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Prove an inequality for the integral of 1 / (1 + x4)

by RoRi

Prove that

    \[ 0.493948 < \int_0^{\frac{1}{2}} \frac{1}{1+x^4} \, dx < 0.493958. \]

(Note: I cannot get the bounds Apostol asks for. I prove a different set below. I cannot figure out if it is a mistake in the book or not.)

Proof. Using the algebraic identity, valid for 0 < x < 1,

    \[ \frac{1}{1-x} = 1+x+x^2+\cdots + x^n + \frac{x^{n+1}}{1-x}, \]

we obtain

    \[ \frac{1}{1+x^4} = 1 - x^4 + x^8 - x^{12} + \cdots + (-1)^n x^{4n} + (-1)^{n+1} \frac{x^{4n+4}}{1+x^4}. \]

Therefore, integrating term by term,

    \begin{align*} \int_0^{\frac{1}{2}} \frac{1}{1+x^4} \, dx &= \int_0^{\frac{1}{2}} \left( 1 - x^4 + x^8 + \cdots + (-1)^n x^{4n} + (-1)^{n+1} \frac{x^{4n+4}}{1+x^4} \right) \, dx \\[9pt] &= \left( x - \frac{x^5}{5} + \frac{x^9}{9} - \cdots + (-1)^n \frac{x^{4n+1}}{4n+1} \right)\Bigr \rvert_0^{\frac{1}{2}} + (-1)^{n+1} \int_0^{\frac{1}{2}} \frac{x^{4n+4}}{1+x^4} \, dx. \end{align*}

Furthermore, we have

    \[ \int_0^{\frac{1}{2}} \frac{x^{4n+4}}{1+x^4} \, dx < \int_0^{\frac{1}{2}} x^{4n+4} \, dx = \left( \frac{1}{2} \right)^{4n+5} \left( \frac{1}{4n+5} \right). \]

Taking n = 2, we then have

    \[ \int_0^{\frac{1}{2}} \frac{1}{1+x^4} \, dx = \left( \frac{1}{2} \right) - \frac{(1/2)^5}{5} + \frac{(1/2)^9}{9} - \int_0^{\frac{1}{2}} \frac{x^{4n+4}}{1+x^4} \, dx.\]

From the inequality for this integral we then have

    \begin{align*} &\frac{1}{2} - \frac{(1/2)^5}{5} + \frac{(1/2)^9}{9} - \left( \frac{1}{2} \right)^{4n+5} \left( \frac{1}{4n+5} \right) < \int_0^{\frac{1}{2}} \frac{1}{1+x^4} \, dx\\ \intertext{and} &\int_0^{\frac{1}{2}} \frac{1}{1+x^4} \, dx < \frac{1}{2} - \frac{(1/2)^5}{5} + \frac{(1/2)^9}{9} \\[9pt] \implies & 0.493958 < \int_0^{\frac{1}{2}} \frac{1}{1+x^4} \, dx < 0.49367. \qquad \blacksquare \end{align*}

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