An improper integral. Prove an inequation.

Question

$$0 < \int\limits_0^\infty \frac{x^{20} + 1}{x^{40} + 1}\,dx - \frac{20}{19} < 0.05$$

I tried to use that $$ \frac{x^{20}}{x^{40}} < \frac{x^{20} + 1}{x^{40} + 1} < \frac{x^{20} + 1}{x^{40}}$$ but it didn't help because the lower bound of the integral $=0$

Answer 1

$$\begin{eqnarray*}\int_{0}^{+\infty}\frac{x^n+1}{x^{2n}+1}\,dx &=& \int_{0}^{1}\frac{x^n+1}{x^{2n}+1}\,dx + \int_{0}^{1}\frac{x^{n-2}+x^{2n-2}}{x^{2n}+1}\,dx\\&=&\int_{0}^{1}\frac{(x^n+1)(x^n+x^2)}{x^2(x^{2n}+1)}\,dx \end{eqnarray*}$$

but over $(0,1)$ we have $x^{2n}<x^n$, hence: $$ \int_{0}^{+\infty}\frac{x^n+1}{x^{2n}+1}\,dx > \int_{0}^{1}\frac{x^n+x^2}{x^2}\,dx = 1+\frac{1}{n-1}$$ but since: $$ \frac{(x^n+1)(x^n+x^2)}{x^2(x^{2n}+1)}-\frac{x^n+x^2}{x^2}=\frac{(x^n-x^{2n})(x^n+x^2)}{x^2(x^{2n}+1)}=x^n\cdot\frac{(1-x^n)(1+x^{n-2})}{1+x^{2n}}$$ is less than $x^n(1-x^n)(1+x^{n-2})$, it is not difficult to get tight upper bounds for $\int_{0}^{+\infty}\frac{x^n+1}{x^{2n}+1}\,dx$, too, like: $$ \int_{0}^{+\infty}\frac{x^n+1}{x^{2n}+1}\,dx < \frac{n}{n-1}+\frac{2}{3n}.$$

An even tighter inequality is:

$$\forall n\geq 1,\qquad 1+\frac{\pi}{2n}\leq \int_{0}^{+\infty}\frac{x^n+1}{x^{2n}+1}\,dx \leq 1+\frac{\pi}{2n-2}.$$

It can be proved by combining the previous approach with the Cauchy-Schwarz inequality.

Answer 2

Write $I = \int_{0}^{1}\frac{x^n+1}{x^{2n}+1}\,dx + \int_{1}^{+\infty}\frac{x^n+1}{x^{2n}+1}\,dx$ and consider the two contributions separately: For the integral over $0, 1$ use $\frac{1+ x^n}{1 + x^{2n}}\,\le 1+ x^n$ so that $\int_{0}^{1}\frac{1 + x^n}{1 + x^{2n}}\,dx \lt \int_{0}^{1} 1 + x^n\,dx = 1 + \frac{1}{n+1}$.

For the second integral over $1, \infty$: $\int_{1}^{+\infty}\frac{x^n+1}{x^{2n}+1}\,dx \lt \int_{1}^{+\infty}\frac{x^n+1}{x^{2n}}\,dx = \frac{1}{n-1} + \frac{1}{2 n-1}$.

Collecting all the results and rearranging gives $I \lt 1 + \frac{1}{n-1} + \frac{1}{n+1} + \frac{1}{2n-1}$. The first two terms is equal to $\frac{n}{n-1}$ so that $I \lt \frac{n}{n-1} + \frac{1}{n+1} + \frac{1}{2n-1} \lt \frac{n}{n-1} + \frac{1}{n} + \frac{1}{2n-1}$.