Application of the operator $\exp\left( \alpha \frac{\partial^2}{\partial q^2}\right)$ I need to apply the operator
$$\exp\left( \alpha \frac{\partial^2}{\partial q^2}\right) \tag{1} \label{1}$$
To the function
$$M(x) N(y +C_{1}p)\mathcal{F}[f(q)](p) \tag{2} \label{2}$$
where $M(x)$ and $N(y + C_{1}p)$ are arbitrary functions, $C_{1}$ is a real constant and $$\mathcal{F}[f(q)](p)=\frac{1}{\sqrt{2\pi}} \int_{\infty}^{\infty}f(q)e^{-iqp}dq \tag{3} \label{D.1} $$ 
is the Fourier transform of the arbitrary function $f(q)$; is important to note that $q$ and $p$ is a conjugate pair (analogous to the position and moment variables of quantum mechanics), such that, in a quantum mechanical sense, the $y$ and $p$ variables are entangled. Specifically I need to know if I am correctly applying the operator $(\ref{1})$ to the fuction $(\ref{2})$, my process is as follows: 
First, I write the function $(\ref{2})$ with the explicit Fourier transform and I expand the operator ($[\ref{1}]$) as an infinite sum of derivatives
$$\sum_{n=0}^{\infty} \frac{1}{n!}\alpha^n \frac{\partial^{2n}}{\partial q^{2n}} M(x) N(y +C_{1}p)\mathcal{F}[f(q)](p) $$
$$=\frac{1}{\sqrt{2\pi}} M(x)N(y+C_{1}p) \quad \sum_{n=0}^{\infty} \frac{1}{n!}\alpha^n \frac{\partial^{2n}}{\partial q^{2n}} \int_{-\infty}^{\infty}f(q)e^{-iqp}dq \tag{4}\label{4}  $$
Second, I introduce the derivative inside the integral
$$=\frac{1}{\sqrt{2\pi}} M(x)N(y+C_{1}p) \quad \sum_{n=0}^{\infty} \frac{1}{n!}\alpha^n  \int_{-\infty}^{\infty} \frac{\partial^{2n}}{\partial q^{2n}} f(q)e^{-iqp}dq \tag{5}\label{5}  $$
and then I use the derivative theorem of the Fourier transform:
$$\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \frac{\partial^{2n}}{\partial q^{2n}} f(q)e^{-iqp}dq =\frac{1}{\sqrt{2\pi}} (ip)^{2n} \int_{-\infty}^{\infty}  f(q)e^{-iqp}dq \tag{D.2}$$
such that the line (\ref{5}) is
$$\frac{1}{\sqrt{2\pi}} M(x)N(y+C_{1}p) \quad \sum_{n=0}^{\infty} \frac{1}{n!}\alpha^n (ip)^{2n} \int_{-\infty}^{\infty} f(q)e^{-iqp}dq $$
$$= M(x)N(y+C_{1}p)e^{-\alpha p^2}  \quad  \mathcal{F}[f(q)](p) \tag{7}\label{7}  $$
where I have reversed the infinite sum, then, the last line is my final result.
So, my question is: 
Am I correctly applying operator ($\ref{1}$) to the function ($\ref{2}$)?
 A: You need to convert the operator to its representation in p-space before applying it. This can be done by setting $\frac{1}{i} \partial/\partial q = p $.
The steps you take from line $(4)$ to line $(5)$ are not correct. The integral, 
$$ \int_{-\infty}^{\infty}  f(q) e^{-iqp} \ dq,$$
is a function of $p$ not $q$. Any partial derivative of this with respect to $q$ would result in $0$.
$$ \frac{\partial}{\partial q }\int_{-\infty}^{\infty}  f(q) e^{-iqp} \ dq =0 .$$
To convince yourself of this consider a more elementary expression. 
$$ \int_0^1 x \ dx = \frac12 $$
Does it make sense to take a partial derivative of both sides with respect to $x$? Following your procedure I would get, 
$$ \frac{\partial}{\partial x} \int_0^1 x \ dx = \frac{\partial}{\partial x} \frac12 $$
$$ \int_0^1 \frac{\partial}{\partial x} x \ dx = 0 $$
$$ \int_0^1 1 \ dx = 0 $$
$$ 1 = 0 $$ 
The correct procedure is to express the operator in terms of the basis that the state vector is represented in. 

To derive the form of the operator in $p$-space we would do the following. 
(1) Act on a generic $q$-state with the operator. 
(2) Apply the Fourier transform to the result of (1)
(3) Determine what how the result of (2) is related to the Fourier transform of the initial state we acted on. 
Step 1:
Let $g(q)$ be a function in our Hilbert space. Then we apply our operator $\exp(\alpha \frac{\partial^2}{\partial q^2} )$ to $g(q)$. 
$$e^{\alpha \frac{\partial^2}{\partial q^2} } g(q) \quad \textbf{(a)}$$
Step 2:
We take the Fourier transform of the result. 
$$ \int e^{-ipq} e^{\alpha \frac{\partial^2}{\partial q^2} } g(q)  \ dq \quad \textbf{(b)}$$
Step 3: 
Using integration by parts it is possible to show that, 
$$ \int e^{-ipq} e^{\alpha \frac{\partial^2}{\partial q^2} } g(q)  \ dq  =  \int e^{-ipq} e^{-\alpha p^2} g(q)  \ dq $$
$$= e^{-\alpha p^2}  \int e^{-ipq}  g(q)  \ dq $$
$$= e^{-\alpha p^2}  \hat{g}(p) \quad \textbf{(c)}$$
We now see that our operator acts on functions in $p$-space as multipication by  $e^{-\alpha p^2} $. We can now act on any $p$-space state by simply multiplying by $e^{-\alpha p^2} $. 
Notice that at no point did I pull a derivative operation inside an integral $\frac{\partial }{\partial q} \int \ dq  \rightarrow \int \frac{\partial}{\partial q} \ dq  $; which would be an invalid step. 

One last thing I would like to point out is that the integration variable in a definite integral is always a dummy variable. It doesn't take on a particular value, and never "survives" the integration process. This is what user Jacky Chong was referring to when he said that you "[...] integrated out the $q$ variable." One feature of dummy variables is that you can change their labels without changing the value of the mathematical expression. 
For example, 
$$ \int f(q) e^{-iqp} \ dq = \int f(y) e^{-iyp} \ dy  = \int f(k) e^{-ikp} \ dk ,$$
in the same way that, 
$$ \int_{0}^1 x^2 \ dx = \int_0^1 y^2 \ dy = \int_0^1 k^2 \ dk = \frac13.$$
Dummy variables also occur in sums, e.g. $\sum_{i=1\dots 10} i^2 = \sum_{j=1\dots 10} j^2$. 
It is a definite red flag to be differentiating an integral with respect to its integration variable because that variable is a dummy variable. 
