Integral evaluation $\int_{-\infty}^{\infty}\frac{\cos (ax)}{\pi (1+x^2)}dx$ 
Possible Duplicate:
Calculating the integral $\int_{0}^{\infty} \frac{\cos x}{1+x^2}\mathrm{d}x$ without using complex analysis 

I need help to evaluate the integral of
$$\int_{-\infty}^{\infty}\frac{\cos (ax)}{\pi (1+x^2)}dx$$ where $a$ is a real number. I know the answer is $e^{-|a|}$. 
I think contour integration can do the job by evaluating $e^{iaz}/(z^2+1)$ but I don't remember the details from complex analysis in college. 
My question is "Is there a way to compute this integration WITHOUT using complex analysis method?" For example, by change of variables with trigonometric function, integration by part, or something like that.

Let $\displaystyle f(a) = \int_{0}^{\infty}\frac{\cos(ax)}{1+x^2}\;{dx}$. Consider the Laplace transform of $f(a)$.
$$\begin{aligned}\mathcal{L}(f(a)) & = \int_{0}^{\infty}\int_{0}^{\infty}\frac{\cos(ax)}{1+x^2}e^{-as}\;{da}\;{dx} \\&= \int_{0}^{\infty}\frac{s}{(1+x^2)(s^2+x^2)}\;{dx} \\& = \frac{\pi}{2(s+1)}.\end{aligned} $$
Thus $ \displaystyle f(a) =\mathcal{L}^{-1}\left(\frac{\pi}{2(s+1)}\right) =\frac{\pi}{2}e^{-|a|} $ and your integral is  $\displaystyle \frac{2}{\pi}f(a) = e^{-|a|}.$
