$\def\d{\mathrm{d}}$After I was studying variations of the integral representation for $\zeta(3)$ due to Beukers, see the section More complicated formulas from this Wikipedia, I've thought an exercise. The motivation is this antiderivative $$-\int_0^1e^y\frac{\log(xy)}{1-xy}\,\d x=-\frac{e^y}{y}\operatorname{Li}_2(1-xy)+\mathrm{constant}.$$ For real numbers $x\geq 2$, $$f(x):=-\int_0^1\frac{e^y}{y}(\operatorname{Li}_x(1-y)-\zeta(x)) \,\d y,$$ I've calculated with Wolfram Alpha online calculator some particular values at integer values $x=n\geq 2$.


A) Is it possible to prove that $f(x)$ is decreasing for $x\geq 2$?

B) Is it possible to calculate $$\lim_{x\to\infty}f(x)?$$ Thanks in advance.

Also I know Euler's Zeta function $\zeta(x)$ tends to $1$ as $x\to\infty$. With respect the definite integral I believe that is (impossible) difficult to get a closed-form for such values.

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    $\begingroup$ en.wikipedia.org/wiki/Polylogarithm ... look at the section on Integral representations ... $\endgroup$ Apr 26, 2017 at 19:52
  • $\begingroup$ Many thanks then I try read it @DonaldSplutterwit . I don't know exactly what formula 1., 2., ... at the section that you are saying. $\endgroup$
    – user243301
    Apr 26, 2017 at 19:55
  • $\begingroup$ Yeah, I think you need to turn $x$ into a parameter (a variable that does not need to be a whole number) ... I guess the calculation will get very complicated ... I will try, but don't expect an answer from me $\ddot \smile$ $\endgroup$ Apr 26, 2017 at 20:03
  • $\begingroup$ The variable $x\geq 2$ is real, I've calculated the first cases with Wolfram Alpha for $x=2,3,4$. I've created this Question when I've thought that get a closed-form for the integral is the (impossible) difficult exercise. Many thanks for your attention @DonaldSplutterwit $\endgroup$
    – user243301
    Apr 26, 2017 at 20:26
  • $\begingroup$ I have completed the proof. :-) $\endgroup$
    – user90369
    May 11, 2017 at 16:33

1 Answer 1


Answer for question $(A)$ :

$\displaystyle -\int\limits_0^1 e^y\frac{Li_x(1-y)-\zeta(x)}{y}dy =\sum\limits_{k=1}^\infty\frac{1}{k^x}\int\limits_0^1 e^y\frac{1-(1-y)^k}{y}dy$

With $\enspace\displaystyle 0\leq (e^y-1)\frac{1-(1-y)^k}{y}-y<1\enspace$ for $\enspace 0\leq y\leq 1$

(see Value range for $a$ when $\enspace 0\leq (e^x-1)\frac{1-(1-x)^a}{x}-x<1\enspace$ with $\enspace 0\leq x\leq 1$ .) we get

$\displaystyle 0<y+\frac{1-(1-y)^k}{y}\leq e^y\frac{1-(1-y)^k}{y}<1+y+\frac{1-(1-y)^k}{y}\enspace$ and therefore

$\displaystyle \int\limits_0^1 e^y\frac{1-(1-y)^k}{y}dy<\int\limits_0^1 (1+y+\frac{1-(1-y)^k}{y})dy=\frac{3}{2}+H_k\enspace$ with $\enspace\displaystyle H_k:=\sum\limits_{v=1}^k\frac{1}{v}$ .

It follows $\enspace\displaystyle 0<-\int\limits_0^1 e^y\frac{Li_x(1-y)-\zeta(x)}{y}dy<\sum\limits_{k=1}^\infty\frac{1}{k^x}(\frac{3}{2}+H_k)=\frac{3}{2}\zeta(x)+\sum\limits_{k=1}^\infty\frac{H_k}{k^x}$ .

With $\enspace\gamma\enspace$ as the Euler–Mascheroni constant and $\enspace H_{k-1}<\gamma+\ln k\enspace$ for $\enspace k\in\mathbb{N}\enspace$ follows

$\displaystyle 0<-\int\limits_0^1 e^y\frac{Li_x(1-y)-\zeta(x)}{y}dy<(\frac{3}{2}+\gamma)\zeta(x)+\zeta(x+1)-\zeta'(x)$

so that the integral convergies for $\enspace x>1$ .

The integral is decreasing because of $\enspace\displaystyle \frac{1}{k^a}>\frac{1}{k^b}\enspace$ for $\enspace k>1\enspace$ and $\enspace 0<a<b$ ,

together with $\enspace\displaystyle\int\limits_0^1 e^y\frac{1-(1-y)^k}{y}dy>0\enspace$ for all $\enspace k$ .

Answer for question (B):

$\displaystyle \lim\limits_{x\to\infty}(-\int\limits_0^1 e^y\frac{Li_x(1-y)-\zeta(x)}{y}dy) =\lim\limits_{x\to\infty}(\sum\limits_{k=1}^\infty\frac{1}{k^x}\int\limits_0^1 e^y\frac{1-(1-y)^k}{y}dy)=$

$\displaystyle =\int\limits_0^1 e^y\frac{1-(1-y)^1}{y}dy=e-1$

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    $\begingroup$ Many thanks for your contribution. $\endgroup$
    – user243301
    May 4, 2017 at 10:16
  • $\begingroup$ @user243301 : You are welcome, but unfortunately it's just like a drop on a hot stone. :-) $\endgroup$
    – user90369
    May 4, 2017 at 11:15
  • $\begingroup$ Don't worry. In any case now I, and all users, can read your remarks, many thanks. $\endgroup$
    – user243301
    May 4, 2017 at 11:22
  • $\begingroup$ @user243301 : Thanks for the credits - but sorry, I cannot give you a complete answer. I tried also to compare with $\sum\limits_{v=1}^k\binom{k}{v}\frac{(-1)^{v-1}}{v}=$$\sum\limits_{v=1}^k\frac{1}{v}$ to get a senseful transformation for $\sum\limits_{v=1}^k\binom{k}{v}(-1)^{v-1}\int\limits_0^1 e^y y^{v-1}dy$ with $\int\limits_0^1 e^y y^{v-1}dy=(-1)^{v-1}(e \cdot !(v-1) - (v-1)!)=(-1)^v(v-1)!(1-e \sum\limits_{j=0}^{v-1}\frac{(-1)^j}{j!})= $$\frac{e}{v}\sum\limits_{j=0}^\infty\frac{(-1)^j v!}{(v+j)!}$, but I (still) haven't got a sum which I can compare with $2\sqrt{k}$. $\endgroup$
    – user90369
    May 7, 2017 at 6:12
  • $\begingroup$ Thanks for your words. I did it in the grace period and still I've not accept an answer @user90369 . I pay the bounty for the answer that showed more merit. $\endgroup$
    – user243301
    May 7, 2017 at 8:35

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