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?

  • By how you've defined it, $I_2(z) = -i I_1(-iz)$, so there is no analytic continuation being made. By definition $I_1(z)$ is analytic on $\operatorname{Re} z > 0$, so $I_2(z)$ is analytic on $\operatorname{Im} z > 0$. The first asymptotic in $(2)$ is valid as $\operatorname{Re} z \to +\infty$, and the second in $(4)$ is valid as $\operatorname{Im} z \to +\infty$. Does this answer your question? – Antonio Vargas Feb 25 '15 at 19:46
  • @AntonioVargas nonetheless the asymptotic expansions of the two are equal. Shouldn't this mean something? After all, I found another function which (apparently at least) is equal to $I_1$ in some region, so why shouldn't this be a valuable candidate for analytic continuation? – glS Feb 25 '15 at 22:24
  • No, $I_2$ is simply $I_1$ rotated $90^\circ$ in the complex plane (and multiplied by $-i$). It doesn't give you any extra information. – Antonio Vargas Feb 25 '15 at 22:47
  • @AntonioVargas suppose we didn't know the actual form of $I_2$, but only that its asymptotic expansion is (4). Doesn't this tell us anything about the relations between $I_1$ and $I_2$? – glS Feb 25 '15 at 23:00
  • Unfortunately no. Different functions may have the same asymptotic expansions. – Antonio Vargas Feb 25 '15 at 23:03

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