Q9Advanced Engineering Mathematics

Question

Prove \int_{0}^{\infty} e^{- x} cos x d x = \frac{2}{π} \int_{0}^{\infty} \frac{( λ ^{2} + 2 ) cos λ x}{λ ^{4} + 4} d λ

Answer 1

Fourier transforms are indispensable mathematical tools for solving integral equations and evaluating complex definite integrals over infinite domains. The problem presents an identity that can be elegantly derived by exploiting the symmetrical relationship between a function and its Fourier Cosine Transform.

1. Defining the Function and its Transform

Let us define a simple exponential decay function, $f (x) = e^{- x}$ (for $x \geq 0$ ). We will calculate its Fourier Cosine Transform, denoted as $F_{c} (λ)$ . The fundamental mathematical definition is:

F_{c} (λ) = \int_{0}^{\infty} f (x) cos (λ x) d x = \int_{0}^{\infty} e^{- x} cos (λ x) d x

This is a standard integral that appears frequently in engineering mathematics. It can be solved using repeated integration by parts, but its solution is a well-known standard result:

\int_{0}^{\infty} e^{- a x} cos (b x) d x = \frac{a}{a ^{2} + b ^{2}}

Applying this formula with $a = 1$ and $b = λ$ , we find the Fourier Cosine Transform of our exponential function:

F_{c} (λ) = \frac{1}{1 + λ ^{2}}

2. Reversing the Process: The Inverse Transform

The true power of Fourier analysis lies in its invertibility. If we know the frequency-domain representation $F_{c} (λ)$ , we can perfectly reconstruct the original time/spatial-domain function $f (x)$ using the Inverse Fourier Cosine Transform formula:

f (x) = \frac{2}{π} \int_{0}^{\infty} F_{c} (λ) cos (λ x) d λ

We substitute our known functions into this inverse formula: $f (x) = e^{- x}$ and $F_{c} (λ) = \frac{1}{1 + λ ^{2}}$ .

e^{- x} = \frac{2}{π} \int_{0}^{\infty} \frac{1}{1 + λ ^{2}} cos (λ x) d λ

By simply rearranging the terms (multiplying both sides by $\frac{π}{2}$ ), we immediately derive a profound integral identity:

\int_{0}^{\infty} \frac{cos ( λ x )}{1 + λ ^{2}} d λ = \frac{π}{2} e^{- x} (for x > 0)

This demonstrates that the integral of $\frac{c o s ( λ x )}{1 + λ ^{2}}$ over all positive frequencies $λ$ yields an exponential decay curve in the spatial domain $x$ .

3. Extensions and Applications

This fundamental identity serves as a stepping stone for solving more complex integrals. For example, by differentiating both sides of this identity with respect to the parameter $x$ (using Leibniz's rule for differentiating under the integral sign), we can instantly evaluate the integral of $\frac{λ s i n ( λ x )}{1 + λ ^{2}}$ without performing any further complex integration. This technique of exploiting transform symmetries is a cornerstone of advanced applied mathematics.

Q9Advanced Engineering Mathematics

Question

Answer

1. Defining the Function and its Transform

2. Reversing the Process: The Inverse Transform

3. Extensions and Applications