Let m(x) be the Lebesgue measure. I want to show that there exists C∈R such that
lim
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!
Answer
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.
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