Show that $\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2} \, dx <\frac{\pi}{4}$

Question

Show that $$\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2}\,dx <\frac{\pi}{4}$$

I want to use if $f<g<h$ then $\int f<\int g<\int h$ formula for Riemann integration.

$1+x^2<1+x+x^2$ and it will give RHS as $$\frac{1}{1+x+x^2}<\frac{1}{1+x^2}$$ How to choose function $f$ and $h.$

Answer 1

For $0< x< 1$ it holds: $1+x^2<1+x+x^2<3$. Therefore $$\frac{1}{3}=\int^{1}_0\frac{1}{3}\,dx<\int^1_0\frac{1}{1+x+x^2}\,dx<\int^1_0\frac{1}{1+x^2}\,dx=\arctan(x)\Big|^1_0=\frac{\pi}{4}$$

Answer 2

You want to show $$\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2}dx <\frac{\pi}{4}$$

Note that for $0<x<1$ we have $$ 1+x^2 < 1+x+x^2 <3 $$ Thus $$ \frac {1}{3} < \frac {1}{1+x+x^2}< \frac {1}{1+x^2}$$

Upon integration we get $$\frac{1}{3}=\int^{1}_0\frac{1}{3}\,dx<\int^1_0\frac{1}{1+x+x^2}\,dx<\int^1_0\frac{1}{1+x^2}\,dx=\arctan(x)\Big|^1_0=\frac{\pi}{4}$$

Answer 3

Note that for $x\ge0$, $1+x\le x^2+x+1\le(x+1)^2$.

Hence, we have

$$\frac13<\frac12=\int_0^1 \frac{1}{(1+x)^2}\,dx\le \int_0^1 \frac{1}{x^2+x+1}\,dx\le \int_0^1 \frac{1}{1+x}\,dx=\log(2)<\frac\pi4$$

which provide tighter bounds then those that were requested.

Answer 4

A better upper bound: by the convexity of the integrand in $[0,1]$: $$\forall x\in[0,1]:\qquad\frac1{1 + x + x^2}\le 1 - \frac{2x}3,$$ and this implies $$\int_0^1\frac{1}{1+x+x^2}dx\le\int_0^1(1 - 2x/3)\,dx = \frac23.$$

Answer 5

Alternative approach: for any $x\in(0,1)$ we have $$ \frac{1}{1+x+x^2}=\frac{1-x}{1-x^3}=(1-x)\sum_{n\geq 0}x^{3n}=\sum_{n\geq 0}x^{3n}-x^{3n+1} \tag{1}$$ hence $$ \mathcal{J}=\int_{0}^{1}\frac{dx}{1+x+x^2} = \sum_{n\geq 0}\frac{1}{(3n+1)(3n+2)} \tag{2}$$ and the series in the RHS of $(2)$ can be approximated through the Hermite-Hadamard inequality, since $g(x)=\frac{1}{(3x+1)(3x+2)}$ is convex on $\mathbb{R}^+$. We get $$ \frac{1}{2}+\frac{1}{20}+\int_{2}^{+\infty}g(x)\,dx\leq \mathcal{J}\leq \frac{1}{2}+\frac{1}{20}+\int_{3/2}^{+\infty}g(x)\,dx \tag{3}$$ $$\frac{11}{20}+\frac{1}{3}\log\left(1+\frac{1}{7}\right)\leq \mathcal{J}\leq \frac{11}{20}+\frac{1}{3}\log\left(1+\frac{2}{11}\right)\tag{4}$$ so the distance between $\mathcal{J}$ and $\frac{3}{5}$ is at most $\frac{1}{100}$.
(2) and the reflection formula for the $\psi$ function imply $\mathcal{J}=\frac{\pi}{3\sqrt{3}}$, which can be proved by just completing the square, too. Another interesting approach is creative telescoping: $$\begin{eqnarray*}\sum_{n\geq 0}\frac{1}{(3n+1)(3n+2)}&=&\frac{1}{2}+\sum_{n\geq 0}\left(\frac{1}{9(n+1)}-\frac{1}{9(n+2)}-\frac{2}{(3n+3)(3n+4)(3n+5)(3n+6)}\right)\\&=&\frac{1}{2}+\frac{1}{9}-\frac{1}{180}-2\sum_{n\geq 0}\frac{1}{(3n+6)(3n+7)(3n+8)(3n+9)}\end{eqnarray*} $$ also leading to the accelerated series $\mathcal{J}=\sum_{n\geq 1}\frac{1}{n\binom{2n}{n}}$.