Computation of $\lim_{n \rightarrow \infty} n(\int_{0}^{\infty} \frac{1}{1+x^4+x^n} \mathrm{d}m(x)-C)$ 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!
 A: 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 $2.$
