The given integral is:

$f(a) = \int_0^\infty \frac{e^{-ax} \sin x}{x} \, dx$ where $a \geq 0$.

This is a known standard integral, which can be evaluated using techniques from complex analysis or by using a standard result for this form of improper integral. The solution for this integral is:

$f(a) = \arctan\left(\frac{1}{a}\right)$

This result holds for $a \geq 0$. Thus, the function $f(a)$ for a continuous function defined by the given integral is:

$f(a) = \arctan\left(\frac{1}{a}\right), \quad \text{for } a \geq 0.$

Would you like further details or explanations on the solution?

Here are 5 related questions to further explore:

1. How is the integral of $\frac{\sin x}{x}$ over infinite limits typically handled?
2. How does one derive the standard result for this integral using contour integration?
3. What is the behavior of $f(a)$ as $a \to 0$ and $a \to \infty$?
4. Can you express the integral using Laplace transforms and what are the advantages of doing so?
5. How does this integral relate to Fourier transforms in signal processing?

Tip: When solving improper integrals, always check the convergence of the integral by examining the behavior at both the lower and upper limits.