# Compute $\lim_{n\to\infty}n^4\int_0^1\frac{x^n\ln^3x}{1+x^n}\ln(1-x)\,dx$

Compute

$$\lim_{n\to\infty}n^4\int_0^1\frac{x^n\ln^3x}{1+x^n}\ln(1-x)\,dx$$

According to Wolfram Alpha, the limit is zero. I tried to make substitution $x=\frac{t}{n}$ and I got $$\lim_{n\to\infty}\int_0^{\large\frac{1}{n}}\frac{t^n\ln^3t}{n^n+t^n}\ln\left(1-\frac{t}{n}\right)\,dx \to 0$$ but I am not sure this approach is correct.

I also tried to make substitution $t=x^n$ and I got $$\lim_{n\to\infty}n^4\int_0^1\frac{x^n\ln^3x}{1+x^n}\ln(1-x)\,dx=\lim_{n\to\infty}\int_0^1\frac{t^{\large\frac{1}{n}}\ln^3t}{1+t}\ln\left(1-t^{\large\frac{1}{n}}\right)\,dt$$ I used the bound $$\left|\int_0^1\frac{t^{\large\frac{1}{n}}\ln^3t}{1+t}\ln\left(1-t^{\large\frac{1}{n}}\right)\,dt\right|\leq\int_0^1\left|t^{\large\frac{1}{n}}\ln^3t\ln\left(1-t^{\large\frac{1}{n}}\right)\right|\,dt$$ because $1+t\ge1$, so $\frac{1}{1+t}\leq1$. I can compute the integral using Taylor series for the logarithm but after taking the limit I got the result was infinity. \begin{align} \lim_{n\to\infty}n^4\int_0^1\frac{x^n\ln^3x}{1+x^n}\ln(1-x)\,dx&=\lim_{n\to\infty}\int_0^1t^{\large\frac{1}{n}}\ln^3t\ln\left(1-t^{\large\frac{1}{n}}\right)\,dt\\ &=-\lim_{n\to\infty}\sum_{k=1}^\infty\frac{1}{k}\int_0^1t^{\large\frac{k+1}{n}}\ln^3t\,dt\\ &=\lim_{n\to\infty}\sum_{k=1}^\infty\frac{6}{k\left(\frac{k+1}{n}+1\right)^4}\to\infty \end{align} Could anyone here please help me? Any help would be greatly appreciated. Thank you.

• This is not the problem you submitted to Wolfram Alpha. – Claude Leibovici Aug 23 '14 at 13:31
• Mr. @ClaudeLeibovici, yeah I know. But WA cannot compute it if I use the integral sign, I thought it will be the same if I removed the sign (╥﹏╥) – Anastasiya-Romanova 秀 Aug 23 '14 at 13:34

We will prove that the limit is $+\infty$.

Let \eqalign{I_n&=n^4\int_0^1\frac{x^n}{1+x^n}\ln^3 x\ln(1-x)dx\cr J_n&=n^4\int_0^1x^n\ln^3 x\ln(1-x)dx} Clearly $$\frac{1}{2}J_n\leq I_n\leq J_n$$ because $\frac{1}{2}\leq\frac{1}{1+x^n}\leq1$ and $\ln^3x\ln(1-x)\geq0$ for $0<x<1$. So, let us consider $J_n$. We have $$-\ln(1-x)=\sum_{k=1}^\infty\frac{x^k}{k}$$ and $$\int_0^1x^{m}(-\ln x)^3dx=\frac{6}{(m+1)^4}$$ Combining these two properties we see that $$J_n=\sum_{k=1}^\infty\frac{6n^4}{k(k+1+n)^4}$$ for a given $m\geq1$ we have $$J_n\geq\sum_{k=1}^m\frac{6n^4}{k(k+1+n)^4}$$ hence $$\liminf_{n\to\infty}J_n\geq 6 \sum_{k=1}^m\frac{1}{k}$$ but $m$ is arbitrary and $\sum \frac{1}{k}$ is divergent, So $$\liminf_{n\to\infty}J_n=+\infty$$ that is $\lim\limits_{n\to\infty}J_n=+\infty$, and consequently $\lim\limits_{n\to\infty}I_n=+\infty$.

• @V-Moy, You are welcome. – Omran Kouba Aug 23 '14 at 15:27
• For $J_n$ we could make $t=x^n$ which will make $n^4$ disappear and then continue. – aziiri Aug 23 '14 at 16:43
• @OmranKouba: Thanks for a nice answer. May I ask you please elaborate how does one obtain the steps: Property 2 as well as expression for $J_{n}$ and the one with $\sum inf J_{n}$? – Abhimanyu Arora Aug 23 '14 at 20:29

For the time being, the only thing I can tell you is that $$$$I_n=n^4\int_0^1\frac{x^n\ln^3x}{1+x^n}\ln(1-x)\,dx$$$$ seems to be an increasing function of $n$ (this has been done numerically): $$I_{10}=4.92397$$ $$I_{100}=18.2376$$ $$I_{1000}=31.7992$$ $$I_{10000}=44.97982$$ $$I_{100000}=58.0781$$

• Thanks for your answer Sir. +1 as always to the one who answer my question (✿◠‿◠) – Anastasiya-Romanova 秀 Aug 23 '14 at 14:12
• You are very welcome ! I did very little. Cheers :-) – Claude Leibovici Aug 23 '14 at 14:15

Omran Kouba has correctly answered the OP question.

Let's just see how the integral behaves as $n$ is great.

As $n \rightarrow +\infty$, we have

$$\lim_{n \to +\infty} \,\color{blue}{\frac{n^4}{\ln n}}\!\int_0^1\frac{x^n\ln^3x}{1+x^n}\ln(1-x)\:\mathrm dx = \color{blue}{\frac{7\pi^4}{120}}. \tag1$$

Moreover,

$$\int_0^1\frac{x^n\ln^3x}{1+x^n}\ln(1-x)\:\mathrm dx = \frac{7\pi^4}{120}\frac{\ln n}{n^4}-\frac{45\:\zeta(5)}{2}\frac{\ln n}{n^5}+\mathcal{O}\left(\frac{\ln n}{n^6}\right). \tag2$$

Recall that $$\int_0^1x^s\ln(1-x) \:\mathrm d x =- \frac{1}{s+1}\left(\gamma+\psi(s+2)\right), \, s>-1, \tag3$$ where $\psi:=\Gamma'/\Gamma$ is the digamma function, as may be easily seen by expanding the logarithmic term and integrating termwise.

Then, using $\displaystyle \ln^3x=\partial_s^3 \left. \left(x^{s}\right) \right|_{s=0}$, one may write

\begin{align} \int_0^1\frac{x^{n}\ln^3x}{1+x^n}\ln(1-x)\:\mathrm dx & = \partial_s^3 \left(\left. \int_0^1\frac{x^{n+s}}{1+x^n}\ln(1-x)\:\mathrm dx\right)\right|_{s=0} \\\\ & =\partial_s^3 \left(\left.\sum_{k=0}^{\infty}(-1)^k\int_0^1x^{n+s}x^{kn}\ln(1-x)\:\mathrm dx\right)\right|_{s=0} \\\\ & =\partial_s^3 \left(\left.\sum_{k=1}^{\infty}\frac{(-1)^k}{nk+s+1}\left(\gamma+\psi(nk+s+2)\right)\right)\right|_{s=0} \\\\ & =\partial_s^3 \left.\left(-\frac{\ln n}{n^4}\sum_{k=1}^{\infty}\frac{(-1)^k}{k^4}+\frac{4\ln n}{n^5}\sum_{k=1}^{\infty}\frac{(-1)^k}{k^5}+\mathcal{O}\left(\frac{\ln n}{n^6}\right) \right)s^3\right|_{s=0} \end{align} where we have used the uniform convergence on $s \in [0,1]$, and the asymptotic expansion of the digamma function, as $M \rightarrow +\infty$, $$\psi(M) = \ln M-\frac{1}{2M}-\frac{1}{12M^2}+\frac{1}{120M^4}-\frac{1}{252M^6}+\mathcal{O}\left(\frac{1}{M^8}\right) ,$$ to obtain $(2)$, deducing $(1)$.

• @V-Moy Maybe you have noticed that, with the answer I posted, you get $$\lim_{n \to +\infty} \frac{n^4}{\ln n}\int_0^1\frac{x^n\ln^3x}{1+x^n}\ln(1-x)\:\mathrm dx = \frac{7\pi^4}{120}.$$ – Olivier Oloa Aug 24 '14 at 6:47
• Thanks Olivier for your answer, +1. I like the way you answer my question, by directly integrating it. But I think to answer question like this we have to use a special bound like Prof. Kouba did. Also thank you for the comment below your answer, I also like it (* ˘⌣˘)◞[]♥[]ヽ(•‿• ) – Anastasiya-Romanova 秀 Aug 24 '14 at 8:04

The comment box does not seem to like this equation! So if this is completely taking you in the wrong direction then please comment below and I will remove.

Would this be of benefit here

$$\lim_{n\rightarrow \infty}\left[\lim_{\alpha\rightarrow 0,\beta\rightarrow 0}\dfrac{\partial^3}{\partial \alpha^3}\dfrac{\partial}{\partial \beta}n^4\int_{0}^{1}\dfrac{x^{\alpha}\left(1-x\right)^{\beta}x^n}{1+x^n}dx\right]$$

• The integral seems hard to calculate. BTW, thanks for your answer Sir. +1 as always to the one who answer my question (✿◠‿◠) – Anastasiya-Romanova 秀 Aug 23 '14 at 14:13
• Thank you. But there is still a lot of work to make this work. Good luck – Chinny84 Aug 23 '14 at 14:40