How to integrate $ \int_{-\infty}^{+\infty} \frac{\sin(x)}{x} \,dx $? 
Possible Duplicate:
Solving the integral $\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$? 

How can I do this integration using only calculus? 
(not Laplace transforms or complex analysis)
$$
\int_{-\infty}^{+\infty} \frac{\sin(x)}{x} \,dx
$$
I searched for solutions not involving Laplace transforms or complex analysis but I could not find.

Putting rigor aside, we may do like this:
$$\begin{align*}
\int_{-\infty}^{\infty} \frac{\sin x}{x} \; dx
&= 2 \int_{0}^{\infty} \frac{\sin x}{x} \; dx \\
&= 2 \int_{0}^{\infty} \sin x \left( \int_{0}^{\infty} e^{-xt} \; dt \right) \; dx \\
&= 2 \int_{0}^{\infty} \int_{0}^{\infty}  \sin x \, e^{-tx} \; dx dt \\
&= 2 \int_{0}^{\infty} \frac{dt}{t^2 + 1} \\
&= \vphantom{\int}2 \cdot \frac{\pi}{2} = \pi.
\end{align*}$$
The defects of this approach are as follows:


*

*Interchanging the order of two integral needs justification, and in fact this is the hardest step in this proof. (There are several ways to resolve this problem, though not easy.)

*It is nothing but a disguise of Laplace transform method. So this calculation contains no new information on the integral.

