in Civil Engineering, many physical phenomena such as heat conduction, vibrations in structures, and signal transmission are best described using functions that may not be periodic. While Fourier Series is a powerful tool for representing periodic functions, it becomes inadequate when dealing with non-periodic functions. To handle such cases, Fourier Integrals are introduced. A Fourier Integral allows us to express a non-periodic function as a continuous superposition of sines and cosines (or exponential functions), making it indispensable in solving engineering problems involving non-periodic boundary conditions and transient phenomena.

Fourier Series represents a function f(x) defined on a finite interval [−L,L] as an infinite sum of sines and cosines:

∞
f(x)=a_0 + ∑(a_n cos(nπx/L) + b_n sin(nπx/L))

But for non-periodic functions or when the interval becomes infinitely large (L→∞), the discrete nature of the Fourier coefficients becomes continuous, and the sum becomes an integral. This transition leads us to Fourier Integrals.

Let f(x) be a piecewise continuous function on (−∞,∞) that satisfies the Dirichlet conditions and is absolutely integrable, i.e.,

∫(−∞ to ∞) |f(x)|dx < ∞. We begin with the Fourier series of f(x) defined on [−L,L]:

f(x)= ∑(a_n cos(nπx/L) + b_n sin(nπx/L))

Let ω = nπ. As L → ∞, ω → ω becomes a continuous variable. The sum becomes an integral:

f(x)= ∫(0 to ∞) [A(ω)cos(ωx) + B(ω)sin(ωx)] dω

Where,

A(ω)= (1/π) ∫(−∞ to ∞) f(t)cos(ωt)dt, B(ω)=(1/π)∫(−∞ to ∞) f(t)sin(ωt)dt. This is the Fourier Integral Representation of f(x).

Let f(x) be a function such that:
• f(x) is piecewise continuous on (−∞,∞)
• f(x) is absolutely integrable over (−∞,∞)

Then,

f(x)= ∫(0 to ∞) [A(ω)cos(ωx) + B(ω)sin(ωx)] dω,

Where:
A(ω)= (1/π) ∫(−∞ to ∞) f(t)cos(ωt)dt, B(ω)=(1/π)∫(−∞ to ∞) f(t)sin(ωt)dt.

Alternatively, combining cosine and sine terms, we can express the function using the complex Fourier integral:

f(x)= (1/(2π)) ∫(−∞ to ∞) fb(ω)e^(iωx)dω,

Where:
fb(ω)= ∫(−∞ to ∞) f(t)e^(-iωt)dt. This is the Fourier Transform of f(x).

If f(x) is even (i.e., f(−x)=f(x)), then its Fourier integral contains only cosine terms:

f(x)= ∫(0 to ∞) A(ω)cos(ωx)dω.

If f(x) is odd (i.e., f(−x)=−f(x)), then its Fourier integral contains only sine terms:

f(x)= ∫(0 to ∞) B(ω)sin(ωx)dω.

This simplification is useful in boundary value problems involving symmetric domains.

For a function f(x) to possess a valid Fourier Integral representation:
1. f(x) must be piecewise continuous in every finite interval of R.
2. f(x) must be absolutely integrable over (−∞,∞).
3. Discontinuities must be finite and of finite magnitude.

If these conditions are satisfied, then:
lim (f(x+ϵ) + f(x−ϵ)) = 2f(x)
ϵ→0
at all points of continuity.

Fourier Integrals are particularly important in Civil Engineering for solving:
• Heat conduction problems in infinite or semi-infinite rods
• Vibration analysis of continuous beams or plates
• Dynamic analysis of structures subject to non-periodic loading
• Soil mechanics for propagation of stress waves
• Ground motion analysis during earthquakes

For example, temperature distribution in a long concrete beam due to an instantaneous point source can be solved using the Fourier integral method.

Example 1:
Evaluate the Fourier sine integral representation of the function:

f(x) = {
1, 0<x<a
0, x≥a
}

Solution:
Since f(x) is odd over (0,∞), use the sine integral:

f(x)= ∫(0 to ∞) B(ω)sin(ωx) dω.

Where:
B(ω)= (1/π) ∫(0 to a) sin(ωt) dt = (1−cos(ωa))/(πω).

Thus,
f(x)= ∫(0 to ∞) (1−cos(ωa))sin(ωx) dω/πω.

Example 2:
Find the Fourier cosine integral of f(x)=e^(-ax), a>0, for x>0.

Use:
f(x)= ∫(0 to ∞) A(ω)cos(ωx) dω.

Where:
A(ω)= (1/(π)) ∫(0 to ∞) e^(-at)cos(ωt) dt = 2a/(a^2 + ω^2).
Thus,
f(x)= ∫(0 to ∞) (2a/(a^2 + ω^2))cos(ωx)/π dω.

A useful result in Fourier Integrals is:

∫(0 to ∞) (sin(ωx)/ω) dω = {π/2, x>0; 0, x=0; −π, x<0}.

This integral frequently appears in solving boundary value problems using Fourier methods.

So far, we have discussed the Fourier integral in terms of sine and cosine functions. However, it is often more convenient and elegant to express it using complex exponentials. Let f(x) be an absolutely integrable function over (−∞,∞). Then the complex Fourier integral representation of f(x) is:

f(x)= (1/(2π)) ∫(−∞ to ∞) fb(ω)e^(iωx) dω,

Where the Fourier transform fb(ω) is defined as:

fb(ω)= ∫(−∞ to ∞) f(t)e^(-iωt) dt.

The inverse Fourier transform is:
f(x)= (1/(2π)) ∫(−∞ to ∞) fb(ω)e^(iωx) dω.

Understanding the properties of the Fourier Transform helps in efficiently solving various problems. Let f(x) ↔ fb(ω) denote the Fourier transform pair.

1. Linearity:
   F[af(x) + bg(x)] = a fb(ω) + b g b(ω).
2. Translation (Shift):
   • In the time domain:
   F[f(x−a)] = e^(−iωa) fb(ω).
   • In the frequency domain:
   F[e^(iaxf(x))] = fb(ω−a).
3. Scaling:
   F[f(ax)] = (1/|a|) fb(ω/a).
4. Differentiation:
   If f(x) is differentiable,
   F[(df(x)/dx^n)] = (iω)^n fb(ω).

This is particularly useful for solving PDEs.

Let’s look at how the Fourier Integral is applied in a real engineering scenario.

Problem: Heat Diffusion in a Semi-Infinite Rod.
A rod of infinite length initially at zero temperature receives an instantaneous unit heat source at x=0 at t=0. The governing equation is the one-dimensional heat equation:

∂u/∂t = α² ∂²u/∂x²
With initial condition:
u(x,0) = δ(x)
And boundary condition:
u(±∞,t) = 0

Taking the Fourier Transform in x, we convert the PDE to an ODE:
∂u(ω,t)/∂t = −α²ω²u(ω,t).
Solving gives:
u(ω,t) = e^(−α²ω²t).
Taking the inverse Fourier transform:

u(x,t) = (1/(2π)) ∫(−∞ to ∞) e^(−α²ω²t)e^(iωx) dω.
Using the standard Gaussian integral identity:
u(x,t) = (1/√(4πα²t)) e^(−(x²/4α²t)).

This is the heat kernel, a fundamental solution showing how heat diffuses through the rod — a crucial result for civil engineers analyzing thermal effects in structures.

Parseval's identity relates the energy of a signal in time domain to that in frequency domain. Let f(x) and g(x) be absolutely integrable functions. Then:

∫(−∞ to ∞) f(x)g(x) dx = (1/(2π)) ∫(−∞ to ∞) fb(ω)g b(ω) dω.

When f = g, it becomes:
∫(−∞ to ∞) |f(x)|² dx = (1/(2π)) ∫(−∞ to ∞)|fb(ω)|² dω.

This has practical importance in energy calculations, error estimation, and signal processing in structural monitoring systems.

Feature Fourier Series Fourier Integral
Applicable to Periodic functions Non-periodic functions
Domain Finite interval Infinite interval
Representation Discrete sum of Continuous integral
sines/cosines
Coefficients a_n,b_n A(ω),B(ω) or fb(ω)
Use in Engineering Vibrations of bounded Heat transfer, infinite
structures domain analysis.

To aid problem-solving, here are standard Fourier integral forms:
1. Fourier Sine Transform of f(x)=1, 0<x<a:
∫(0 to ∞) sin(ωx)dx = (1−cos(ωa))/ω.
2. Fourier Cosine Transform of e^(-ax):
∫(0 to ∞) e^(-ax)cos(ωx)dx = a/(a² + ω²).
3. Fourier Sine Transform of e^(-ax):
∫(0 to ∞) e^(-ax)sin(ωx)dx = ω/(a² + ω²).

1. Derive the Fourier cosine integral of f(x)=xe^(-x) for x>0.
2. Evaluate the Fourier sine integral of the step function:
   f(x)={
   1, 0<xL
   }
3. Use the complex Fourier integral to represent f(x)= 1/(1+x²).
4. Apply Fourier integral methods to solve the initial value problem for the one-dimensional wave equation.
5. Show that the function f(x)=e^(-a|x|) has a Fourier transform and compute it.