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Playing with Mathematica, I found that $$\Re\left[\Gamma\left(-\frac{n}{2},x\right)\right]=\Gamma\left(-\frac{n}{2}\right)$$ for $x<0$ and $n$ any positive integer.

Would it be possible to prove that?

On Wikipedia, I found a formula for the incomplete Gamma function when the first argument is a negative integer: $$\Gamma(-n,z)=\frac{1}{n!}\left(\frac{e^{-z}}{z^n}\sum_{k=0}^{n-1}(-1)^k(n-k-1)!z^k+(-1)^k+(-1)^n\Gamma(0,z)\right)$$

Maybe there is a similar formula when the first argument is a negative half-integer!

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    $\begingroup$ So the real part of $\Gamma(-n/2. x)$ does not depend on $x$? Seems unlikely. $\endgroup$ – marty cohen Dec 2 '17 at 20:23
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    $\begingroup$ That's why it's intriguing... It should depend on both variables but it doesn't. $\endgroup$ – Jennifer Dec 2 '17 at 23:47
  • $\begingroup$ @ marty Cohen: You doubing Thomas! $\endgroup$ – Jennifer Dec 2 '17 at 23:47
  • $\begingroup$ This is the incomplete gamma function and depends on both the order $n$ and the lower integration bound $x$. $\endgroup$ – Jessica Dec 2 '17 at 23:50
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    $\begingroup$ @Jennifer: The following seems to confirm your conjecture (emphasis mine): Ian Thompson, Algorithm 926: "Incomplete Gamma Functions with Negative Arguments", ACM Trans. Math. Software 39 (2), Article 14. "This article is concerned with the computation of $\gamma(n + 1/2,−x)$, where $n$ is an integer and $x > 0$. It turns out that these are purely imaginary quantities, and so, once the lower function is computed, $\Gamma(n + 1/2,−x)$ can be obtained from (3) without the loss of any significant digits." where equation (3) is $\gamma(a, t) = \Gamma(a) - \Gamma(a,t)$. $\endgroup$ – njuffa Dec 4 '17 at 22:52
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By dlmf 8.4.6 $$ \Gamma\left(\tfrac{1}{2},z\right)=2\int\limits_{\sqrt{z}}^{\infty}e^{-t^{2}}dt. $$ For negative arguments we have $$ \Gamma\left(\tfrac{1}{2},-|z|\right)=2\int\limits_{i\sqrt{|z|}}^{\infty}e^{-t^{2}}dt=2\int\limits_{i\sqrt{|z|}}^{i\sqrt{|z|}\,+\infty}e^{-t^{2}}dt=2\int_{0}^{+\infty}e^{-\left(t+i\sqrt{|z|}\right)^{2}}dt. $$ $$ \text{Re}\left\{\Gamma\left(\tfrac{1}{2},-|z|\right)\right\}=2\int_{0}^{+\infty}e^{-\left(t^2-|z|\right)}\cos\left(2t\sqrt{|z|}\right)dt=2e^{|z|}\cdot\frac{\sqrt{\pi}}{2}e^{-|z|}=\Gamma\left(\frac12\right). $$ By dlmf 8.8.2 $$ \Gamma\left(a+1,z\right)=a\Gamma\left(a,z\right)+z^{a}e^{-z}. $$ From this one can easily deduce by induction that the claim is true for all $n\in\mathbb{Z}$: $$\text{Re}\left\{\Gamma\left(n+\tfrac{1}{2},-|z|\right)\right\}=\Gamma\left(n+\tfrac{1}{2}\right).$$

Numerical checked confirmed this formula for non-negative $n$ too.

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This is a partial answer I hope will provide a useful starting point. I confirmed the asker's numerical observations with the web version of Mathematica function evalution and the Keisan high-precision calculator. I found a recent paper which gives information in its introduction that also confirms the asker's conjecture:

Ian Thompson, "Algorithm 926: Incomplete Gamma Functions with Negative Arguments", ACM Transactions on Mathematical Software, Volume 39, Issue 2, February 2013, Article No. 14

This article is concerned with the computation of $\gamma (n+1/2,−x)$, where $n$ is an integer and $x>0$. It turns out that these are purely imaginary quantities, and so, once the lower function is computed, $\Gamma(n+1/2,−x)$ can be obtained from (3) without the loss of any significant digits.

The referenced equation (3) is the well-known identity $\gamma(a,t)=\Gamma(a)−\Gamma(a,t)$. The paper doesn't go into details as to why $\gamma(n+1/2,-x)$ is always a purely imaginary quantity (at least not in any form that I, as a non-mathematician, recognize as such). It provides the following expansion which may be useful:

$$ \begin{align} S_{n}(x) = e^{-x} \sum_{j=0}^{\infty} \frac{x^{j}}{j!(n+\frac{1}{2}+j)}, \enspace \enspace n \in{}\mathbb{Z}, \enspace x \ge 0, \\ \gamma(n+\frac{1}{2},-x) = i(-1)^{n}e^{x}x^{n+1/2}S_{n}(x). \end{align} $$ It also notes the recurrence relation $$(n+\frac{1}{2})S_{n}(x) + xS_{n+1}(x)=1.$$

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    $\begingroup$ Thank you, you are the only person who understands me. $\endgroup$ – Jennifer Dec 5 '17 at 0:48
  • $\begingroup$ @Jennifer You might want to consider contacting Dr. Thompson at the University of Liverpool as to the derivation of the above formula for $\gamma(n+1/2,-x)$. His paper makes reference to frequently cited works by D.E. Amos and W. Gautschi on the incomplete gamma functions, as well as chapter 8 of DLMF but I have not yet found anything there that motivates the formula in a straightforward manner and illuminates why the results are always purely imaginary. $\endgroup$ – njuffa Dec 5 '17 at 3:06
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    $\begingroup$ If you accept the expansion with $S_n$ the result is trivial: Because the $S_n$ are real, $\gamma(n+\frac{1}{2},-x)$ is purely imaginary. This is also stated in the introduction of the manual included in the Algorithm archive calgo.acm.org/926.zip $\endgroup$ – gammatester Dec 5 '17 at 9:31
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$\Gamma(-\frac{1}{2}, -1)$ has real part $-0.8232\dots$; the real part of $\Gamma(-\frac{1}{2}, -50)$ is about $2.1663\times 10^{18}$.

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    $\begingroup$ @-Anomaly: WolframAlpha gives values that are completely different from yours. $\endgroup$ – Jessica Dec 4 '17 at 0:25
  • $\begingroup$ @Jessica: Those values of from Mathematica, which I would trust more than WolframAlpha. You also may be using a different definition of the incomplete gamma function than I am (which would affect the raw numbers, but it's not constant for any reasonable definition I've seen). $\endgroup$ – anomaly Dec 4 '17 at 3:43
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    $\begingroup$ For $\Gamma\left(-\frac{1}{2}, -1\right)$, online Mathematica function evaluation gives $-3.54490770181103 + 0.41404332671064 i$, and Keisan hgih-precision calculator gives $-3.544907701811 +0.41404332671064 i$ $\endgroup$ – njuffa Dec 4 '17 at 21:15
  • $\begingroup$ For $\Gamma\left(-\frac{1}{2}, -50\right)$, online Mathematica function evaluation gives $-3.544907701811032 + 1.5128191390659211439491117751052003 \times 10^{19} i$, and Kaisan high-precision calculator gives $-3.544907701811 +1.5128191390659 \times 10^{19} i$ $\endgroup$ – njuffa Dec 4 '17 at 21:24
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    $\begingroup$ I don't understand why Anomaly wouldn't trust WolframAlpha. $\endgroup$ – Jennifer Dec 5 '17 at 0:49

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