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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.

Consider a continuous function f(a) = ∫[0, ∞] (e^(-ax) * sin(x) / x) dx ; a >= 0, then

Learn how to solve the improper integral for the continuous function f(a) = ∫[0, ∞] (e^(-ax) * sin(x) / x) dx where a >= 0. This problem applies advanced calculus concepts, including the use of standard integral results and improper integrals.

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