The problem asks to evaluate the definite integral $I = \int_{0}^{\infty} \frac{\sin(x)}{x} dx$. This is a well-known improper integral.

1. Problem Description

The problem asks to evaluate the definite integral $I = \int_{0}^{\infty} \frac{\sin(x)}{x} dx$. This is a well-known improper integral.

2. Solution Steps

We can evaluate this integral using Laplace transforms. Define $I(t)$ as follows:

$I(t) = \int_0^{\infty} e^{-tx} \frac{\sin x}{x} dx$

Differentiate with respect to $t$:

$\frac{dI}{dt} = \int_0^{\infty} \frac{\partial}{\partial t} \left( e^{-tx} \frac{\sin x}{x} \right) dx = \int_0^{\infty} -xe^{-tx} \frac{\sin x}{x} dx = - \int_0^{\infty} e^{-tx} \sin x dx$

We know that:

$\int e^{ax} \sin(bx) dx = \frac{e^{ax}}{a^2 + b^2} (a \sin(bx) - b \cos(bx)) + C$

In our case, $a = -t$ and $b = 1$, so

$\int e^{-tx} \sin(x) dx = \frac{e^{-tx}}{t^2 + 1} (-t \sin(x) - \cos(x)) + C$

Then,

$\frac{dI}{dt} = - \left[ \frac{e^{-tx}}{t^2 + 1} (-t \sin(x) - \cos(x)) \right]_0^{\infty} = - \left( 0 - \frac{1}{t^2 + 1} (-t \cdot 0 - 1) \right) = -\frac{1}{t^2 + 1}$

Integrate with respect to $t$:

$I(t) = \int -\frac{1}{t^2 + 1} dt = -\arctan(t) + C$

Now, as $t \to \infty$, $I(t) \to 0$, so

$0 = -\arctan(\infty) + C = -\frac{\pi}{2} + C$, which implies $C = \frac{\pi}{2}$.

Therefore, $I(t) = -\arctan(t) + \frac{\pi}{2}$.

We want to find $I(0) = \int_0^{\infty} \frac{\sin x}{x} dx$, so we evaluate $I(t)$ at $t=0$:

$I(0) = -\arctan(0) + \frac{\pi}{2} = -0 + \frac{\pi}{2} = \frac{\pi}{2}$.

3. Final Answer

$\frac{\pi}{2}$