Details on the definition of the upper incomplete Gamma function

Question

Consider the Gamma function $$ \Gamma(z)=\int_0^{\infty}e^{-t}t^{z-1}\,dt $$ and the function is well defined for $\Re(z)>0$. If we integrate from another point we get the so called incomplete Gamma function $$ \Gamma(z,x)=\int_x^{\infty}e^{-t}t^{z-1}\,dt $$ I want to know what $x$ can be in this formula. Every reference I check is sloppy on the details. can $x$ take complex values? if so what path am I integrating in?

Answer 1

Choosing the usual branch of $\log t$ analytic on $ \mathbb{C} \setminus (-\infty,0]$, so that $t^{s-1} = e^{(s-1)\log t}$ is analytic on $ \mathbb{C} \setminus (-\infty,0]$

If $z=a+ib \in \mathbb{C} \setminus (-\infty,0]$ and $s \in \mathbb{C}$ then $$\Gamma(s,z) = \int_z^{+\infty} t^{s-1} e^{-t}dt = \int_a^{+\infty} (t+ib)^{s-1} e^{-t-ib}dt \tag{1}$$

Note that $f(t) = t^{s-1} e^{-t}$ is analytic on $\mathbb{C} \setminus (-\infty,0]$ so it has a primitive $\lambda(t)$ analytic on $\mathbb{C} \setminus (-\infty,0]$. Check that $\displaystyle\lim_{\Re(t) \to \infty}\lambda(t) =\lambda(+\infty)$ has a well-defined unique value, so taking $C_z$ to be any curve going from $z$ to $+\infty$ we have $$\oint_{C_z} f(t)dt = \lambda(+\infty)-\lambda(z) = \Gamma(s,z)$$ and in particular we can choose $C_z$ to be the straight horizontal line $[a+ib,+\infty+ib)$ as in $(1)$

Choosing another branch of $\log t$ and $t^{s-1} = e^{(s-1)\log t}$ or equivalently continuing analytically $\Gamma(s,z)$ beyond the branch cut $(-\infty,0]$ you'll get another definition of the incomplete Gamma function, analytic on a different domain. This also proves that $\Gamma(s,z)$ has a branch point at $z=0$.

Answer 2

If $z$ is a whole number, then one can use a different formula for the incomplete Gamma function:

$$\Gamma(z,x)=(z-1)!e^{-x}\sum_{k=0}^{n-1}\frac{x^k}{k!}$$

Other such extensions follow easily through series expansions for any $x,z\in\mathbb C$:

$$\Gamma(z,x)=\Gamma(z)-z^{-1}x^ze^{-x}\ _1F_1(1;1+z;x)$$

Where $_1F_1$ is the Confluent Hypergeometric Function of the First Kind.

Taken from Wolfram.

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As far as the actual integral itself, one usually takes any path between the two points that does not cross any singularities or branch cuts, and as $e^{-t}t^{z-1}$ has no singularities for $z>1$ (and no branch cuts if you choose your cuts right), then

$$\int_x^{+\infty}e^{-t}t^{z-1}\ dt=\int_Ce^{-t}t^{z-1}\ dt$$

for any path $C$ from $x$ to $+\infty$.

For example, take $C:e^{i\theta},\theta\in[0,\pi]$:

$$\begin{align}-2&=\int_{+1}^{-1}(x+1)\ dx\\&=\int_C(z+1)\ dz\\&=\int_0^\pi(e^{i\theta}+1)\ d(e^{i\theta})\\&=i\int_0^\pi(e^{2i\theta}+e^{i\theta})\ d\theta\\&=i\int_0^\pi(\cos(2\theta)+i\sin(2\theta)+\cos(\theta)+i\sin(\theta))\ d\theta\\&=i\left[2i\right]\\&=-2\end{align}$$

Try any other path starting at $1$ and going to $-1$ and you will get the same result.

Supposing that $\Re(x)>0$, one could use the following:

$$\Gamma(z,x)=\int_{\Re(x)}^{+\infty}e^{-t}t^{z-1}\ dt+i\int_0^{\Im(x)}e^{-(\Re(x)-it)}(\Re(x)-it)^{z-1}\ dt$$

which would be a path from $x$ to $\Re(x)$, then to $+\infty$.