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I tried to find the limit $$\mathop {\lim }\limits_{n \to \infty } \ln (n) \cdot \int_0^1 \ln ( n^{-t} + 1 ) \, \mathrm dt$$, but I couldn't find a conclusive argument.

By assuming that the sequence $\displaystyle x_n = \ln (n) \cdot \int_0^1 \ln ( n^{-t} + 1 ) \, \mathrm dt$ is convergent to $k$, I managed to do the following:
It is well known that $T_{2q,0} (x) \le \ln(1+x) \le T_{2q+1,0} (x) \; , \; \forall x>0 , \forall q \in \mathbb{N} ^{*}$. (I used the Taylor polynom here)
Using this inequality, we obtain a bounding for $x_n$. By making $n \to \infty$, we obtain:

$$\sum\limits_{p = 1}^{2q} { \frac{ (-1)^{p+1} }{ p^2 } } \le k \le \sum\limits_{p = 1}^{2q+1} { \frac{ (-1)^{p+1} }{ p^2 } } \, , \, \forall q \in \mathbb{N} ^{*} $$

Then, by making $q \to \infty$, we have that $ \sum\limits_{p = 1}^{ \infty } { \frac{ (-1)^{p+1} }{ p^2 } } = \sum\limits_{p = 1}^{ \infty } { \frac{ 1 }{ p^2 } } - 2 \cdot \sum\limits_{p = 1}^{ \infty } { \frac{ 1 }{ (2p)^2 } } = \sum\limits_{p = 1}^{ \infty } { \frac{ 1 }{ p^2 } } - \frac{1}{2} \cdot \sum\limits_{p = 1}^{ \infty } { \frac{ 1 }{ p^2 } } = \frac{1}{2} \cdot \sum\limits_{p = 1}^{ \infty } { \frac{ 1 }{ p^2 } } = \frac{ \pi ^2}{12} $, which implies that $k = \frac{ \pi ^2 }{12}$.

Naren
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meowy03
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  • I would substitute $t\ln n=x$. Then it is a simple matter of $\lim\limits_{n\to\infty}\int_0^{\ln n}=\int_0^\infty$. – metamorphy Aug 13 '20 at 13:07

2 Answers2

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We have $\int ln(1+n^{-t})dt=\int \sum_{k=1}^{\infty}(-1)^{k+1}\frac{n^{-tk}}{k}dt=\sum_{k=1}^{\infty}(-1)^{k+1}\frac{1}{k} \int n^{-tk}dt$ by Fubini/Tonelli theorems (and using the Taylor series $ln(1+x)=\sum_{k=1}^{\infty}\frac{x^k}{k}$ provided $x\in(-1,1]$).

Then $\int_{0}^{1}n^{-tk}dt=[-\frac{1}{kln(n)n^{kt}}]|_{t=0}^{t=1}=\frac{1}{kln(n)}-\frac{1}{n^{k}ln(n)k}$ by using the substitution $u=-tk$ and applying the exponential rule $\int a^{x}dx=\frac{a^{x}}{ln(a)}+C.$

So it follows that $ln(n)\int_{0}^{1} ln(1+n^{-t})dt=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^2}-\sum_{k=1}^{\infty}(-1)^{k+1}\frac{1}{k^{2}n^{k}}$.

Finally taking the limit we have $lim_{n\rightarrow\infty}ln(n)\int_{0}^{1} ln(1+n^{-t})dt=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^2}=\frac{\pi^{2}}{12}.$

Alessio K
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Substitute $x=t\ln(n)$, then the integral becomes

$$\lim_{n\to\infty}\int_0^{\ln(n)}\ln\left(1+e^{-x}\right)\,dx=\int_0^{\infty}\ln\left(1+e^{-x}\right)\,dx.$$

Now

$$\int_0^\infty \ln(1+e^{-x})dx=\int_0^\infty\left(e^{-x}-\frac{e^{-2x}}{2}+\frac{e^{-3x}}{3}-\frac{e^{-4x}}{4}+\cdots\right)dx $$ $$=\int_0^\infty e^{-x}dx-\frac{1}{2}\int_0^\infty e^{-2x}dx+\frac{1}{3}\int_0^\infty e^{-3x}dx-\frac{1}{4}\int_0^\infty e^{-4x}dx+\cdots$$ $$=1-\frac{1}{2}\cdot\frac{1}{2}+\frac{1}{3}\cdot\frac{1}{3}-\frac{1}{4}\cdot\frac{1}{4}+\cdots $$ $$=\eta(2)=\frac{\pi^2}{12},$$

where $\eta(s)$ is the Dirichlet eta function. The last equality follows from the Basel Problem by the amazing

$$\eta(2)=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}=\frac{\pi^2}{12}.$$

Axion004
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