I have to solve this partial equation using Fourier transform: $$\frac{\partial u_{(x,t)}}{\partial t} - \frac{\partial u_{(x,t)}}{\partial x} = 0$$ $$u_{(x,0)}=\frac{1}{1+x^2}$$ Fourier transforming on both sides with respect to x:
$\mathcal{F} [\frac{\partial u_(x,t)}{\partial t}] = \frac{\partial \mathcal{F}[u_{(x,t)}]}{\partial t}$ and $\mathcal{F} [\frac{\partial u_(x,t)}{\partial x}] = (i\omega ) \partial \mathcal{F}[u_{(x,t)}]$ Calling $U=\mathcal{F}[u_{(x,t)}]$ I have $$(i\omega)U_{(\omega, t)}=\frac{dU_{(\omega, t)}}{dt} \rightarrow U_{(\omega, t)}= C e^{i\omega t} \rightarrow C= U_{(\omega,0)}=\mathcal{F}[\frac{1}{1+x^2}]=\pi e^{|\omega|} $$(from tables)
Then $U_{(\omega,t)}=\pi e^{|\omega|} e^{i\omega t}$, so the transform of a function equals the product of the transform of two functions, so this tells me that the original funcition $u_{(x,t)}$ is the convolution of, say, f and g, where $\mathcal{F}[f]= \pi e^{|\omega|}$ and $\mathcal{F}[g]=\pi e^{i\omega t}$ $f(x)= \frac{1}{1+x^2}$ and for g(x) I use the definition of the anti-Fourier transform: $$g(x)=\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{i\omega t} e^{i\omega x} d\omega = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{i\omega (x+t)} d\omega = \delta _{(t+x)}$$
Is this correct? If that's the case, then $$u_{(x,t)}=(f*g)=\int_{-\infty}^{\infty} \frac{\delta _{[t+(x-y)]}}{1+y^2} dy$$
I'm not sure how to go from here. I'd appreciate any help.
EDIT: Based on Gonçalo's answer my guess is that I can make a change of varibles as $(t+x-y)=v \rightarrow dv=-dy$ so the integral becomes
$$- \int_{-\infty}^{\infty} \frac{\delta _{(v)} dv}{1+(t+x-v)^2}= \frac{-1}{1+(t+x)^2}$$