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Find the value of the limit $$\lim_{n \to \infty}n\left(\left(\int_0^1 \frac{1}{1+x^n}\,\mathrm{d}x\right)^n-\frac{1}{2}\right)$$

I can't solve the integral $\int_0^1 \mathrm{\frac{1}{1+x^n}}\,\mathrm{d}x$. But maybe the questions doesn't require solving the integral.
Apparently the $\lim_{n \to \infty}(\int_0^1 \mathrm{\frac{1}{1+x^n}}\,\mathrm{d}x)^n$ should be $\frac{1}{2}$ for the question to make sense. That's all I know.

Did
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idpd15
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3 Answers3

12

Let $I(n)$ be given by the integral

$$\begin{align} I(n)&=\int_0^1 \frac{1}{1+x^n}\,dx \tag 1\\\\ \end{align}$$

Then, expanding the integrand of the integral on the right-hand side of $(1)$ in the Taylor series $ \frac{1}{1+x^n}=\sum_{k=0}^\infty (-1)^kx^{nk}$ reveals

$$\begin{align} I(n)&=\sum_{k=0}^\infty \frac{(-1)^k}{nk+1}\\\\ &=1+\frac1n \sum_{k=1}^\infty\frac{(-1)^k}{k+1/n} \tag 2 \end{align}$$

Next, using the fact that $\log(2)= \sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}$ and that $\frac{\pi^2}{12}=-\sum_{k=1}^\infty \frac{(-1)^k}{k^2}$, we can write the series in $(2)$ as

$$\begin{align} \sum_{k=1}^\infty\frac{(-1)^k}{k+1/n} &=-\log(2)+\sum_{k=1}^\infty (-1)^k\left(\frac{1}{k+1/n}-\frac1k\right)\\\\ &=-\log(2)-\frac1n \sum_{k=1}^\infty \frac{(-1)^k}{k(k+1/n)}\\\\ &=-\log(2)-\frac1n \sum_{k=1}^\infty\frac{(-1)^k}{k^2}-\frac1n\sum_{k=1}^\infty (-1)^k\left(\frac{1}{k(k+1/n)}-\frac{1}{k^2}\right)\\\\ &=-\log(2)+\frac{\pi^2}{12n}+O\left(\frac1{n^2}\right) \tag 3 \end{align}$$

Using $(1)-(3)$ yields

$$I(n)=1-\frac{\log(2)}{n}+\frac{\pi^2}{12n^2}+O\left(\frac1{n^3}\right) \tag 4$$

Next, using $(4)$, we can write

$$\begin{align} \left(I(n)\right)^n&=e^{n\log\left(1-\frac{\log(2)}{n}+\frac{\pi^2}{12n^2}+O\left(\frac1{n^3}\right)\right)}\\\\ &=e^{-\log(2)+\frac{\pi^2}{12n}-\frac{\log^2(2)}{2n}+O\left(\frac{1}{n^2}\right)}\\\\ &=\frac12 \left(1+\frac{\pi^2}{12n}-\frac{\log^2(2)}{2n}+O\left(\frac{1}{n^2}\right)\right) \end{align}$$

Finally, we have

$$\begin{align} \lim_{n\to \infty}\left(n\left(\left(I(n)\right)^n-\frac12\right)\right)&=\lim_{n\to \infty}\left(\frac{\pi^2}{24}-\frac{\log^2(2)}{4}+O\left(\frac1n\right)\right)\\\\ &=\bbox[5px,border:2px solid #C0A000]{\frac{\pi^2}{24}-\frac{\log^2(2)}{4}} \end{align}$$

And we are done!

Mark Viola
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  • exzellent! (+1) – tired Nov 15 '16 at 16:25
  • @tired Thank you! Much appreciative. – Mark Viola Nov 15 '16 at 16:28
  • You have a typo - the result is $\frac{\pi^2}{24} - \frac12 \log^2{2} $. Otherwise, what I did. – Ron Gordon Nov 15 '16 at 19:57
  • @RonGordon Yes; good catch. And the coefficient on $\log^2(2)$ is actually $\frac14$. – Mark Viola Nov 15 '16 at 20:11
  • You seem to have used Abel's Thm in showing $I(n)$ equals that series. – zhw. Nov 15 '16 at 20:34
  • @zhw. The DCT is a quick way to go here. – Mark Viola Nov 15 '16 at 20:41
  • I don't see the DCT argument. – zhw. Nov 15 '16 at 20:54
  • @zhw. I didn't embed the argument in the solution. I believe that the legitimacy is well known. -Mark – Mark Viola Nov 15 '16 at 21:01
  • The legitimacy of what, the result or the DCT argument? It's the latter I question (and I doubt it works). The result is fine by Abel's theorem. But that is worth pointing out for sure. – zhw. Nov 15 '16 at 22:34
  • @zhw. $$\left|\frac{1-(-x^n)^{K+1}}{1+x^n}\right| \le \frac{2}{1+x^n}$$This gives an integrable dominating function independent of $K$. The DCT applies. – Mark Viola Nov 15 '16 at 22:41
  • Dr. what you need is $|\sum_{k=0}^{K}(-1)^kx^{nk} - 1/(1+x^n)|$ is dominated by some $g\in L^1(0,1),$ independent of $K.$ – zhw. Nov 15 '16 at 22:57
  • @zhw. We wish to know if $$\lim_{K\to \infty}\int_0^1 \sum_{k=1}^K (-x^n)^k=\lim_{K\to \infty}\int_0^1\frac{1-(-x^n)^{K+1}}{1+x^n},dx=\int_0^1 \lim_{K\to \infty}\frac{1-(-x^n)^{K+1}}{1+x^n},dx=\int_0^1\frac{1}{1+x^n},dx$$To apply the DCT, we need to find an integrable dominating function $f(x;n)$ such that $$\left|\frac{1-(-x^n)^{K+1}}{1+x^n} \right|\le f(x;n)$$Certainly, $f(x;n)=\frac{2}{1+x^n}$ qualifies as such. – Mark Viola Nov 15 '16 at 23:10
  • !!!! Finally stupid me can see you are right. I learned $\sum_{n=1}^{\infty} (-1)^{n+1}/n = \ln 2$ from Abel decades ago, and I've always thought you needed Abel here. Didn't know this. Thank you for your patience. I learned something new. +1 – zhw. Nov 15 '16 at 23:46
  • @zhw. Well, first of all, you are not stupid. In fact, I've been highly impressed with your knowledge and expertise. I've even wondered whether you're a professor or a high-level graduate student. In any case, thank you for putting a second pair of eyes on things. And I'm really pleased that I was "able" to help ... pun intended. – Mark Viola Nov 15 '16 at 23:53
  • Former professor, left academia a while back to do other things. I like to keep my hand in the game, so I come to MSE. My comments here were a case of "I know I'm right" without bothering to entertain the idea I could be mistaken - until my boat sank. – zhw. Nov 15 '16 at 23:57
  • @zhw. I am also a former professor (electromagnetic field theory, a branch of mathematical physics) and left academia in 1996 to pursue a career in financial derivatives trading. No worry about your comments. It forced me to post the justification in the comment, which might be useful to others. -Mark – Mark Viola Nov 16 '16 at 00:07
  • @Dr. MV: Instructive and concise derivation! ... and nice comment! ( +1) – Markus Scheuer Nov 16 '16 at 06:13
  • @markusscheur Thank you Markus. Much appreciative. -Mark – Mark Viola Nov 16 '16 at 14:58
5

Hint. If one knows the digamma function $\psi=\Gamma'/\Gamma$, one may write, as $n \to \infty$, $$ \begin{align} \int_0^1\frac1{1+x^n}\:dx&=\int_0^1\frac{1-x^n}{1-x^{2n}}\:dx \\\\&\stackrel{u=x^{2n}}=\frac1{2n}\int_0^1\frac{(1-u^{1/2})u^{1/(2n)-1}}{1-u}\:du \\\\&=\frac1{2n}\left[\psi\left(\frac1{2n}+\frac12 \right)-\psi\left(\frac1{2n} \right) \right] \\\\&=1-\frac{\ln 2}{n}+\frac{\pi ^2-6 \ln^2 2}{12n^2}+O\left(\frac{1}{n^3}\right) \end{align} $$ giving, as $n \to \infty$, $$ n\ln \left(1-\frac{\ln 2}{n}+\frac{\pi ^2-6 \ln^2 2}{12n^2}+O\left(\frac{1}{n^3}\right) \right)=-\ln 2+\frac{\pi ^2-6 \ln^2 2}{24n}+O\left(\frac{1}{n^2}\right) $$ that is

$$ n\left[\left(\int_0^1\frac1{1+x^n}\:dx \right)^n-\frac12\right]\to \frac{\pi ^2-6 \ln^2 2}{24}. $$

Olivier Oloa
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5

My approach is a little different. I would sub $x=u^{1/n}$ in the integral and get

$$I(n) = \int_0^1 \frac{dx}{1+x^n} = \frac1n \int_0^1 du \frac{u^{\frac1n-1}}{1+u} = \frac1n \int_0^1 du \, u^{1/n} \left (\frac1u - \frac1{1+u} \right ) \\= 1-\frac1n \int_0^1 du \frac{u^{1/n}}{1+u}$$

We then note that $u^{1/n} = e^{\log{u}/n}$ and that $n$ is large enough for the following series expansion to be valid:

$$I(n) = 1- \sum_{j=0}^{\infty} \frac1{n^{j+1}} \int_0^1 du \frac{\log^j{u}}{1+u} $$

Due to the nature of the limit we are posed, we go out to second order; thus

$$I(n) = 1-\frac{\log{2}}{n} + \frac{\pi^2}{12 n^2} +O \left ( \frac1{n^3} \right ) $$

Then

$$\begin{align}I(n)^n &= e^{n \log{\left [1-\frac{\log{2}}{n} + \frac{\pi^2}{12 n^2} +O \left ( \frac1{n^3} \right )\right ]} } \\ &= e^{n\left [-\frac{\log{2}}{n} + \frac{\pi^2}{12 n^2} - \frac{\log^2{2}}{2 n^2}+O \left ( \frac1{n^3} \right ) \right ] } \\ &= \frac12 \left [1+\left (\frac{\pi^2}{12} - \frac12 \log^2{2} \right ) \frac1n +O \left ( \frac1{n^2} \right ) \right ]\end{align}$$

The sought after limit is then

$$\lim_{n \to \infty} n \left [I(n)^n-\frac12 \right ] = \frac{\pi^2}{24} - \frac14 \log^2{2} $$

Ron Gordon
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