Simplifying a monoplist's objective function under incomplete information I am working through Rochet and Stohle (2003) chapter on multi-dimensional screening, and I am struggling filling in the blanks between equation (2.1) p. 154, and its simplified form on page 155. In particular, my question concerns:

Consumers are of type $\theta$ $\in$ $\Theta$ = [$\underline\theta$, $\bar\theta$] with associated cdf $F(\theta)$, which is absolutely continuous, and associated density function $f(\theta)$ = $F'(\theta)$. This is the distribution of types.

There also is a preference relation for each consumer for $q \in Q = [0,\bar q]$: $u = v(q, \theta) - P$ with the single crossing property $v_{q\theta} &gt; 0 $. I am adding this for context, but all information may not be necessary for my particular question.

This renders an indirect utility function defined by:

$u(\theta) = \max_{\rm {q\in Q}} \{v(q, \theta) - P(q)\}$

The monopoly firm is using a non-linear tariff, $P(q)$, and wants to maximize its expected payoff (I am skipping some information not needed for my question on what renders this equation): 

$E(\pi) = \int^\bar\theta_\underline\theta [S(q(\theta), \theta) - u(\theta)] dF(\theta)$ equation (2.1) in text

subject to:

$du/d\theta = v_\theta(q(\theta), \theta)$ 

$dq(\theta)/d\theta\geq0 $

$IR\space constraint$

Note that the function $S(q(\theta), \theta)$ is not important for my question, hence, I am not including its definition here.

Now to my question. The authors makes a simplification of the problem:

$E(\pi) = \int^\bar\theta_\underline\theta [S(q(\theta), \theta) - \frac {1-F(\theta)}{f(\theta)}v_\theta(q(\theta), \theta) - u(\underline\theta)]dF(\theta)$

subject to:

$dq(\theta)/d\theta\geq0 $

$IR\space constraint$

Now, this to me is the same as saying:

$\int^\bar\theta_\underline\theta u(\theta) dF(\theta) = \int^\bar\theta_\underline\theta [\frac {1-F(\theta)}{f(\theta)}v_\theta(q(\theta), \theta) + u(\underline\theta)]dF(\theta) $

And here is where I need help to fill in the steps. As the authors explain, this is done by integration by parts, and using one of the constraints. I get something like this:

$\int^\bar\theta_\underline\theta u(\theta) dF(\theta) = u(\theta)F(\theta)|^\bar\theta_\underline\theta - \int^\bar\theta_\underline\theta F(\theta)v_\theta(q(\theta),\theta)d\theta$

I can simplify this a bit further, but it does not simplify to the authors' expression. Hence, can anybody help me fill in the blanks?

I hope that I have left sufficient information, otherwise, please let me know what I need to clarify.
I'll start from your last equation.

\begin{align*}
\int_{\underline{\theta}}^{\bar{\theta}}{u(\theta) dF(\theta)} &amp; = \Big[ u(\theta)F(\theta) \Big]_{\underline{\theta}}^{\bar{\theta}} - \int_{\underline{\theta}}^{\bar{\theta}}{F(\theta) v_{\theta}(q(\theta),\theta) d\theta} \\
&amp; = u(\bar{\theta})F(\bar{\theta})-u(\underline{\theta})F(\underline{\theta})-\int_{\underline{\theta}}^{\bar{\theta}}{F(\theta) v_{\theta}(q(\theta),\theta) d\theta} \\
&amp; = u(\bar{\theta})-\int_{\underline{\theta}}^{\bar{\theta}}{F(\theta) v_{\theta}(q(\theta),\theta) d\theta} \text{ since } F(\bar{\theta})=1, F(\underline{\theta})=0\\
&amp; = u(\underline{\theta})+(u(\bar{\theta})-u(\underline{\theta}))-\int_{\underline{\theta}}^{\bar{\theta}}{F(\theta) v_{\theta}(q(\theta),\theta) d\theta} \\
&amp; = u(\underline{\theta})+\int_{\underline{\theta}}^{\bar{\theta}}{u'(\theta) d\theta}-\int_{\underline{\theta}}^{\bar{\theta}}{F(\theta) v_{\theta}(q(\theta),\theta) d\theta} \\
&amp; = u(\underline{\theta})+\int_{\underline{\theta}}^{\bar{\theta}}{v_{\theta}(q(\theta),\theta) d\theta}-\int_{\underline{\theta}}^{\bar{\theta}}{F(\theta) v_{\theta}(q(\theta),\theta) d\theta} \text{ since } u'=v_{\theta}\\
&amp; = u(\underline{\theta})+\int_{\underline{\theta}}^{\bar{\theta}}{(1-F(\theta)) v_{\theta}(q(\theta),\theta) d\theta} \\
&amp; = u(\underline{\theta})+\int_{\underline{\theta}}^{\bar{\theta}}{\dfrac{1-F(\theta)}{f(\theta)} v_{\theta}(q(\theta),\theta) dF(\theta)} \\
&amp; = \int_{\underline{\theta}}^{\bar{\theta}}{u(\underline{\theta})dF(\theta)}+\int_{\underline{\theta}}^{\bar{\theta}}{\dfrac{1-F(\theta)}{f(\theta)} v_{\theta}(q(\theta),\theta) dF(\theta)} \\
&amp; = \int_{\underline{\theta}}^{\bar{\theta}}{\Big[u(\underline{\theta})+\dfrac{1-F(\theta)}{f(\theta)} v_{\theta}(q(\theta),\theta) \Big]dF(\theta)}
\end{align*}