A way of evaluating integrals without doing anything? The user known as sos440 posted this:
$$\begin{align*} \sum_{n=0}^\infty \frac{r^n}{n!} \int_0^\infty x^n e^{-x} \; dx & = \int_{0}^\infty \sum_{n=0}^\infty \frac{(rx)^n}{n!} e^{-x} \; dx = \int_0^\infty e^{-(1 - r)x} \; dx \\ & = \frac{1}{1 - r} = \sum_{n=0}^\infty r^n \end{align*}$$
(citing Tonelli's theorem to justify interchanging the sum and the integral),
and concluded that
$$
\frac{1}{n!}\int_0^\infty x^n e^{-x}\,dx=1.
$$
Is this a very isolated thing or just an instance of something generally useful for some much broader class of integrals? (Obviously, what is seen below is generally true, but how generally is it useful?)
$$
\sum_{n=0}^\infty r^n \int_A f_n(x)\,dx = \int_A \sum_{n=0}^\infty \big( r^nf_n(x) \big) \, dx = \int_A g(r,x) \, dx = h(r) = \sum_{n=0}^\infty c_n r^n
$$
$$
\therefore \int_A f_n(x)\,dx = c_n.
$$
 A: Rather than try and address this question generally here we give another---hopefully sufficiently different---example of the use of this type of machinery.
Suppose we are interested in integrals of the form 
$$\begin{equation*}
\int_0^{2\pi} x^n \sin x\, dx
\quad\textrm{ or }\quad
\int_0^{2\pi} x^n \cos x \, dx,
\end{equation*}$$
where $n\in\mathbb{N}$. 
It is natural to consider the more general integral 
$$\begin{equation*}
I_n = i\int_0^{2\pi} x^n e^{i x}\, dx.
\end{equation*}$$
(A factor of $i$ has been introduced for convenience.) 
Then 
$$\begin{eqnarray*}
I(r) &\equiv& \sum_{n=0}^\infty \frac{(i r)^n}{n!}I_n \\
&=& i\int_0^{2\pi} e^{i (r+1) x}dx \\
&=& \frac{e^{2\pi i r}-1}{1+r} \\
&=& \sum_{n=0}^\infty
\left(\sum_{k=0}^{n-1}(-1)^{k}\frac{(2\pi i)^{n-k}}{(n-k)!}\right)r^n.
\end{eqnarray*}$$
(The last series is the Cauchy product of the Taylor series for 
$1/(1+r)$ and the Taylor series for $e^{2\pi i r}-1$.) 
Therefore, 
$$\begin{equation*}
\int_0^{2\pi} x^n e^{i x}\, dx
= \frac{n!}{i^{n+1}}\sum_{k=0}^{n-1} (-1)^{k} 
    \frac{(2\pi i)^{n-k}}{(n-k)!}.\tag{1}
\end{equation*}$$
By taking the real and imaginary part of (1) we can find explicit formulas for the original integrals: 
$$\begin{eqnarray*}
\int_0^{2\pi} x^n \cos x\, dx
&=& n!\sum_{m=0}^{\lfloor\frac{n-2}{2}\rfloor} 
    (-1)^m\frac{(2\pi)^{n-2m-1}}{(n-2m-1)!} \\
\int_0^{2\pi} x^n \sin x\, dx
&=& n!\sum_{m=0}^{\lfloor\frac{n-1}{2}\rfloor}
    (-1)^{m+1}\frac{(2\pi)^{n-2m}}{(n-2m)!}.
\end{eqnarray*}$$
A: Nothing magic :-) (if I understand the question correctly.)
If $f(r)$ is an analytic function around $r_0=0$ then the Taylor expansion is unique. It yields,
$$
f(r)=\sum_{n=1}^{\infty}c_nr^n,
$$
where $r\in(-a,a)$, $a>0$. So, if 
$$
f(r)=\sum_{n=1}^{\infty}c_n d_n r^n
$$
is another power series representation of $f$, then $c_n=c_n d_n$, i.e., $d_n=1$ if $c_n\neq 0$, anything is $d_n$ (integral or something else).
A: $$\sum_{n=0}^\infty \frac{r^n}{n!} \int_0^\infty x^n e^{-x} \; dx = \sum_{n=0}^\infty \frac{r^n}{n!} \left(\int_0^\infty x^n e^{-x} \; dx\right) =  \sum_{n=0}^\infty \frac{r^n}{n!} \Gamma(n+1) =  \sum_{n=0}^\infty \frac{r^n}{n!} n! =  \sum_{n=0}^\infty r^n$$
Where $\Gamma(n+1)$ is the Gamma function. I hope this is right!
