Prove that
![\[ 0.493948 < \int_0^{\frac{1}{2}} \frac{1}{1+x^4} \, dx < 0.493958. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-9ca2da2e832eae718820b45dccaf67b4_l3.png "Rendered by QuickLaTeX.com")
(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 
,
![\[ \frac{1}{1-x} = 1+x+x^2+\cdots + x^n + \frac{x^{n+1}}{1-x}, \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-3e7e8a85dc56d188155362b5b0644d5f_l3.png "Rendered by QuickLaTeX.com")
we obtain
![\[ \frac{1}{1+x^4} = 1 - x^4 + x^8 - x^{12} + \cdots + (-1)^n x^{4n} + (-1)^{n+1} \frac{x^{4n+4}}{1+x^4}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-c4eb7c5e1458b38b9176e669fa762319_l3.png "Rendered by QuickLaTeX.com")
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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-4695e424608726f67a7afd9967efbc05_l3.png "Rendered by QuickLaTeX.com")
Furthermore, we have
![\[ \int_0^{\frac{1}{2}} \frac{x^{4n+4}}{1+x^4} \, dx < \int_0^{\frac{1}{2}} x^{4n+4} \, dx = \left( \frac{1}{2} \right)^{4n+5} \left( \frac{1}{4n+5} \right). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-03518a39741185959477118827aa632c_l3.png "Rendered by QuickLaTeX.com")
Taking 
, we then have
![\[ \int_0^{\frac{1}{2}} \frac{1}{1+x^4} \, dx = \left( \frac{1}{2} \right) - \frac{(1/2)^5}{5} + \frac{(1/2)^9}{9} - \int_0^{\frac{1}{2}} \frac{x^{4n+4}}{1+x^4} \, dx.\]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0f509993aae879a8c5d3387e73985232_l3.png "Rendered by QuickLaTeX.com")
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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-e1fcaeafd7b52718901919245ae49491_l3.png "Rendered by QuickLaTeX.com")
The integral evaluates to a number slightly higher than the upper bound in the book, so I guess he rounded to 6 decimals. If you take the first 4 terms up till x^12 and round the result, you get the upper bound in the book.