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This is gamma function: $\Gamma (n) = \int_0^\infty x^{n-1}e^{-x}\,dx$ What will be Result if I add Imaginary Number to Exponential of Euler Gamma Function?

$$? = \int_0^\infty x^{n-1}e^{-ix}\,dx$$

where the $i^2=-1$

isn't it a new function!? it will and will not converge?

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2 Answers 2

Looks related to Fourier Transform: $$ \hat{f}(\xi)=\mathcal{F}f(x) = \int_{-\infty}^{\infty} f(x)\ e^{- 2\pi i x \xi}\,dx $$ A scaled version of your integral with limits $]-\infty,\infty[$ is given here:

308 | $\mathcal{F}x^n \rightarrow \left(\frac{i}{2\pi}\right)^n \delta^{(n)} (\xi)\,$| Here, $n$ is a natural number and $\textstyle \delta^{(n)}(\xi)$ is the $n$-th distribution derivative of the Dirac delta function. This rule follows from rules 107 and 301. Combining this rule with 101, we can transform all polynomials.

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No this is not a "new function", because the integral diverges for every $n$ (as $x\to0$ if $n\leqslant0$ and as $x\to+\infty$ if $n\geqslant0$).

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It diverges for every real number n, but not necessarily for complex numbers. –  Marcks Thomas Nov 24 '12 at 11:49
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@MarcksThomas Did you check that the set of complex numbers $n$ such that the integral converges was not... empty, before writing this comment? –  Did Nov 24 '12 at 11:53
    
I did check. It's not empty, but I'm still figuring out when it will and will not converge. –  Marcks Thomas Nov 24 '12 at 13:18
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@MarcksThomas I would be grateful if you could provide an explicit example of such a complex number $n$. –  Did Nov 24 '12 at 17:14
    
I think so, too. I'm interested. I did a computer plot of the integral with finite, variable upper bound of integration and $n = i - 0.25$. It does at least appear like it's homing in on something, but the convergence is very slow. That's not a proof, but it's a hint, at least. Also, the integral with finite upper integration bound can be expressed via the lower incomplete gamma function as $(-i)^n \gamma(n, ia)$, where $a$ is the upper bound. Not sure how the $\gamma$ behaves toward imaginary infinity in its second argument, though. –  mike4ty4 Dec 4 '12 at 0:27

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