# Adding imaginary number to exponential of Euler Gamma function

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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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
@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
@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

It converges by contour integration (quarter cake slice in the lower right quadrant, no residues):

$\int_0^\infty dz\, z^{n-1} e^{-iz} + \int_{-i \infty}^0 dz\, z^{n-1} e^{-iz} +$ $+ \lim_{a \rightarrow 0^-}\lim_{R \rightarrow \infty}\int_a^{-\pi/2}d\phi \,i R e^{i \phi} R^{n-1} e^{i(n-1) \phi} e^{-i R \cos \phi} e^{R \sin \phi} = 0$

so

$\int_0^\infty dz\, z^{n-1} e^{-iz} = \int_0^{-i \infty} dz\, z^{n-1} e^{-iz} = (-i)^n \int_0^\infty dy\, y^{n-1} e^{-y} = (-i)^n \Gamma(n)$ .

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How do you justify the interchange of the limits $a \to 0^-$ and $R \to +\infty$? From the contour integration, you get $$\lim_{R\to \infty} \int_0^{-\pi/2} iRe^{i\phi} R^{n-1} e^{i(n-1)\phi} e^{-iR\cos \phi} e^{R\sin \phi}\,d\phi,$$ and it is far from clear that what you wrote is correct. For real $n$, it is definitely incorrect. – Daniel Fischer Nov 7 '15 at 12:52