I was reading a physics paper and the author made a statement that I didn't follow.
Suppose $p(t)$ is given by the Fourier transform of $\omega(E)$
\begin{equation} p(t) = \int_{-\infty}^{+\infty} e^{-iEt}\omega(E)dE, \end{equation}
where $\omega(E)$ is a function that is zero for $E \leq 0$. Now lets make the variable $t$ complex: $t = t_1 + it_2$ and define $p_1(t) \equiv p(-t)$. So:
\begin{equation} p_1(t) = \int_{-\infty}^{+\infty} e^{iEt_1}e^{-Et_2}\omega(E)dE. \end{equation}
Now he claims that $p_1(t)$ is an analytical function in the upper plane ($t_2 > 0$) because $\omega(E)$ is semi-finite (i.e. $\omega(E) = 0 $ for $E \leq 0$). Why is that? Isn't $p_1(t)$ analytical in the whole plane?
In my understanding for a complex function $f(x+iy) = u(x,y) + iv(x,y)$ to be analytic it must obey the Cauchy-Riemann equations:
\begin{equation}
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}
\end{equation}
and
\begin{equation} \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x} \end{equation}
Which the function $p_1(t)$ definitively obeys for all $t_1$ and $t_2$. Why is $E$ even important in this context?
For anyone interested, the paper is: Khalfin, L. A., Sov. Phys. JETP 6, 1053 (1958). My question is in the paragraph right after equation (2.5).
Thanks
$\omega(E)=0$ for $E<0$ means that the lower integration limit $E\to-\infty$ can be replaced with $E\to0$. So, in what remains of the integral, $E$ is positive. That means $-Et_2$, in the exponential, has the opposite sign of $t_2$; if $t_2$ is negative, then you're integrating $e^{+E}dE$ to infinity, which diverges (in general; it may converge for specific $\omega$).