We need to evaluate the definite integral $I = \int_{0}^{\infty} \frac{\sin(x)}{x} dx$. This is a classic improper integral, often called the Dirichlet integral.

1. Problem Description

We need to evaluate the definite integral $I = \int_{0}^{\infty} \frac{\sin(x)}{x} dx$. This is a classic improper integral, often called the Dirichlet integral.

2. Solution Steps

We introduce a parameter $t$ and consider the integral

$I(t) = \int_{0}^{\infty} e^{-tx} \frac{\sin(x)}{x} dx$.

Note that $I(0) = \int_{0}^{\infty} \frac{\sin(x)}{x} dx$ which is the integral we want to find, and $I(\infty) = 0$.

We differentiate $I(t)$ with respect to $t$:

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

We can evaluate this integral using integration by parts twice.

Let $u = e^{-tx}$ and $dv = \sin(x) dx$, so $du = -te^{-tx} dx$ and $v = -\cos(x)$.

Then $\int_{0}^{\infty} e^{-tx} \sin(x) dx = [-e^{-tx}\cos(x)]_{0}^{\infty} - \int_{0}^{\infty} t e^{-tx} \cos(x) dx = [0 - (-1)] - t\int_{0}^{\infty} e^{-tx} \cos(x) dx = 1 - t\int_{0}^{\infty} e^{-tx} \cos(x) dx$.

Now, let $u = e^{-tx}$ and $dv = \cos(x) dx$, so $du = -te^{-tx} dx$ and $v = \sin(x)$.

Then $\int_{0}^{\infty} e^{-tx} \cos(x) dx = [e^{-tx}\sin(x)]_{0}^{\infty} - \int_{0}^{\infty} -t e^{-tx} \sin(x) dx = 0 + t \int_{0}^{\infty} e^{-tx} \sin(x) dx$.

Substituting this back into the first integration by parts:

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

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

Then $\frac{dI}{dt} = - \frac{1}{1+t^2}$.

Integrating with respect to $t$, we have $I(t) = -\arctan(t) + C$.

As $t \to \infty$, $I(t) \to 0$, so $0 = -\arctan(\infty) + C = -\frac{\pi}{2} + C$, which gives $C = \frac{\pi}{2}$.

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

We want $I(0) = \int_{0}^{\infty} \frac{\sin(x)}{x} dx$, so we evaluate $I(0) = -\arctan(0) + \frac{\pi}{2} = 0 + \frac{\pi}{2} = \frac{\pi}{2}$.

3. Final Answer

$\frac{\pi}{2}$