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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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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.2012-11-24
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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?2012-11-24
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    I did check. It's not empty, but I'm still figuring out when it will and will not converge.2012-11-24
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    @MarcksThomas I would be grateful if you could provide an explicit example of such a complex number $n$.2012-11-24
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    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.2012-12-04
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    @mike4ty4 Let me be more explicit: there is **NO** such complex number. None. Nada.2012-12-04
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    I'm sorry, but I don't think that is true. Looks like I made some mistakes in my comment, there. If we use $n = 0.75 + i$ instead of $n = i - 0.25$, then the incomplete gamma seems to approach a lim. as $a$ approaches both $0$ and $\infty$ -- namely, 0 in the first case, and some non-zero complex number otherwise. You can try it yourself with, say, Wolfram's calculator. The singularity at 0 seems mild enough that the integral does indeed converge. Note, again, not a proof, just an "empirical" result, but still...2012-12-07
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    According to Wolfram's calculator, $\gamma(0.75 + i, 10^{-6}i)$ has magnitude about $10^{-6}$. $\gamma(0.75 + i, 10^{9}i)$ is approximately $0.425960 - 0.291359i$. $\gamma(0.75 + i, 10^{16}i)$ is approximately $0.426703 - 0.290448i$. Veeery slow convergence, but convergence nevertheless, at least, so it seems. Now, perhaps there is something wrong with Wolfram's calculator and it explodes. Perhaps it settles to a tiny periodic cycle. Note that I don't claim this result proven, btw.2012-12-07
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    @mike4ty4 Your faith in technology is touching... :-) What W|A may compute, at best, is a limit of the integral from $a$ to $b$ when $a\to0^+$ and $b\to+\infty$ (in a unspecified way). This may be of interest but is not what people call the integral on $(0,+\infty)$.2012-12-07
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    @did: Well, to be fair, it wasn't an _unreserved_ faith as I did mention that it could have an error. But I always thought an improper integral was defined as the _limit_ of the proper definite integral as the bound goes to infinity, when there's an infinite bound.2012-12-07
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    At least, that's how it's done for Riemann integrals. For the Lebesgue integral, you can directly integrate over the infinite interval. Perhaps this is what you're getting at? Since it looks that the Lebesgue integral over $(0, \infty)$ does not exist.2012-12-07
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    This is analogous to how $\int_{0}^{\infty} \frac{\sin(x)}{x} dx$ exists an an improper Riemann integral, but not when interpreted as a Lebesgue integral.2012-12-07
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    Is this what you mean, though, when you say the integral doesn't exist? Because in every calc. reference I've seen, an improper (Riemann) integral is defined _by that limit_. Though I think in this case you should split it into two integrals, one for the "singularity" part and one for the "infinity" part, and take the lims (to $0$ and to $\infty$ respectively). But it gives the same result either way. But the point is, the limit is what _defines_ the improper integral. So am I right in guessing you're talking about Lebesgue, or is it something else?2012-12-08
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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.2015-11-07