Derivation of Fokker Planck forward equation I am studing the derivation of the Fokker-Planck equation. My source is this link http://www.pnas.org/content/pnas/suppl/2007/03/16/0608270104.DC1/08270Appendix.pdf. it is very clear, but
I am struggling with one passage: I have the equation
\begin{equation}
\int h(y)\frac{\partial p(y, t|x, 0)}{\partial t}dy- \int p(z, t|x,0)\sum_{n=1}^{\infty}D_{(n)}(z)h^{(n)}(z)dz=0.
\end{equation}
and I have to obtain, integrating by parts,
\begin{equation}
\int h(z)\left(\frac{\partial p(y, t|x, 0)}{\partial t} -\sum_{n=1}^{\infty}\left[\left(-\frac{\partial}{\partial z}\right)^n \left(D_{(n)} p(z,t|x,0)\right)\right]  \right) dz = 0.
\end{equation}
Could you please explain me what does the author intend with integrating by parts? How can I delete the extra parts that come out integrating by parts? Does that come from that fact that $p$ is a probability distribution so it is very small at infinity?
Thank you!
 A: By integrating by parts $n$ times the second term, you are essentially transferring the $n$ derivatives from the term $h^{(n)}(z)$ to the term $D_{(n)} p(z,t|x,0)$, obtaining $\left(-\frac{\partial}{\partial z}\right)^n D_{(n)} p(z,t|x,0)$. The minus terms comes from the formula for integration by parts, $\int_a^b u dv=uv\Big|_a^b-\int_a^b v du$ (notice how the new integral on the right hand side is multiplied by $-1$ compared to the integral on the left hand side). Doing this $n$ times will give you a factor of $(-1)^n$, in addition to the $n$ derivatives from $\left(\frac{\partial}{\partial z}\right)^n$ because of the derivative transfer from $h^{(n)}(z)$ to $D_{(n)} p(z,t|x,0)$. 
Now you may be wondering why the author ignores the extra $h^{(n-k)}(z)\left(-\frac{\partial}{\partial z}\right)^k D_{(n)} p(z,t|x,0)\Biggr|_{-\infty}^\infty$ term, for $k=0,1,...n$, when using integration by parts. This is because the function $h$ is defined to be smooth and have compact support by the author, between equations 4 and 5. As a result, that term will always vanish.
I think that the author integrated by parts $n$ times like this in order to be able to factor out the $h(z)$ function from both terms, to show that, because $h(z)$ is arbitrary, the other term must always be $0$:
$$\frac{\partial p(y,t|x,0)}{\partial t}-\sum_{n=1}^\infty \left(-\frac{\partial}{\partial z}\right)^n D_{(n)} p(z,t|x,0)=0$$
Or,
$$\frac{\partial p(y,t|x,0)}{\partial t}=\sum_{n=1}^\infty \left(-\frac{\partial}{\partial z}\right)^n D_{(n)} p(z,t|x,0)$$
