Very nifty limit

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

Solution: I chose this problem because the answer is highly nontrivial and just out of left field. But the process of getting there seems so straightforward; it is not really.

Substitute $x=u^{1/n}$ 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:

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

Is it really straightforward that this approach would work, never mind be valid? Not sure. It took me about four failed attempts before I stumbled upon this approach. I guess when you write things up, it all looks so easy. Trust me, it isn’t. (And there are other ways to solve this problem as well.)

OK, enough ranting. 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

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

5 Comments

  • CodeLabMaster wrote:

    Unless I’m looking at it wrong, I think you’re missing a $j!$ term in the denominator of the series formula for $I(n)$. It ends up having no impact on the answer, though, because $0! = 1! = 1$, and the $j \geq 2$ terms end up inside the big O.

    Interesting problem choice, and nice solution as always!

  • Thanks for pointing that out – yes, it was a typo. I appreciate the feedback!

  • Bennett Gardiner wrote:

    Beautiful!

  • Lanier Freeman wrote:

    I realize this is a poor place to put this but seeing as I couldn’t find a way to DM you, here we are. I’ve followed you on Math SE for a while and admire your work (if you don’t believe me, my name should pop up under user search there). I’m aware this is out of nowhere, but I’m supposed to interview someone who does work in my field or related fields for a college project and was wondering if you’d be willing to let me interview you tonight or tomorrow morning. I realize that’s a super late notice but I’ve been cramming before the break and thought I’d at least try.

    Thanks.

  • Hi Lanier, I would be honored to be interviewed by you. You can send me an email at rlgordonma at gmail dot com.

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