Calculating Incomplete Gamma Function for Complex Arguments

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SUMMARY

The integral \(\int_{ix}^{i\infty} e^{-t} t^{-s-1}dt\) for \(x>0\) and \(s>0\) can be calculated using the incomplete gamma function \(\Gamma(-s, ix)\) as provided by Mathematica. The substitution \(u = -it\) transforms the integral's limits from \(ix\) to \(i\infty\), aligning it with the definition of the incomplete gamma function \(\Gamma(s,x) = \int_{x}^\infty dt~t^{s-1}e^{-t}\). This formal approach allows for the identification of the integral with the incomplete gamma function, leveraging its analytic continuation for complex arguments to justify the result.

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bruno67
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How do I calculate the integral

\int_{ix}^{i\infty} e^{-t} t^{-s-1}dt,
where x>0, s>0? Mathematica gives \Gamma(-s,ix), where \Gamma(\cdot,\cdot) is the incomplete gamma function, but I am not sure how to justify this formally.
 
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Substitute u = -it, so the integral is from x to inf.
 
The incomplete Gamma function is defined by the integral

\Gamma(s,x) = \int_{x}^\infty dt~t^{s-1}e^{-t}.

Replacing x with ix formally gives \Gamma(s,ix). However, the meaning of the integral with lower bound ix is really just formal, I think. You identify the integral with the incomplete Gamma function, and then you determine the "integral's" value by using the analytic continuation of the incomplete Gamma function for complex arguments.
 
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