# Laplace Transform Integral

1. Jan 1, 2014

### pierce15

I posted this in the homework section, but I haven't received any help, so hopefully putting it in this section won't be an issue. I'm trying to compute the integral

$$\int_0^ \infty \frac{ \cos xt}{1 + t^2} \, dt$$

using the Laplace transform. The first thing that catches my eye is the 1 /(1 + t^2) factor, which is equal to the Laplace transform of sin x:

$$= \int_0^\infty \cos xt \, L[ \sin x ] \, dt$$

Any ideas?

2. Jan 2, 2014

### vanhees71

Why do you want to use the Laplace transformation here? I'd rather use the Fourier transformation. It's simpler to put it first in the exponential form. Your Integral is
$$F(x)=\frac{1}{2}\int_0^{\infty} \frac{\exp(\mathrm{i} x t)+\exp(-\mathrm{i} x t)}{1+t^2}.$$
Substituting $t'=-t$ in the second integral you get after some algebra
$$F(x)=\frac{1}{2} \int_{-\infty}^{\infty} \frac{\exp(\mathrm{i} x t)}{1+t^2}. \qquad (1)$$
Now we use the fact that
$$\int_{-\infty}^{\infty} \frac{\mathrm{d} x}{2 \pi} \exp(-|x|) \exp(-\mathrm{i} t x)=\frac{1}{\pi (1+t^2)}.$$
Thus (1) is (up to a factor $\pi/2$) the inverse of this Fourier transform. This gives
$$F(x)=\frac{\pi \exp(-|x|)}{2}.$$

3. Jan 2, 2014

### pierce15

Ok, thanks for that. Do you also see any way to do it with the Laplace transform?

4. Jan 3, 2014

### vanhees71

Hm, I've no idea. Perhaps you can somehow use the convolution theorem with some clever trick?