Consider the asymptotic $z \to \infty$ behaviour of the function $$ \tag 1 I_1(z) \equiv \int_0^\infty \frac{e^{-zt}}{1+t^4} dt.$$ This converges for $\Re(z) > 0$, and the asymptotic expansion $$ \tag 2 I_1(z) \sim \sum_{n=0}^\infty (-1)^n \frac{\Gamma(4n+1)}{z^{4n+1}}, \quad z \to \infty, \,\, \Re(z) > 0 $$ is easily obtained.
I now want to extend the function $I_1$ to the region of the complex plane with $\Im( z) > 0$, and to do this I define another function $I_2$ as $$ \tag 3 I_2(z) \equiv \int_0^{-i\infty} \frac{e^{-zt}}{1+t^4} dt = -i \int_0^\infty \frac{e^{izy}}{1+y^4} dy, $$ which converges for $\Im(z)>0$, as wanted. The asymptotic expansion of (3) is again easily shown to be $$\tag 4 I_2(z) \sim \sum_{n=0}^\infty (-1)^n \frac{\Gamma(4n+1)}{z^{4n+1}}, \quad z \to \infty, \,\, \Im(z) > 0 $$ which is identical to (2).
But now I'm being told that despite the apparences, $I_2(z)$ is not the analytic continuation of (2) in the region $\Im(z) > 0$. I don't understand how can this be the case: (2) and (4) being equal, shouldn't $I_1$ and $I_2$ coincide in the region $\Re(z)>0, \Im(z)>0$, where both are well defined?