transforming ordinary generating function into exponential generating function I have seen a post here that says that you can convert an exponential generating function into an ordinary one with the aid of the Laplace transform. Is it possible to do the reverse transformation? i.e. I want to convert an ordinary generating function into an exponential one. 
 A: There may be  some cases where complex variables,  the residue theorem
and the  residue at infinity are  helpful. Suppose your  OGF is $f(z)$
and the desired EGF is $g(w).$ Then we have
$$g(w) = \sum_{n\ge 0} \frac{w^n}{n!}
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
f(z) \; dz.$$
This will simplify together with some conditions on convergence to give
$$g(w) =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{f(z)}{z}
\sum_{n\ge 0} \frac{1}{n!} \frac{w^n}{z^n} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{f(z)}{z} \exp(w/z) \; dz.$$
Example I. Suppose $$f(z) = \frac{1}{1-z},$$ which yields
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{1-z} \frac{1}{z} \exp(w/z) \; dz.$$
Now for $z=R\exp(i\theta)$ with $R$ going to infinity we have
$$2\pi R\times \frac{1}{R^2} \times \exp(|w|/R) \rightarrow 0$$
as $R\rightarrow \infty$ so this integral is
$$- \mathrm{Res}_{z=1} \frac{1}{1-z} \frac{1}{z} \exp(w/z)$$
and we get $$g(w) = \exp(w)$$
which is the correct answer.

Example II. This time suppose that
$$f(z) = \frac{z}{(1-z)^2}$$
so that we should get
$$g(w) = \sum_{n\ge 1} n \frac{w^n}{n!} =
w \sum_{n\ge 1} \frac{w^{n-1}}{(n-1)!} = w\exp(w).$$
The integral formula yields
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{z}{(1-z)^2} \frac{1}{z} \exp(w/z) \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{(1-z)^2} \exp(w/z) \; dz.$$
The residue at infinity is zero as before and we have
$$\exp(w/z)
= \sum_{n\ge 0}
\left. (\exp(w/z))^{(n)}\right|_{z=1} \frac{(z-1)^n}{n!}$$
The coefficient on $(z-1)$ is
$$\left. -\frac{1}{z^2} w\exp(w/z)\right|_{z=1} = - w\exp(w)$$
which is the  correct answer taking into account the  sign flip due to
$z=1$ not being inside the contour.
Remark. Good news. The sum in the integral converges everywhere.

Addendum: somewhat more involved example. The OGF of Stirling
numbers of the second kind  for set partitions into $k$ non-empty sets
is
$$\sum_{n\ge 0} {n\brace k} z^n
= \prod_{q=1}^k \frac{z}{1-qz}.$$
We thus have that $$g(w) =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z} \exp(w/z)
\prod_{q=1}^k \frac{z}{1-qz}
 \; dz
\\ = \frac{(-1)^k}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z} \exp(w/z)
\prod_{q=1}^k \frac{z}{qz-1}
 \; dz
\\ = \frac{(-1)^k}{k! \times 2\pi i}
\int_{|z|=\epsilon} \frac{1}{z} \exp(w/z)
\prod_{q=1}^k \frac{z}{z-1/q}
 \; dz.$$
Computing the  sum of the residues  at the finite  poles not including
zero we get
$$\frac{(-1)^k}{k!}
\sum_{q=1}^k q\exp(qw) \times \frac{1}{q}
\prod_{m=1}^{q-1} \frac{1/q}{1/q-1/m}
\prod_{m=q+1}^k \frac{1/q}{1/q-1/m}
\\ = 
\frac{(-1)^k}{k!}
\sum_{q=1}^k \exp(qw)
\prod_{m=1}^{q-1} \frac{m}{m-q}
\prod_{m=q+1}^k \frac{m}{m-q}
\\ = 
\frac{(-1)^k}{k!}
\sum_{q=1}^k \exp(qw) \frac{k!}{q}
\prod_{m=1}^{q-1} \frac{1}{m-q}
\prod_{m=q+1}^k \frac{1}{m-q}
\\ = 
\frac{(-1)^k}{k!}
\sum_{q=1}^k \exp(qw) \frac{k!}{q}
\frac{(-1)^{q-1}}{(q-1)!}
\frac{1}{(k-q)!}
\\ = 
- \frac{1}{k!}
\sum_{q=1}^k \exp(qw) (-1)^{k-q} 
{k\choose q}
\\ = -\left(\frac{(\exp(w)-1)^k}{k!} - \frac{(-1)^k}{k!}\right).$$
This is a case where the residue at infinity is not zero.
We have the formula for the residue at infinity
$$\mathrm{Res}_{z=\infty} h(z)
= \mathrm{Res}_{z=0}
\left[-\frac{1}{z^2} h\left(\frac{1}{z}\right)\right]$$
This yields for the present case
$$- \mathrm{Res}_{z=0} 
\frac{1}{z^2} z \exp(wz)
\prod_{q=1}^k \frac{1/z}{1-q/z}
= - \mathrm{Res}_{z=0} 
\frac{1}{z} \exp(wz)
\prod_{q=1}^k \frac{1}{z-q}
\\ = - \frac{1}{k!} \mathrm{Res}_{z=0} 
\frac{1}{z} \exp(wz)
\prod_{q=1}^k \frac{1}{z/q-1}
\\ = - \frac{(-1)^k}{k!} \mathrm{Res}_{z=0} 
\frac{1}{z} \exp(wz)
\prod_{q=1}^k \frac{1}{1-z/q}
= - \frac{(-1)^k}{k!}.$$
Adding  the residue  at infinity  to the  residues from  the  poles at
$z=1/q$ we finally obtain
$$-\left(\frac{(\exp(w)-1)^k}{k!} - \frac{(-1)^k}{k!}\right)
- \frac{(-1)^k}{k!}
= - \frac{(\exp(w)-1)^k}{k!}.$$
Taking into account  the sign flip we have indeed  computed the EGF of
the Stirling numbers of the second kind
$$\sum_{n\ge 0} {n\brace k} \frac{z^n}{n!}$$
as can be seen from the combinatorial class equation
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}\textsc{SET}(\mathcal{U}\times\textsc{SET}_{\ge 1}(\mathcal{Z}))$$
which gives the bivariate generating function
$$G(z, u) = \exp(u(\exp(z)-1)).$$
A: Let $f(z)$ be the ordinary generating function and $\mathcal{L}$ be  the Laplace transform.  Then  the EGF  $g(z)$  has the form 
$$
g(z)=\mathcal{L}^{-1}\left(\frac{1}{z} f(z)\Big |_{z=\frac{1}{z}}\right)
$$
A: See also the integral formula given in this post for more concrete real integral versions of the contour integrals expressed in the previous posts. This integral formula is stated in the appendix (page 566) of Concrete Mathematics by Graham, Knuth and Patashnik.
