Approximating the integral $\int_0^1 \frac{x}{1+x^8} dx$

Question

Integral: $J=\int_0^1 \frac{x}{1+x^8}dx$

Consider the following assertions:

$I:J> \frac{1}{4}$ and $II:J< \frac{\pi}{8}$

A. Both are true

B. Only $I$ is true

C. Only $II$ is true

D. Both are wrong

I tried using the Trapezoidal rule and got $J \approx \frac {1}{4}$ but I'm not sure how to prove the inequality/ find out which one is correct.

Answer 1

$$\int_0^1\frac x{1+x^8}dx=\frac12\int_0^1\frac 1{1+z^4}dz>\frac12\int_0^1\frac 1{1+z^2}dz=\frac\pi8>\frac14.$$

(By WA, $J\approx0.433486$.)

Answer 2

A different approach is to recognize $\frac{x}{1+x^8}$ as the sum of an infinite geometric series with $r=-x^8$ and $a_0=x$, which converges given the bounds of the integral. \begin{align*} \int_0^1 x-x^9+x^{17}\;\mathrm{d}x&>\int_0^1 \frac{x}{1+x^8}\; \mathrm{d}x>\int_0^1 x-x^9\; \mathrm{d}x \\ \frac{41}{90}&>J> \frac{2}{5} \\ 0.4555&>J>0.4 \\ \end{align*}

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Albeit the question is looking for an $\underline{approximation}$, it's good to know that there's a closed expression for the integral in terms of the Digamma Function $\ds{\Psi}$. Namely, \begin{align} &\bbox[5px,#ffd]{\int_{0}^{1}{x \over 1 + x^{8}}\,\dd x} = \int_{0}^{1}{x - x^{9} \over 1 - x^{16}}\,\dd x = {1 \over 16}\int_{0}^{1}{x^{-7/8} - x^{-3/8} \over 1 - x}\,\dd x \\[5mm] = &\ {1 \over 16}\bracks{\int_{0}^{1}{1 - x^{-3/8} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{-7/8} \over 1 - x}\,\dd x} \\[5mm] = &\ {1 \over 16}\bracks{\Psi\pars{5 \over 8} - \Psi\pars{1 \over 8}} \\[5mm] = &\ \bbx{{\root{2} \over 16}\bracks{\pi + \ln\pars{3 + 2\root{2}}} \approx 0.4335} \\ & \end{align} The Digamma Functions evaluation were performed with the Gauss Digamma Theorem.