analytic-continuationcomplex-analysisintegrationsequences-and-series
Given
$$f_1(z)=\int_0^\infty t^{z-1}e^{-t}dt$$
where $\operatorname{Re}(z)>0$
$$f_2(z)=\sum_{n=0}^\infty\frac{(-1)^n}{(n+z)(n!)}+\int_1^\infty t^{z-1}e^{-t}dt$$
except for the values $z=0,-1,\ldots$.
How to prove that $f_2$ is an analytic continuation of $f_1$?
I have been simply trying to represent function $f_2$ as a Taylor series of the $f_1$'s to show that $f_1$ is equal to $f_2$ in the intersecting domains. But dont know how to do this.
Analytic continuation is not as universally applicable as we might sometimes want. The identity theorem says that if $f$ and $g$ are analytic on a open connected domain $D$ and agree on an open subset of $D$ (or even on a set of points that has am accumulation point in $D$) then $f=g$ throughout $D$. But notice that $f$ and $g$ must first both be analytic on $D$. You cannot always guarantee that you can extend $f$ from a smaller to larger domain unless you first know such an extension will work.
An example is as follows. We know one version of $\log z$ can be defined that is analytic on the domain
\begin{align}
D_1 = \{ z: 1 < |z| < 3, |\arg(z)| > \pi/8\}.
\end{align}
Using a power series centred at $z = 2$ we can create a second function analytic in the disc $D_2 = \{z: |z-2| < 2\}$ and which agrees with our $\log z$ on the part of $D_1 \cap D_2$ for which $\Im z > 0$. But this will not then agree with our $\log z$ on the intersection $D_1 \cap D_2$ where $\Im z < 0$. That is because there is no analytic function on a domain that circles the origin and at the same time matches our $\log z$ throughout $D_1$. In this example the identity theorem cannot be used as such an extension does not exist.
$D_1$ and $D_2$" />
Turning to your example, you can extend your power series into a larger domain, but you may end up with different branches depending on how your extension progresses around the singularity, which in your case is at $z=1$. But as soon as you try to make your extension circle back to the first domain, the identity theorem ceased to apply.
Think of the composite expression $w=\sqrt{z^2+1}$ as being constructed in two stages: $w=\sqrt{\zeta}$ with $\zeta = z^2+1$. Since the square root always has two values, the expression $w$ is "double-valued". To eliminate the ambiguity (i.e., to create a single-valued continuous solution) one starts initially by constructing a local solution that is single-valued and continuous in a small region around a reference point $z_0$. This local solution is called a germ. Next one tries to extend this local solution to a single-valued continuous function on as large a region in the $z$ plane as possible. Usually this is performed by what is called analytic continuation along a path. (If one gets too greedy, and chooses too large a region, then as one traces the values of $w$ in its plane as $z$ travels on a closed loop path, the values of $w$ need not return to their original value. In such cases we have failed to eliminate the ambiguity, and failed to obtain a single-valued function.
To construct a single-valued function defined on a large region $G$ in the $z$ plane, one usually constructs $G$ by introducing branch cuts.
The standard branch cut for $\sqrt{\zeta}$ is the negative real axis of the $\zeta $plane, say $\zeta = -t$ with $t>0$. To find the corresponding branch cuts in the $z$ plane find all solutions of $z^2+1= -t$. The solutions are $z=\sqrt{-(1+t)} =\pm i \sqrt{1+t}$ which can be sketched as two mirror-symmetric rays in the $z$ plane that both lie on the imaginary axis. One ray travels up from $i$ toward $\infty$ and the other ray travels downward from $-i$. When you delete these branch cuts from the $z$ plane, what remains is a simply-connected region $G$. This ensures that the local solution must extend to a global single-valued solution on $G$. Thus there are two global solutions that we can denote by $w= g_1(z)$ and $w= g_2(z)$ defined on $G$. They satisfy $g_1(z)= - g_2(z)$. Intuitively, we have "disentangled" the two solutions by splitting the solution set into two large pieces called sheets.
It is an interesting advanced topic to explore how the two sheets fuse together along the branch cuts. As you cross a branch cut, one solution smoothly matches up with its opposite solution $g_1(z) \to g_2(z)$. You can make a physical model of this process by making two sheets of clear plastic that are copies of $G$ (slit). Stack them over one another and imagine drawing level contours of $Re(w)$ and $Im(w)$ on these sheets. One set of contours will smoothly join with those of the other sheet as you cross each slit.
Best Answer
Hints: $$\int_1^\infty t^{z-1}e^{-t}dt$$ is an entire function. When $\Re z >0$ we can write $$\int_0^1 t^{z-1}e^{-t}dt$$ as $\sum \frac {(-1)^{n}} {(n+z)n!}$ by expanding $e^{-t}$ in its Taylor series and integrating term by term. But now we can see that this series converges for any $z$ which is not of the form $-n$ with $n \in\{0,1,2...\}$. The sum is analytic in $\mathbb C \setminus \{0,-1,-2,...\}$ because the series converges uniformly on compact subsets of this region. Thus we have found an analytic continuation of $$\int_0^\infty t^{z-1}e^{-t}dt$$ to $\mathbb C \setminus \{0,-1,-2,...\}$.