Let $m(x)$ be the Lebesgue measure. I want to show that there exists $C\in \mathbb R$ such that $$\lim_{n \rightarrow \infty} n\left(\int_{0}^{\infty} \frac{1}{1+x^4+x^n} \mathrm{d}m(x)-C\right)$$ exists as a finite number and then compute the limit. Rewriting this gives:

$$\lim_{n \rightarrow \infty}\int_{0}^{\infty}n \left(\frac{1-C(1+x^4+x^n)}{1+x^4+x^n}\right)\mathrm{d}m(x)$$

And I thought I should use the Lebesgue dominated convergence theorem somehow to put the limit inside the integral (because calculating an integral of a rational function like this doesn't seem like a good idea) but I can't find any dominating function. Also I tried splitting into the intervals $(0,1)$ and $(1,\infty)$ but this didn't help me either... Any help is appreciated!

  • $\begingroup$ for the limit to be finite, the second factor inside the limit must approach zero, so $C = \int_0^\infty dx/(1+x^4)$. $\endgroup$ – Stefan Smith Jan 6 '14 at 0:55

Write the integral as $$\int_0^1 \frac{dx}{1+x^4 + x^n} + \int_1^\infty\frac{dx}{1+x^4 + x^n}.$$ The second integral goes to $0,$ while the first goes to $\int_0^1 \frac{dx}{1+x^4},$ so $$C = \int_0^1 \frac{dx}{1+x^4} = \frac{\pi + 2 \mathop{acoth}(\sqrt{2})}{4\sqrt{2}}.$$ Now that we know what $C$ is, we need to estimate the error.

The second integral is bigger than $\int_1^\infty x^{-n} dx = \frac{1}{n+1},$ and smaller than $(1+\epsilon) \int_1^\infty x^{-n} dx,$ for any $x,$ so the second integral contributes $1$ to the limit.

The first integral less $C$ equals $$\int_0^1 \frac{x^n d x}{1+x^4 + x^n},$$ which can be explicitly evaluated as $$ \frac{1}{8} \left(\psi ^{(0)}\left(\frac{n+5}{8}\right)-\psi ^{(0)}\left(\frac{n+1}{8}\right)\right), $$ where $\psi^{(0)}$ is the digamma function, and is asymptotic to $1/2n,$ so the limit is $3/2.$ If the digamma does not make you happy, it is easy to see that the limit is $\frac{1}{2n}$ for the first integral. This is so because for any $\epsilon,$ the integral from $0$ to $1-\epsilon$ decreases exponentially in $n,$ and near $1$ the estimate is easy to get, by approximating $1+x^4$ by $2.$

  • $\begingroup$ Doesn't this only work when $\frac{x^n}{1+x^4}<1$? In the range $(0,\infty)$, this is not always the case. $\endgroup$ – Clayton Jan 6 '14 at 21:02
  • $\begingroup$ @Clayton you are right, fixed. $\endgroup$ – Igor Rivin Jan 7 '14 at 0:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.