Find the Fourier Cosine transform 𝐹𝑐𝑒-𝑎𝑥 of f(x) = 𝑒-𝑎𝑥 where a>0

The Fourier Cosine transform is defined as:

F_c{f(x)} = √(2/π) ∫ from 0 to ∞ [f(x) cos(wx) dx]

We want to find the Fourier Cosine transform of f(x) = e^-ax.

So, we substitute f(x) = e^-ax into the Fourier Cosine transform formula:

F_c{f(x)} = √(2/π) ∫ from 0 to ∞ [e^-ax cos(wx) dx]

This integral can be solved using the method of integration by parts, where we let u = e^-ax and dv = cos(wx) dx.

Then, du = -a e^-ax dx and v = (1/w) sin(wx).

Using the formula for integration by parts ∫ u dv = uv - ∫ v du, we get:

F_c{f(x)} = √(2/π) [(1/w) e^-ax sin(wx) - ∫ (1/w) sin(wx) (-a e^-ax dx)] from 0 to ∞

Solving this integral gives:

F_c{f(x)} = √(2/π) [(1/w) e^-ax sin(wx) + (a/w) ∫ e^-ax sin(wx) dx] from 0 to ∞

This is another integral that can be solved by integration by parts.

After solving, we get:

F_c{f(x)} = √(2/π) [(1/w) e^-ax sin(wx) + (a/w) e^-ax cos(wx) - (a^2/w^2) ∫ e^-ax cos(wx) dx] from 0 to ∞

This integral is the same as the one we started with, so we can solve for it and get:

F_c{f(x)} = √(2/π) [(1/w) e^-ax sin(wx) + (a/w) e^-ax cos(wx) - (a^2/w^2) F_c{f(x)}] from 0 to ∞

Solving for F_c{f(x)} gives:

F_c{f(x)} = [√(2/π) (a/w) e^-ax cos(wx)] / [1 + (a^2/w^2)] from 0 to ∞

This is the Fourier Cosine transform of f(x) = e^-ax.