The following series came up, and I'm wondering (hoping!) it can be solved. It is the same as the well-known series for $e^x$, except it has a logarithm term added:$$ \sum_{k=0}^{\infty} \frac{x^k}{k!}\log(k) $$How, if possible, can I evaluate this in closed-form?
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I doubt this has a closed form. Most series with logarithmic terms don't. – Chappers Sep 08 '17 at 22:24
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3I suppose that it starts at $k=1$. – José Carlos Santos Sep 08 '17 at 22:24
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Might be something from this, but I wouldn't hold out... Mother Goddess Mathematics tends to not like closed forms... – Sep 08 '17 at 22:24
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The point is: can you think of a function $g(x)$ such that $g^{(k)}(0)=\log k$ for all $k\ge1$? Personally, nothing comes to mind. – Sep 08 '17 at 22:25
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A similar question was discussed in the comments here concerning $x\to\infty$ – Paul Enta Sep 08 '17 at 22:41
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1You have to start the sum at 1 because the log of 0 is not defined. – marty cohen Sep 08 '17 at 22:42
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@imallett Wolfram Alpha says that the series converges by the ratio test, so you can calculate a numerical approximation. – Toby Mak Sep 08 '17 at 23:27
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See here for the $x=1$ case. A similar formula can be worked out as the derivative of the Bell Polynomials using (12). As to what constitutes a "closed form" is a different matter. Also known as the Touchard Polynomials, they have a non-integral order $n$ generalization, which would be needed for differentiation. – Paul LeVan Sep 09 '17 at 05:22
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Regarding the $k=0,1$ error, I actually should have written $\log^k(A)$, but a copying error made me think it was $\log(k) A$. – geometrian Sep 09 '17 at 06:02
1 Answers
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Note that
$$\log(k)=\int_0^\infty\frac{e^{-t}-e^{-kt}}t~\mathrm dt$$
And so, assuming you intend to start at $k=1$,
$$f(x)=\int_0^\infty\sum_{k=1}^\infty\frac{x^k}{k!}\frac{e^{-t}-e^{-kt}}t~\mathrm dt$$
The sum isn't hard to evaluate and comes out to
$$f(x)=\int_0^\infty\left(\frac{e^{x-t}-e^{xe^{-t}}}t-1\right)~\mathrm dt$$
That $e^{xe^{-t}}$ makes it fairly clear that I would doubt on any closed form.
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