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I have the following expression for $\alpha,z>0$: \begin{equation} \pi\mathrm i-\Gamma(\alpha)(-1)^\alpha\Gamma(1-\alpha,-z). \end{equation} In the context of the problem I am looking at this expression should be real-valued. If we plot it as a function of $\alpha$, (I used $z=\pi$), we see that the imaginary part of the second term appears to be equal to $\pi\mathrm i$ (which it should be). If true, the imaginary terms cancel out and the value of the expression is real.

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Is there any way I can use the analytic continuation of the incomplete gamma function to separate the real and imaginary parts of the second term? I was able to do this for the case when $\alpha\in\Bbb N$ using this identity. Put another way, I am looking for the expression for \begin{equation} \Re\{\Gamma(\alpha)(-1)^\alpha\Gamma(1-\alpha,-z)\} \end{equation} when $\alpha,z>0$.

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  • $\begingroup$ Do you mean you want to get the imaginary part of the second term? $\endgroup$ – Szeto Jun 19 '18 at 14:46
  • $\begingroup$ @Szeto I know that the imaginary part of the second term must be $\pi\mathrm i$. I want to write the second term with its real and imaginary components separated so that I can cancel out the $\pi\mathrm i$'s. $\endgroup$ – Aaron Hendrickson Jun 19 '18 at 14:52
  • $\begingroup$ @Szeto The reason why I want to do this is because when I do numerical evaluation, round-off error results in a small imaginary term being in the result even though the solution should always be real-valued. $\endgroup$ – Aaron Hendrickson Jun 19 '18 at 14:54
  • $\begingroup$ Which branch of the incomplete gamma function did you choose? If you choose $$\Gamma(s,x)=x^s E_{1-s}(x)$$ and take the principal value, there could be no imaginary part at all! $\endgroup$ – Szeto Jun 19 '18 at 15:45
  • $\begingroup$ @Szeto I have assumed the principal-valued branch cut. There should be an imaginary part of $\pi\mathrm i$ as the plot in my post shows. Unless I am mistaken, $E_{\nu}(-x)$ for $x>0$ should have a non zero imaginary component. $\endgroup$ – Aaron Hendrickson Jun 19 '18 at 15:59
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I am going to provide the solution for \begin{equation} \Re\left\{(-z)^{\alpha-1}\Gamma(1-\alpha,-z)\right\}. \end{equation} All that is needed to get the solution for the original question is a bit of algebra.


Now, we have to evaluate the real component of the second term for the cases of $\alpha\in\Bbb N$ and $\alpha\notin\Bbb N$ separately.

Starting with the noninteger $\alpha$ case, we us DLMF $8.7.3$ to rewrite the incomplete gamma term. After a bunch of algebraic simplification we find \begin{equation} \Re\left\{(-z)^{\alpha-1}\Gamma(1-\alpha,-z)\right\}=% -\sum_{k=0}^\infty \frac{z^k}{k!(1-\alpha+k)}% -z^{\alpha-1}\Gamma(1-\alpha)\cos\pi\alpha. \end{equation} Then using $(1-\alpha+k)^{-1}=\frac{(1-\alpha)_k}{(1-\alpha)(2-\alpha)_k}$ we put the series term into the form of ${_1F_1}(a;b;z)$ to get

\begin{equation} \Re\left\{(-z)^{\alpha-1}\Gamma(1-\alpha,-z)\right\}=% \frac{{_1F_1}(1-\alpha;2-\alpha;z)}{\alpha-1}% -z^{\alpha-1}\Gamma(1-\alpha)\cos\pi\alpha,\quad\alpha\notin\Bbb N. \end{equation}

For the integer $\alpha$ case we employ this identity and note that $z>0$, $\log(-z)=\log z+\pi\mathrm i$. Again, with a good amount of algebraic simplification we find

\begin{equation} \Re\left\{(-z)^{\alpha-1}\Gamma(1-\alpha,-z)\right\}=% \frac{z^{\alpha-1}}{\Gamma(\alpha)}% \left(% e^{z}\sum_{k=0}^{\alpha-2}\frac{k!}{z^{k+1}}% -\operatorname{Ei}(z)% \right),\quad \alpha\in\Bbb N. \end{equation}

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