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Note first that $x^{x}=e^{{x\log x}}$. (We define $x\log x$ to be $0$ when $x=0$ so that it is consistent with the limits of both $x\log x$ and $x^{x}$ as $x$ approaches $0$ from the right.) The Maclaurin series for $e^{x}$ yields that
$x^{x}=\sum_{{n=0}}^{{\infty}}\frac{(x\log x)^{n}}{n!}.$ 
Thus,
$\int\limits_{0}^{1}x^{x}\,dx=\sum_{{n=0}}^{{\infty}}\frac{1}{n!}\int\limits_{0% }^{1}(x\log x)^{n}\,dx.$ 
Making the substitution $u=\log x$, we have that $x=e^{u}$ and $dx=e^{u}\,du$. Therefore,
$\int\limits_{0}^{1}x^{x}\,dx=\sum_{{n=0}}^{{\infty}}\frac{1}{n!}\int\limits_{{% \infty}}^{0}(e^{u}\cdot u)^{n}e^{u}\,du=\sum_{{n=0}}^{{\infty}}\frac{1}{n!}% \int\limits_{{\infty}}^{0}u^{n}e^{{(n+1)u}}\,du.$ 
Making the substitution $v=(n+1)u$, we have that $dv=(n+1)\,du$. Hence,
$\begin{array}[]{rl}\displaystyle\int\limits_{0}^{1}x^{x}\,dx&\displaystyle=% \sum_{{n=0}}^{{\infty}}\frac{1}{(n+1)!}\int\limits_{0}^{{\infty}}\left(\frac{% v}{n+1}\right)^{n}e^{{v}}\,dv\\ \\ &\displaystyle=\sum_{{n=0}}^{{\infty}}\frac{1}{(n+1)!}\left(\frac{1}{n+1}% \right)^{n}\int\limits_{0}^{{\infty}}v^{n}e^{{v}}\,dv\\ \\ &\displaystyle=\sum_{{n=1}}^{{\infty}}\frac{1}{n!}\left(\frac{1}{n}\right)^{{% n1}}\int\limits_{0}^{{\infty}}v^{{n1}}e^{{v}}\,dv\\ \\ &\displaystyle=\sum_{{n=1}}^{{\infty}}\frac{1}{n!}\left(\frac{1}{n}\right)^{{% n1}}\Gamma(n)\\ \\ &\displaystyle=\sum_{{n=1}}^{{\infty}}\frac{1}{n!}\left(\frac{1}{n}\right)^{{% n1}}(n1)!\\ \\ &\displaystyle=\sum_{{n=1}}^{{\infty}}\frac{1}{n}\left(\frac{1}{n}\right)^{{n% 1}}\\ \\ &\displaystyle=\sum_{{n=1}}^{{\infty}}\frac{(1)^{{n1}}}{n^{n}}\end{array}$
The above summation is an alternating series that converges extremely rapidly, but to what???
The theorem presented here relates to constructing infinite locally cyclic rings.
Theorem 1.
Let all of the following be true:

$n$ is a positive integer;

$p_{1},\dots,p_{{n+1}}$ are distinct primes;

$\displaystyle P=\prod_{{j=1}}^{n}p_{j}$;

$Q=p_{{n+1}}P$;

$K$ is a positive divisor of $P$.
Then there exists a $c\in\mathbb{Z}$ such that $\varphi\colon K\mathbb{Z}_{{KP}}\to K\mathbb{Z}_{{KQ}}$ defined by $\varphi(m)=p_{{n+1}}cm$ is a ring homomorphism (if $K=1$, it is not insisted upon that $\varphi(1)=1$) and $\varphi$ yields a ring isomorphism between $K\mathbb{Z}_{{KP}}$ and $p_{{n+1}}K\mathbb{Z}_{{KQ}}$.
Proof.
It is clear that $\varphi\left(K\mathbb{Z}_{{KP}}\right)=p_{{n+1}}K\mathbb{Z}_{{KQ}}$ and that $\varphi$ preserves the additive structures of these two rings. Since $K$ is a divisor of $P$, the ring $K\mathbb{Z}_{{KP}}$ has order $P$ and behavior $K$. Since
$\frac{KQ}{p_{{n+1}}K}=P,$ 
we also have that $p_{{n+1}}K\mathbb{Z}_{{KQ}}$ has order $P$. To determine the ring’s behavior, let us consider what happens when we square $p_{{n+1}}K$, which is a generator of the ring. Note that $p_{{n+1}}$ is relatively prime to $P$. Since $(p_{{n+1}}K)^{2}=(p_{{n+1}}K)(p_{{n+1}}K)$, we have a generator $r$ of the ring such that $r^{2}=(p_{{n+1}}K)r$. Within the theorem that proves existence of behavior for cyclic rings, it is proven that this fact implies that the behavior of the ring is $\gcd(p_{{n+1}}K,P)=K$, and the $c\in\mathbb{Z}$ as described in the statement of the theorem is constructed. Hence, $\varphi$ yields a ring isomorphism between $K\mathbb{Z}_{{KP}}$ and $p_{{n+1}}K\mathbb{Z}_{{KQ}}$. This completes the proof. ∎
This theorem seems to indicate a way to extend any cyclic ring with squarefree order and any valid behavior $K$ to an infinite locally cyclic ring with behavior $K$. In fact, there are many ways to perform this extension, depending on which primes you choose to “attach”. (It has been wellknown to me for a long time that finite cyclic rings of squarefree order are very nice, so this result is not much of a surprise to me, though it does seem to help out.)
Now, we look at a more general result.
Theorem 2.
Assume that we have a chain of finite cyclic groups $C_{1}\subseteq C_{2}\subseteq C_{3}\dots$ with a multiplication defined on each cyclic group such that, for any $j$, if $a,b\in C_{j}$, then $ab\in C_{{j+1}}$. Then each $C_{j}$ is a cyclic ring in its own right with the order of $C_{j}$ dividing the order of $C_{{j+1}}$ for all $j$.
Proof.
Since $C_{j}$ is a subgroup of $C_{{j+1}}$ for all $j$, we know that the order of $C_{j}$ divides the order of $C_{{j+1}}$.
For all $j$, let $x_{j}$ be a generator of $C_{j}$, $c_{j}$ be the order (which is also the characteristic) of $C_{j}$, and $m_{j},n_{j}\in\mathbb{N}$ such that
$\displaystyle x_{j}$  $\displaystyle=m_{j}x_{{j+1}},$  
$\displaystyle{x_{j}}^{2}$  $\displaystyle=n_{j}x_{{j+1}}.$ 
Without loss of generality, we may assume that $m_{j}$ divides $c_{{j+1}}$. (If not, $x_{{j+1}}$ can be rechosen so that it is still a generator and $m_{j}$ divides $c_{{j+1}}$.) Then
$\displaystyle\left(\frac{c_{{j+1}}}{m_{j}}\right)x_{j}$  $\displaystyle=\left(\frac{c_{{j+1}}}{m_{j}}\right)m_{j}x_{{j+1}}$  
$\displaystyle=c_{{j+1}}x_{{j+1}}$  
$\displaystyle=0.$ 
Thus,
$\displaystyle c_{{j+1}}\left(\frac{n_{j}}{m_{j}}\right)x_{{j+1}}$  $\displaystyle=\left(\frac{c_{{j+1}}}{m_{j}}\right)n_{j}x_{{j+1}}$  
$\displaystyle=\left(\frac{c_{{j+1}}}{m_{j}}\right){x_{j}}^{2}$  
$\displaystyle=0x_{j}$  
$\displaystyle=0.$ 
Since $c_{{j+1}}$ is the characteristic of $C_{{j+1}}$ and $x_{{j+1}}$ is a generator of $C_{{j+1}}$, it follows that
$c_{{j+1}}\mid c_{{j+1}}\left(\frac{n_{j}}{m_{j}}\right).$ 
Therefore, $m_{j}$ divides $n_{j}$. Let $a_{j}\in\mathbb{N}$ such that $n_{j}=a_{j}m_{j}$. Then
$\displaystyle{x_{j}}^{2}$  $\displaystyle=n_{j}x_{{j+1}}$  
$\displaystyle=a_{j}m_{j}x_{{j+1}}$  
$\displaystyle=a_{j}x_{j}.$ 
Hence, each $C_{j}$ is a cyclic ring. ∎
Note that the behavior of $C_{j}$ is $\gcd(a_{j},c_{j})$ for all $j$.
Actually, since the $C_{j}$’s are finite cyclic rings, there is no need to be so abstract anymore. For all $j$, we can define
$C_{j}=k_{j}\mathbb{Z}_{{k_{j}c_{j}}}$ 
with $k_{j}$ dividing $c_{j}$ for all $j$. Note that $k_{j}$ is a generator of $C_{j}$ for all $j$. Thus, with notation as before, we can declare $x_{j}=k_{j}$ for all $j$; however, in doing so, we may no longer have $m_{j}$ dividing $c_{{j+1}}$. Fortunately, since we are dealing exclusively with integers, having $m_{j}$ divide $c_{{j+1}}$ is no longer as crucial.
Since $k_{j}$ divides $c_{j}$ for all $j$, let $s_{j}\in\mathbb{N}$ such that $c_{j}=k_{j}s_{j}$. Since ${k_{j}}^{2}\equiv m_{j}k_{{j+1}}\;\;(\mathop{{\rm mod}}c_{j})$, we have that $c_{{j+1}}$ divides ${k_{j}}^{2}m_{j}k_{{j+1}}$. Therefore, $k_{{j+1}}$ divides ${k_{j}}^{2}m_{j}k_{{j+1}}$. Hence, $k_{{j+1}}$ divides ${k_{j}}^{2}$.
The next thing that I would like to do is divide into cases. The one that is most familiar to me is when $k_{{j+1}}$ divides $k_{j}$. The other case, obviously, will be when $k_{{j+1}}$ does not divide $k_{j}$.
Theorem 3.
If $n_{1},n_{2},k1,k_{2}$ are integers such that $k_{1}n_{1}$ is relatively prime to $k_{2}n_{2}$, then $k_{1}k_{2}\mathbb{Z}_{{k_{1}k_{2}n_{1}n_{2}}}$ is isomorphic to $k_{1}\mathbb{Z}_{{k_{1}n_{1}}}\times k_{2}\mathbb{Z}_{{k_{2}n_{2}}}$.
Proof.
Suppose that $a,b,c,d,e,f$ are integer multiples of $k_{1}k_{2}$ such that $a+b$ is congruent to $c$ modulo $k_{1}k_{2}n_{1}n_{2}$ and $ef$ is congruent to $g$ modulo $k_{1}k_{2}n_{1}n_{2}$. Then, by the Chinese Remainder Theorem, this is equivalent to the statements that $a+b\equiv c$ modulo $k_{1}n_{1}$, $a+b\equiv c$ modulo $k_{2}n_{2}$, $ef\equiv g$ modulo $k_{1}n_{1}$, and $ef\equiv g$ modulo $k_{2}n_{2}$. Hence $k_{1}k_{2}\mathbb{Z}_{{k_{1}k_{2}n_{1}n_{2}}}$ is isomorphic to $k_{1}k_{2}\mathbb{Z}_{{k_{1}n_{1}}}\times k_{1}k_{2}\mathbb{Z}_{{k_{2}n_{2}}}$. However, because $k_{2}$ is relatively prime to $k_{1}n_{1}$, the ring $k_{1}k_{2}\mathbb{Z}_{{k_{1}n_{1}}}$ is isomorphic to $k_{1}\mathbb{Z}_{{k_{1}n_{1}}}$; likewise $k_{1}k_{2}\mathbb{Z}_{{k_{2}n_{2}}}$ is isomorphic to $k_{2}\mathbb{Z}_{{k_{2}n_{2}}}$. Hence, $k_{1}k_{2}\mathbb{Z}_{{k_{1}k_{2}n_{1}n_{2}}}$ is isomorphic to $k_{1}\mathbb{Z}_{{k_{1}n_{1}}}\times k_{2}\mathbb{Z}_{{k_{2}n_{2}}}$. ∎
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