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This question already has an answer here:

The Harmonic numbers $H_x = \sum_{n=1}^x 1/n$ are the sum of the reciprocals of the natural numbers up to a given number. The first few are $0, 1, 3/2, 11/6, \ldots$. $H_x$ can be defined for noninteger $x$ by the integral definition $H_x = \int_0^1 \frac{1-t^x}{1-t} \, dt$, ie $H_{1/2} = 2-2\ln 2$

$H_x$ is not a rational function because a rational function is asymptotic to $x^k$ for some integer $k$, and $H_x$ is asymptotic to $\ln x + \gamma$, where $\gamma\sim 0.577 \dots$.

Can you prove $H_x$ is not an elementary function using the integral definition? Take "elementary" to mean a finite composition of arithmetic operators $(+, -, \div, \times)$, exponents, logarithms, and known constants (algebraic numbers and algebraic compositions of proven transcendental numbers, like $\pi, e, e^\pi \dots$ etc).

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marked as duplicate by Winther, user147263, A.P., Cameron Williams, user223391 Nov 13 '15 at 3:38

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  • $\begingroup$ So basically you're saying $\int_0^1\frac{t^x}{1-t}dt$ has no closed form solution? $\endgroup$ – Gregory Grant Nov 12 '15 at 17:11
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    $\begingroup$ OP is asking if there's a proof $\int_0^1\frac{1-t^x}{1-t}dx$ is not an elementary function of $x$. I think the question is clear. It is indeed possible to prove special functions are not elementary (in differential Galois theory, this is the analogue of proving some algebraic numbers cannot be expressed with radicals in usual Galois theory). $\endgroup$ – anon Nov 12 '15 at 17:23
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    $\begingroup$ Why is this considered a duplicate? The linked thread is about integers values of $x$ not noninteger ones! $\endgroup$ – Raymond Manzoni Nov 12 '15 at 21:17
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Your $H_x$ function is (up to the Euler constant $\gamma$) the digamma function $\psi(x+1)$ : $$\tag{1}H_x = \int_0^1 \frac{1-t^x}{1-t} dt= \psi(x+1)+\gamma$$ There is no closed form in the general case of $x\in\mathbb{R}$ (or $\mathbb{C}$) as we may deduce from the Hölder theorem for the $\Gamma$ function.

Elementary functions and their integrals (recursively) define the "Liouvillian functions" which are all solutions of algebraic differential equations. From the Hölder theorem we deduce that $\Gamma$ and thus $\;\psi(x)=(\log\Gamma(x))'$ are not solutions of any algebraic differential equation and thus can't be elementary ($\psi$ is a special function).

On the other side I must add that a closed form exists for any fixed fraction $x\in \mathbb{Q}$ from a nice formula from Gauss :

For two positive integers $r$ and $m$ with $r<m$ we have $$\tag{2}\psi\left(\frac rm\right)=-\gamma-\log(2m)-\frac {\pi}2\cot\left(\frac{r\pi}m\right)+2\sum_{n=1}^{\lfloor(m-1)/2\rfloor}\cos\left(\frac{2\pi nr}m\right)\log\,\sin \left(\frac{\pi n}m\right)$$

For fractions out of $(0,1)$ we may then use the recurrence $$\tag{3}\psi(x+1)=\psi(x)+\dfrac 1x$$

ADDITION (from the comments by Rahul and Winther) :

I added fixed fraction in the previous text because using the previous formulas you will obtain a closed form for (say) $\psi(179/3)\;$ but I fear this won't be considered as a closed form for a general fraction $\;x=\dfrac rm\;$ since :

  • you will have to add a number of terms depending of $\lfloor x\rfloor$
  • the denominator of $x$ (i.e. $m$) will add further $m$ or $\left\lfloor\frac {m-1}2\right\rfloor$ terms

    Setting an upper bound for $|x|$ and $m$ may be more satisfying for a closed form...

The digamma function could be considered as a mere rewriting of $H_n$ but the link with the $\Gamma$ function allows to get very useful expansions for large $n$ (and $x$) and numerous neat analytical results (polygamma and generalizations).

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  • $\begingroup$ Doesn't Gauss's formula only apply for $x\in\mathbb Q\cap(0,1)$? $\endgroup$ – Rahul Nov 12 '15 at 17:45
  • $\begingroup$ @Rahul: The recurrence formula for $\psi$ should help for this. $\endgroup$ – Raymond Manzoni Nov 12 '15 at 17:46
  • $\begingroup$ You should perhaps add that this formula (of Gauss) is of no use for finding a finite order elementary formula for $H_n$. If you try to apply it you would just end up with the definition $H_n = 1 + 1/2 + \ldots + 1/n$. $\endgroup$ – Winther Nov 12 '15 at 17:48
  • $\begingroup$ @Winther: You are right bul the reference allows to find excellent approximations (using say Euler Maclaurin) and to prove the claim made here (for $x\in \mathbb{R}$). $\endgroup$ – Raymond Manzoni Nov 12 '15 at 17:50
  • $\begingroup$ Could there be an elementary function $f$ such that $\forall n\in\Bbb N,f(n)=H_n$? That is, could there be an elementary function that agrees with it on the integers? $\endgroup$ – Akiva Weinberger Nov 12 '15 at 19:13

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