# On the monotony of $-\int_0^1\frac{e^y}{y}(\operatorname{Li}_x(1-y)-\zeta(x)) \,\mathrm dy$

$$\def\d{\mathrm{d}}$$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 $$x=n\geq 2$$.

Question.

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.

• en.wikipedia.org/wiki/Polylogarithm ... look at the section on Integral representations ... Apr 26, 2017 at 19:52
• Many thanks then I try read it @DonaldSplutterwit . I don't know exactly what formula 1., 2., ... at the section that you are saying.
– user243301
Apr 26, 2017 at 19:55
• 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 $\ddot \smile$ Apr 26, 2017 at 20:03
• 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
– user243301
Apr 26, 2017 at 20:26
• I have completed the proof. :-) May 11, 2017 at 16:33

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 $\enspace 0\leq y\leq 1$

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

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

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

• Many thanks for your contribution.
– user243301
May 4, 2017 at 10:16
• @user243301 : You are welcome, but unfortunately it's just like a drop on a hot stone. :-) May 4, 2017 at 11:15
• Don't worry. In any case now I, and all users, can read your remarks, many thanks.
– user243301
May 4, 2017 at 11:22
• @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 $2\sqrt{k}$. May 7, 2017 at 6:12
• 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.
– user243301
May 7, 2017 at 8:35