How can I tell if the improper integral of $1/(1+u^8)$ converges, and how can I compute that limit? This question piqued my interest in a rather monstrous definite integral: $$F(x)=\int_0^x \frac{du}{1+u^8}$$
When I initially looked at the graph, it seemed that $F(x)\to1$ as $x\to\infty$ and that $F(x)\to-1$ as $x\to-\infty$; however, after looking more closely at the graph and tabulating some values, I realized that this was not the case:
$$\begin{array}{c|c}
x & F(x) \\ \hline
1 & \: \; 0.924\,651\,71 \\
2 & 1.025\,058\,1 \\
3 & 1.026\,106\,8 \\
4 & 1.026\,163\,4 \\
5 & 1.026\,170\,3 \\
6 & 1.026\,171\,6 \\
7 & 1.026\,172\,0 \\
8 & 1.026\,172\,1 \\
9 & 1.026\,172\,1 \\
10 & 1.026\,172\,1 \\
10^2 & 1.026\,172\,2 \\
10^3 & 1.026\,172\,2 \\
10^7 & 1.026\,172\,2 \\
\end{array}$$
My calculator started returning error at $10^7$, so I would make an educated guess that $\lim_{x\to\infty}F(x)\approx1.026$. However, out of sheer curiosity, I’d like to know the following:


*

*How can I (using calculus and/or algebra) demonstrate that $F(x)$ approaches some real value?

*How would I go about computing that value?

*What is that value?


The only method I know would be to use brute force and tackle the integral’s closed form with an $\epsilon$-$\delta$ approach or, as Will Jagy pointed out, standard limit identities.
 A: You use Cauchy's criterion. If $0<x<y$, then
$$
\int_x^y \frac{du}{1+u^8}<\int_x^y \frac{du}{u^8}=\frac17\,\Bigl(\frac{1}{y^7}-\frac{1}{x^7}\Bigr)<\frac{1}{y^7}.
$$
This means that $\int_x^y \frac{du}{1+u^8}$ van be made as small as you wish just by taking $x,y$ large enough, and this implies the existence of $\lim_{x\to\infty}\int_0^x \frac{du}{1+u^8}$.
To compute the value I am afraid you will have to compute the integral.
If you want to know more, google for improper integral and Weierstrass $M$-test.
A: With very little effort, we can compute an upper bound on $$\int_{u=0}^\infty \frac{du}{1+u^8}$$ by observing that on $u \in [0,1]$, $$\frac{1}{1+u^8} \le 1.$$  Then on $u \in (1,\infty)$ we have $$\frac{1}{1+u^8} < \frac{1}{u^8}.$$  Consequently $$\int_{u=0}^\infty \frac{du}{1+u^8} = \int_{u=0}^1 \frac{du}{1+u^8} + \int_{u=1}^\infty \frac{du}{1+u^8} < 1 + \int_{u=1}^\infty \frac{du}{u^8} = 1 + \left[-\frac{1}{7u^7}\right]_{u=1}^\infty = \frac{8}{7}.$$
An exact computation is possible with elementary methods, but is a bit tedious.
A: Since the integrand is meromorphic and has good decay at infinity, this integral can be computed exactly with the machinery of complex analysis. In the interests of your understanding I have attempted to be thorough, and have left a number of links throughout which may help. 
This integral is simultaneously "not hard" and "a little messy", because it only uses the standard tools for integrating functions with residue theory, while also being longer than what I'd expect to see in homework. There are many many more examples of this technique, some very smart, on this website alone. If this is too hard, I would first look there for more examples.
First, a quick check to see that the integral has a limit; The integrand $\frac1{1+u^8}$ is reasonably nice (bounded, continuous) so at least the integral from $0$ to $x$ is well-defined, for any $x$. 
Notice that the integrand is positive. This means that its integral $\int_0^x \frac{du}{1+u^8}$ is an increasing function of $x$, and a simple bound like heropup's tells us that it is bounded. An increasing bounded function has a limit; thus
$$I = \int_0^\infty \frac{du}{1+u^8} := \lim_{x\to\infty} \int_0^x \frac{du}{1+u^8}$$
exists.
To compute the integral, we need to set up the integral so that we  can apply the residue theorem. Notice that the integrand is even in $u$. Thus,
$$ 2I = \int_{-\infty}^\infty \frac{du}{1+u^8}$$
We will need to decide on a closed contour for the integration. Define the points $\vec{E}=-R,\vec F=R$ for $R>0$, and write $\vec{FdE}$ for the semicircular arc going counter clockwise from $F$ to $E$. I have plotted these, together with the roots of $1+u^8 = 0$ (i.e. poles of the integrand),

(I have used the automatic labelling from the program (Geogebra), the letters don't have any special significance other than the order I created the objects in)
The arc $\vec{FdE}$ has counter-clockwise parameterisation
$$ u(t) = R e^{it}, \quad t \in [0,π]$$
with $u'(t) = iR e^{it}$, and we have the bound using the estimation lemma and the triangle inequality in the form $||a|-|b||≤|a+b|$,
$$ \left|\int_{\vec{FdE}} \frac{du}{1+u^8}\right| ≤ \int_0^π \frac{|iRe^{it}|dt}{|1+R^8\exp(8it)|} ≤ 2π R \frac{1}{|R^8-1|} ≤ \frac{C}{R^7} \xrightarrow[R\to\infty]{} 0  $$
But we also know that 
$ \int_{\vec{EF}} \frac{du}{1+u^8} \xrightarrow[R\to\infty]{}  2I$.
Hence the integral over the closed contour $\gamma$, which is the line $\vec{EF}$ followed by the arc $\vec{FdE}$ (traversed in the counter-clockwise direction) is
$$ \int_{\gamma} \frac{du}{1+u^8} = \int_{\vec{EF}} \frac{du}{1+u^8} + \int_{\vec{FdE}} \frac{du}{1+u^8}  \xrightarrow[R\to\infty]{}  2I$$
On the other hand, the residue theorem gives us an alternative characterisation for $R>1$,
$$ \int_{\gamma} \frac{du}{1+u^8} = 2π i \sum_z \operatorname{Res}\left(\frac1{1+u^8},z\right)$$
where the sum is over the poles $z$ that are enclosed in $\gamma$, i.e. $z=B',B'',B''',C$ in the diagram, which are explicitly
$$ B' = e^{iπ/8}=:w,\  B'' =w^3, \ B''' =w^5,\ C =w^7.$$
All the poles of $\frac1{1+u^8}$ are simple; hence we have the easy formula for any pole $z$,
 $$ \operatorname{Res}\left(\frac1{1+u^8}, z\right) = \lim_{u\to z} \frac{u-z}{1+u^8} = \frac{1}{8z^7}$$
where the last equality is by l'Hopital's rule. Hence we can compute 
\begin{align} 2I 
&= 2π i \frac{w^{-7}}{8}\left(1 + w^{-14} + (w^{-14})^2 + (w^{-14})^3  \right) \\
&= \frac{π i }{4} \cdot \frac{1-w^{-56}}{w^{-7}-w^{-7}}
\end{align}
$56 = 3\times16+ 8$ so as $w$ is an $16$th root of unity, $w^{-56} = w^{-8} = -1$, so the numerator is 2. 
As for the denominator, $w^{-7} = \overline{w^7}$ the complex conjugate, and when you subtract a number from its complex conjugate, $x - \bar x = 2 i \operatorname{Im} x$ , i.e. $2i$ times its imaginary part remains. So the denominator is $2i \sin \pi/8$. 
Putting this all together finally gives
$$2I = \frac{π i}{4} \cdot \frac{2}{2i \sin \pi/8} =  \frac{π}{4}\csc\left(\frac{π}{8}\right)$$
so that the integral we originally wanted is
$$ I = \int_0^\infty \frac{du}{1+u^8} = \frac{π}{8}\csc\left(\frac{π}{8}\right) \approx 1.02617215\color{lightgray}{297703088887146778087283}$$
 which agrees with all the digits you found, which is reasonable because $10^7$ is big. In fact, your truncation error can be estimated,
$$ \int_x^\infty \frac{du}{1+u^8} ≤ \int_x^\infty \frac{du}{u^8} = \frac{7}{x^7}$$
So truncating at $x=10^7$ means (if your calculator was perfect) you should have at least 48 correct digits $$  \int_{10^7}^\infty \frac{du}{1+u^8} ≤7 \times 10^{-49} ≤ 10^{-48}$$
Depending on how the number is stored, this could be past the accuracy of your machine.
