Expectation of a function of a normally distributed random variable Consider that I have to produce this result:
$$E[u(W_0+r(\theta))] = u(W_0)+\theta-\frac 12\rho\sigma^2$$
From this:
$$
E[u(W_0+r(\theta))]
    = \int_{-\infty}^\infty u(w_0+r) \frac{1}{\sigma \sqrt{2\pi}}
                            \exp\left(-\frac{(r-\theta)^2}{2 \sigma^2}\right) dr
$$
And:
$$\int_{-\infty}^\infty
          \frac{1}{\sigma \sqrt{2\pi}}
                            \exp\left(-\frac{(r-\theta)^2}{2 \sigma^2}\right) dr = 1 $$
Also:
$u(w) = -\exp(-\rho w) $
And:
$r(\theta) \sim N(\theta, \sigma^2)$
I have reached the point where I removed the constants, and realize that I have both a positive exponent and negative exponent, and I believe no way to combine them.
Help is much appreciated. Thanks.
 A: Hint: Complete the square of the exponent in the integral by using
$$
2\rho\sigma^2 r + r^2-2r\theta = (r+\rho\sigma^2-\theta)^2-(\rho\sigma^2-\theta)^2
$$

An "alternative" way to see the solution without explicitly doing the integral, since you are given a Gaussian random variable $R\sim\mathcal N(\theta, \sigma^2)$, is to use its moment generating function (m.g.f) which is well known, so that for some constant $w_0$, 
$$
\mathbb E[u(w_0+R)]=\mathbb E[(-1)e^{-\rho(w_0+R)}]=-\mathbb E[e^{-\rho(w_0+R)}]\\[3ex]=-e^{-\rho w_0}\mathbb E[e^{-\rho R}]\color{blue}{\stackrel{m.g.f.}{=}}-e^{-\rho w_0}e^{-\rho\theta+1/2\rho^2\sigma^2}=-e^{-\rho(w_0+\theta-1/2\rho\sigma^2)}=\color{red}{u(w_0+\theta-1/2\rho\sigma^2)}
$$
The answer, in red, is what I think your answer should be.

Random variables should be written in capital letters and non-random variables in lower case. When computing an expectation with respect to some random variable, the variable of integration should be lower case. So, for instance, I interpret what you wrote as
$$
\mathbb E[u(w_0+R)]=\frac1{\sigma\sqrt{2\pi}}\int_{-\infty}^{\infty}u(w_0+r)e^{-\frac12(r-\theta)^2/\sigma^2}\mathrm dr
$$
for $R\sim\mathcal N(\theta, \sigma^2)$.
