The problem is to evaluate the definite integral $I = \int_{0}^{\infty} \frac{\sin(x)}{x} dx$.

1. Problem Description

The problem is to evaluate the definite integral $I = \int_{0}^{\infty} \frac{\sin(x)}{x} dx$.

2. Solution Steps

This integral is a classic example and can be solved using various methods, such as Laplace transforms or contour integration. Here, we will use Laplace transforms.

Let $I(t) = \int_{0}^{\infty} \frac{\sin(x)}{x} e^{-tx} dx$. We want to find $I(0)$. Differentiating with respect to $t$, we have:

$\frac{dI(t)}{dt} = \frac{d}{dt} \int_{0}^{\infty} \frac{\sin(x)}{x} e^{-tx} dx = \int_{0}^{\infty} \frac{\partial}{\partial t} (\frac{\sin(x)}{x} e^{-tx}) dx = \int_{0}^{\infty} \frac{\sin(x)}{x} (-x e^{-tx}) dx = -\int_{0}^{\infty} \sin(x) e^{-tx} dx$

Now we need to evaluate the integral $\int_{0}^{\infty} \sin(x) e^{-tx} dx$. We can use integration by parts twice or recall that

$\int_{0}^{\infty} e^{-ax} \sin(bx) dx = \frac{b}{a^2+b^2}$

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

$\int_{0}^{\infty} e^{-tx} \sin(x) dx = \frac{1}{t^2+1}$

Therefore, $\frac{dI(t)}{dt} = -\frac{1}{t^2+1}$.

Integrating with respect to $t$, we get

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

Now, we need to determine the constant $C$. As $t$ approaches infinity, $I(t)$ approaches 0 because the exponential term $e^{-tx}$ makes the integral vanish. Thus, $\lim_{t \to \infty} I(t) = 0$.

Also, $\lim_{t \to \infty} -\arctan(t) = -\frac{\pi}{2}$.

So, $0 = -\frac{\pi}{2} + C$, which means $C = \frac{\pi}{2}$.

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

We want to find $I(0)$, which is $\int_{0}^{\infty} \frac{\sin(x)}{x} dx$.

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

3. Final Answer

$\frac{\pi}{2}$