How does this "integration by differentiation" method work Apparently, the integral of a function f(x) from a to b can be done through differentiation through this method:
$$
  \int_a^b f(x)dx = \lim_{x \rightarrow  \ 0 }  f(\frac{d}{d x} )\frac{e^{bx}-e^{ax}}{x}
$$
Can anyone explain why this works? We can assume that f has a MacLauren representation with an infinite radius of convergence.
An example with f(x) = c, c constant, the integral of f from a to b is:
$$
\lim_{x \rightarrow  \ 0 }  c\frac{e^{bx}-e^{ax}}{x} = 
\lim_{x \rightarrow  \ 0 }  c\frac{1+bx+0.5(bx)^2...-1-ax-0.5(ax)^2...}{x} = c(b-a)
$$ as expected.
However I don't know how to show this is true for all (most?) functions with a MacLauren series. Help? The f(d/dx) and limit is what's being difficult to deal with. If f is something like x^2 then f(d/dx) is d^2/dx^2 but with a general function f I can't establish a pattern.
 A: How should we define $f(\frac{d}{ds})$ when $f$ is not a polynomial?  Probably the standard way to do it is with the Laplace transform.
Let's use this notation for the (bilateral) Laplace transform:  If $f(t)$ is a (nice enough) function, its Laplace transform is a function $F(s)$ defined by
$$
F(s) = \int_{-\infty}^\infty f(t) e^{-ts}\;dt
$$
Write $F(s) = \mathscr{L}\{f(t)\}$ for the Laplace transform; and $f(t) = \mathscr{L}^{-1}\{F(s)\}$ for the inverse Laplace transform.  I will attempt to use variables $s$ and $t$ consistently in this way.
If $\mathscr{L}\{f(t)\} = F(s)$, then compute $\mathscr{L}\{-tf(t)\} = F'(s)$.  By induction, $\mathscr{L}\{(-t)^kf(t\}) = F^{(k)}(s)$ for natural number $k$.  By linearity, $\mathscr{L}\{\phi(-t)f(t)\} = \phi(\frac{d}{ds})F(s)$ for any polynomial $\phi$.  
Define $\phi(\frac{d}{ds})$ for non-polynomials $\phi$ using the same formula.  Thus
$$
\phi\left(\frac{d}{ds}\right)F(s) = \mathscr{L}\big\{\phi(-t)f(t)\big\}
=\mathscr{L}\left\{\phi(-t)\mathscr{L}^{-1}\{F(s)\}\right\}
\tag{*}$$
Of course you need conditions for all these to exist.  When $f$ is bounded, measurable, with compact support, there is no problem at all.
Now to this question.  Fix $a < b$ real numbers.  Let $\phi$ be the function to integrate from $a$ to $b$.  Define function $h(t)$ by
$$
h(t) = \begin{cases}0\quad\text{if } t<-b
\\1\quad\text{if } -b \le t \le -a
\\0\quad\text{if } -a < t\end{cases} .
$$
This is bounded, measurable, with compact support.
Compute
$$
\mathscr{L}\{h(t)\} = \frac{e^{sb}-e^{sa}}{s} .
$$
Thus with definition $(*)$ we have
$$
\phi\left(\frac{d}{ds}\right) \frac{e^{sb}-e^{sa}}{s}= 
\mathscr{L}\{\phi(-t)h(t)\} =
\int_{-\infty}^\infty \phi(-t)h(t)e^{-st}\;dt
\\
=\int_{-b}^{-a}\phi(-t)e^{-st}\,dt = \int_a^b \phi(x) e^{sx}\;dx .
$$
(We changed variables $t=-x$.)  
And thus
$$
\lim_{s \to 0}\phi\left(\frac{d}{ds}\right) \frac{e^{sb}-e^{sa}}{s}= 
\int_a^b\phi(x)\;dx .
$$
A: For polynomials only, so not really an answer, but maybe one can take from here. Also, too long to put in a comment.
Let $f(x)=x^m$.
An expansion of $\frac{e^{bx}-e^{ax}}{x}$ gives
$$
\frac{e^{bx}-e^{ax}}{x}=\sum_{n=1}^{+\infty}\frac{(b^n-a^n)x^{n-1}}{n!}
$$
Thus, there is only one term surviving (when $n=m+1$) when differentiating, and we get
$$
\lim_{x\to 0}\frac{d^m}{dx^m}\frac{e^{bx}-e^{ax}}{x}=\frac{b^{m+1}-a^{m+1}}{m+1}
$$
which agrees with $\int_a^b x^m\,dx$.
By linearity it works for all polynomials. For more general functions I guess one should be very precise what is meant by $f(d/dx)$.
