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harmonic numbers $H_n=\sum_{i=1}^{n}\frac{1}{i}$

Euler's constant

$\gamma=\lim_{n\to \infty} [H_n-\ln(n)]$

Factorial

$n!=n(n-1)(n-2)\cdots2\cdot1$; valid for all non-negative integers

Show that,

$$(s+1)\gamma-\int_0^1\left(\frac{1}{\ln(x)}+\frac{1}{1-x}\right)\sum_{n=0}^{s}x^ndx=\sum_{i=1}^{s}H_i-\ln(s+1)!$$


Inspiring from the integral of Euler's constant; see Wikipedia $$\int_0^1\left(\frac{1}{\ln(x)}+\frac{1}{1-x}\right)=\gamma$$ Mathematical experimental

We did trial and error using wolfram integrator and observe the numerical values of the integral and we was able to come up with a closed form for it.

Unfortunately we are unable to provide a proof for it, can anyone provide us a prove of it?

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    $\begingroup$ By chance, are you and pisquare the same person? $\endgroup$ – Jack D'Aurizio May 10 '16 at 10:54
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    $\begingroup$ @JackD'Aurizio That is very likely, as they format their questions the same way, and both use the odd phrase "mathematical experimental". $\endgroup$ – nospoon May 10 '16 at 11:32
  • $\begingroup$ @nospoon when your eye is open contact me, we talk. I was talking to jack last night and I haven't done with him yet. $\endgroup$ – user335850 May 11 '16 at 9:45
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It is enough to check that: $$\begin{eqnarray*}\color{blue}{\int_{0}^{1}\left(\frac{1}{\log x}+\frac{1}{1-x}\right)x^k}\,dx &=&\int_{0}^{+\infty}\left(\frac{1}{e^t-1}-\frac{1}{t e^t}\right)e^{-kt}\,dt \tag{1}\\[0.2cm]&=&\gamma+\int_{0}^{+\infty}\frac{e^{-kt}-1}{e^t-1}\,dt-\int_{0}^{+\infty}\frac{e^{-kt}-1}{te^t}\,dt\tag{2}\\[0.2cm]&=&\color{blue}{\gamma-H_k+\log(k+1)}\tag{3}\end{eqnarray*} $$ then sum over $k$. Explanation:

  • $(1)$ : substitution $x=e^{-t}$;
  • $(2)$ : we exploit $\int_{0}^{+\infty}\left(\frac{1}{e^t-1}-\frac{1}{te^t}\right)\,dt=\gamma$ and break the remaining integral in two parts;
  • $(3)$ : the first integral can be computed by just expanding the integrand function as a geometric sum and exploiting $\int_{0}^{+\infty}e^{-nt}\,dt=\frac{1}{n}$; the second integral can be computed from Frullani's integral.
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  • $\begingroup$ Are you trying to be banned at all costs? I can give you a hand. $\endgroup$ – Jack D'Aurizio May 11 '16 at 11:19
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Here is my two pence worth and is not a complete answer but paves the way I think.

Theorem [Euler 1731] The limit $$\gamma = \lim_{n \rightarrow \infty}(H_n-\log n)$$ Is given by the convergent series $$\gamma = \sum_{n=2}^{\infty}(-1)^{n}\frac{\zeta(n)}{n}$$

We can evaluate this formula using Mercator's expansion $$\log(1+x) = \sum_{k=1}^{\infty}(-1)^{k}\frac{x^{k}}{k}$$ and in doing so we find \begin{align} \log 2 &= 1-\frac{1}{2}\left(\frac{1}{1}\right)^{2}+\frac{1}{3}\left(\frac{1}{1}\right)^{3}-\ldots\\ \log \frac{3}{2} &= \frac{1}{2}-\frac{1}{2}\left(\frac{1}{2}\right)^{2}+\frac{1}{3}\left(\frac{1}{2}\right)^{3}-\ldots\\ \log \frac{4}{3} &= \frac{1}{3}-\frac{1}{2}\left(\frac{1}{3}\right)^{2}+\frac{1}{3}\left(\frac{1}{3}\right)^{3}-\ldots\\ \log \frac{5}{4} &= \frac{1}{4}-\frac{1}{2}\left(\frac{1}{4}\right)^{2}+\frac{1}{3}\left(\frac{1}{4}\right)^{3}-\ldots\\ \end{align} summing columns of the first $n$ terms gives $$\log(n+1) = H_n-\frac{1}{2}H_{n, 2}+\frac{1}{3}H_{n, 3}-\ldots$$ or $$H_n-\log(n+1) = \frac{1}{2}H_{n, 2}-\frac{1}{3}H_{n, 3}+\ldots$$

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  • $\begingroup$ And so? Apparently, this has nothing to do with OP's question. $\endgroup$ – Jack D'Aurizio May 10 '16 at 11:36
  • $\begingroup$ @JackD'Aurizio And, so? $\endgroup$ – Kevin May 10 '16 at 13:15
  • $\begingroup$ So this is not an answer. $\endgroup$ – Jack D'Aurizio May 10 '16 at 13:30
  • $\begingroup$ @JackD'Aurizio From my definition of 'answer' I direct you tot he answers of this thread meta.math.stackexchange.com/questions/11723/… $\endgroup$ – Kevin May 11 '16 at 7:59

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