Evaluate $\lim_{n\to\infty}\int_1^n\frac{\ln x}{c_n+x\ln x}\,dx$

Question

Let $c_n$ be an unbounded sequence: $c_n\to\infty$ when $n\to\infty$

Find the value of the limit: $$\lim_{n\to\infty}\int_1^n\frac{\ln x}{c_n+x\ln x}\,dx$$

What I managed to find was the fact that $$\lim_{n\to\infty}\frac{\ln x}{c_n+x\ln x}=0$$ I was thinking that the limit is going to be $0$, but I have to prove it still. Maybe it is not and I am wrong I do not know for sure.

I denoted $$I(n)=\int_1^n\frac{\ln x}{c_n+x\ln x}\,dx$$

I was thinking of finding a recurrence formula for this integral, Though, I could not be able to make any significant progress.

What should I do?

Answer 1

As @Ryszard Szwarc mentioned in the comment, if $\frac{c_n}{n\ln n}\to\infty,\,\,\lim_{n→∞} \int _1^n\frac{\ln\:x}{c_n\:+\:x\:\ln x}dx=0$, and the most interesting is the case $c_n=n\ln n$.

So, we want to find $\lim_{n→∞} I(n)=\lim_{n→∞}\int _1^n\frac{\ln\:x}{n\ln n\:+\:x\:\ln x}dx$. Making the substitution $x=e^t$ $$I(n)=\int _1^n\frac{\ln\:x}{n\ln n\:+\:x\:\ln x}dx=\int_0^{\ln n}\frac{te^t}{n\ln n+e^t t}dt$$ Making another substitution $t=s\ln n$ $$I(n)=\frac{\ln n}{n}\int_0^1\frac{sn^s}{1+\frac{s}{n^{1-s}}}ds$$ As $\frac{s}{n^{1-s}}\leqslant1$, we can decompose the denominator into the series: $$I(n)=\frac{\ln n}{n}\int_0^1n^ss\big(1-sn^{s-1}+s^2n^{2s-2}-s^3n^{3s-3}+-...\big)ds$$ Integrating by part, we get the terms with the different rate of convergence, for example: $$\int_0^1sn^sds=\frac{1}{\ln^2n}\int_0^{\ln n}se^sds=\frac{1}{\ln^2n}\big(n\ln n-n\big)=\frac{n}{\ln n}-\frac{n}{\ln^2 n}$$ The same story happens to all the terms; only the first ones will give a non-zero limit at $n\to\infty$. Therefore, keeping only these main terms $$I(n)=\frac{\ln n}{n}\Big(\frac{1}{\ln^2n}\int_0^{\ln n}e^xxdx-\frac{1}{n\ln^3n}\frac{1}{2^3}\int_0^{2\ln n}e^xx^2dx+\frac{1}{n^2\ln^4n}\frac{1}{3^4}\int_0^{3\ln n}e^xx^3dx-+...\Big)$$ The pattern is clear. Integrating by part and keeping only first terms of every integral $$I(n)=\frac{\ln n}{n}\Big(\frac{1}{\ln^2n}n\ln n-\frac{1}{n\ln^3n}\frac{1}{2^3}2^2n^2\ln^2n+\frac{1}{n^2\ln^4n}\frac{1}{3^4}3^3n^3\ln^3n-+...\Big)+O\Big(\frac{1}{\ln n}\Big)$$ $$\Rightarrow\,\,\boxed{\,\,\lim_{n\to\infty} I(n)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+-...=\ln 2=0.693147...\,\,}$$ The numeric evaluation of the limit confirms this conclusion: $$\int _1^n\frac{\ln\:x}{n\ln n\:+\:x\:\ln x}dx\,\Big|_{n=10^{300}}=0.692143...$$ though the rate of convergence, of course, is monstrously low :)

It is not difficult to evaluate in the same way the second and third terms of the asymptotics. I got $$\boxed{\,\,I(n)=\ln2-\frac{\ln2}{\ln n}+\Big(\ln2-\frac{\pi^2}{12}\Big)\frac{1}{\ln^2n}+O\Big(\frac{1}{\ln^3n}\Big)\,\,}$$

$\mathbf{PS}$

I checked numerically the revised asymptotics. It seems it works! $$\int _1^n\frac{\ln\:x}{n\ln n\:+\:x\:\ln x}dx\,\Big|_{n=1000}=0.590148...$$ Asymptotics gives $$I(n)\sim\ln2-\bigg(\frac{\ln2}{\ln n}+\Big(\ln2-\frac{\pi^2}{12}\Big)\frac{1}{\ln^2n}\bigg)\,\bigg|_{n=1000}=0.590093...$$

Answer 2

It turns out that we have the following inequalities:

$$ \log\biggl(1 + \varliminf_{n\to\infty} \frac{n\log n}{c_n}\biggr) \leq \varliminf_{n\to\infty} I(n) \leq \varlimsup_{n\to\infty} I(n) \leq \log\biggl(1 + \varlimsup_{n\to\infty} \frac{n\log n}{c_n}\biggr) $$

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Let $g : [0, \infty) \to [1, \infty)$ be the inverse of $x \mapsto x \log x$ for $x \geq 1$, that is, $g$ solves the functional equation

$$ g(t) \log g(t) = t $$

(In fact, we can write $g(t) = t/W(t)$ where $W(t)$ is the Lambert W-function. However, this observation is not needed here.) Then by substituting $t = x \log x$, or equivalently, $x = g(t)$, and noting that

$$ \frac{\mathrm{d}x}{x} = \frac{g'(t) \, \mathrm{d}t}{g(t)} = \frac{\mathrm{d}t}{g(t)(\log g(t) + 1)} = \frac{\mathrm{d}t}{t + g(t)} $$

we have

\begin{align*} I(n) = \int_{1}^{n} \frac{x \log x}{c_n + x \log x} \, \frac{\mathrm{d}x}{x} = \int_{0}^{n \log n} \frac{t}{c_n + t} \, \frac{\mathrm{d}t}{t + g(t)} \end{align*}

Writing $\rho_n = (n \log n) / c_n$ and substituting $t = c_n u$,

\begin{align*} I(n) &= \int_{0}^{\rho_n} \frac{u}{1 + u} \cdot \frac{1}{u + g(c_n u)/c_n} \, \mathrm{d}u \end{align*}

Now, let $ \displaystyle \alpha = \varliminf_{n\to\infty} \rho_n$ and $ \displaystyle \beta = \varlimsup_{n\to\infty} \rho_n$. Then

$$ \varlimsup_{n\to\infty} I(n) \leq \varlimsup_{n\to\infty} \int_{0}^{\rho_n} \frac{u}{1 + u} \cdot \frac{1}{u} \, \mathrm{d}u = \varlimsup_{n\to\infty} \log(1+\rho_n) = \log(1+\beta) \tag{1} $$

On the other hand, together with the fact that $g(t) \ll t$ and the Fatou's lemma,

\begin{align*} \varliminf_{n\to\infty} I(n) &\geq \int_{0}^{\infty} \varliminf_{n\to\infty} \frac{u}{1 + u} \cdot \frac{1}{u + g(c_n u)/c_n} \mathbf{1}_{[0, \rho_n]}(u) \, \mathrm{d}u \\ &= \int_{0}^{\infty} \frac{u}{u+1} \cdot \frac{1}{u} \mathbf{1}_{[0, \alpha)}(u) \, \mathrm{d}u \\ &= \log(1+\alpha) \tag{2} \end{align*}

Answer 3

This is not an entire solution, but a long comment.

Following the comments of Henry and Ryszard: from $$\frac{\log x}{c_n + x\log x}\mathbb{1}_{(1,n]}(x)\leq \frac{\log n}{c_n}\mathbb{1}_{(1,n]}(x)$$ we have $$\int^n_1\frac{\log x}{c_n + x\log x}\,dx\leq \frac{n\log n}{c_n}$$ Thus, of $\lim_n\frac{c_n}{n\log n}=\infty$, the case discussed in the comments has $0$ limit.

For the general case, the limit if exists, may not be $0$ $$\frac{\log(m-1)}{c_n+m\log m}\leq \int^m_{m-1}\frac{\log x}{c_n + x\log x}\,dx \leq\frac{\log m}{c_n+ (m-1)\log (m-1)}\leq \frac{\log m}{c_n}$$ adding all terms for $2\leq m\leq n$ yields $$\frac{\log((n-1)!)}{c_n+n\log n}\leq\sum^n_{m=2}\frac{\log(m-1)}{c_n+m\log m}\leq \int^n_1\frac{\log x}{c_n+x\log x}\,dx\leq \frac{1}{c_n}\log(n!)$$