# Math Help - Is there an easy way of doing this integral?

1. ## Is there an easy way of doing this integral?

At least for someone who hasn't taken any complex analysis and doesn't know any beyond the really basic stuff, is there an easy way to do the following integral?

$\displaystyle \int_{\mathbb R} \frac 1 \pi \frac{e^{itx}}{1 + x^2} \mbox{ } dx = e^{-|t|}$

I trudged up a proof, but it uses techniques that I don't know. Obviously it is equivalent to evaluating

$\int_{\mathbb R} \frac 1 \pi \frac{\cos(tx)}{1 + x^2} \mbox{ } dx$

but I get the feeling looking at the result that this is something you would want to evaluate using the first expression.

2. Originally Posted by Guy
At least for someone who hasn't taken any complex analysis and doesn't know any beyond the really basic stuff, is there an easy way to do the following integral?

$\displaystyle \int_{\mathbb R} \frac 1 \pi \frac{e^{itx}}{1 + x^2} \mbox{ } dx = e^{-|t|}$

I trudged up a proof, but it uses techniques that I don't know. Obviously it is equivalent to evaluating

$\int_{\mathbb R} \frac 1 \pi \frac{\cos(tx)}{1 + x^2} \mbox{ } dx$

but I get the feeling looking at the result that this is something you would want to evaluate using the first expression.
I don't know what you consider easy and what you know, but you can skirt the complex analysis by using the Laplace transform.

Let

$f(t)=\frac{1}{\pi}\int_{\mathbb{R}}\frac{\cos(tx)} {x^2+1}dx=\frac{2}{\pi}\int_{0}^{\infty}\frac{\cos (tx)}{x^2+1}dx$

And note that f is an EVEN function! This will be important later.

If we take the Laplace transform we get

$\mathcal{L}\{f\}=\frac{2}{\pi}\int_{0}^{\infty} \frac{s}{x^2+s^2} \cdot \frac{1}{x^2+1}dx$

Now if we use partial fraction decomposition with respect to x we get that

$\frac{s}{x^2+s^2}\cdot \frac{1}{x^2+1}=\frac{s}{s^2-1}\cdot \frac{1}{x^2+1}-\frac{s}{s^2-1}\cdot \frac{1}{x^2+s^2}$

So if we take the integral we get

$\mathcal{L}\{f\}=\frac{2}{\pi}\cdot \frac{s}{s^2-1}\int_{0}^{\infty}\frac{1}{x^2+1}- \frac{1}{x^2+s^2}dx$

$\mathcal{L}\{f\}=\frac{2}{\pi}\cdot \frac{s}{s^2-1}\left(\frac{\pi}{2}-\frac{\pi}{2s} \right)=\frac{1}{s+1}$

So if we take the Inverse Laplace transform we get

$f(t)=e^{-t}, \text{ if } t \ge 0$

But since we know f is even we get

$f(t)= \begin{cases}e^{-t}, \text{ if } t \ge 0 \\ e^{t} \text{ if } t < 0\end{cases}=e^{-|t|}$

3. Perfect!