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According to Investment under Uncertainty by Dixit & Pindyck (1993, p.82), the expected present discounted value of a variable following a geometric Brownian motion is defined as: $$E\left[\int_{0}^{\infty}{F(x(t))}\exp[-rt]dt\right]$$ Suppose $$\tag 1 F(x)=x^\theta$$ and $$\tag 2 dx={α}{x}{dt}+{σ}{x}{dz}$$ where $dz$ is the increment of a Weiner process. Equation $(1)$ then follows a geometric Brownian motion. Exapanding equation $(1)$ by Itô's lemma, we can derive $$\tag 3 \theta{x}^\theta({α}{x}{dt}+{σ}{x}{dz})+\frac{1}{2}\theta(\theta-1){x}^\theta\sigma^2dt$$ Substituting equation $(1)$ for ${x}^\theta$ and rearranging gives $$\tag 4 [\theta{α} + \frac{1}{2}\theta(\theta-1)\sigma^2]Fdt + \theta\sigma{F}dz$$ If $x(0)$ is currently ${x_0}^\theta$, the expected value of $F(x)$ is given by equation $(5)$. $$\tag 5 {x_0}^\theta\exp[t(\theta{\alpha}+\frac{1}{2}\theta(\theta-1)\sigma^2)]$$ According to the textbook mentioned above, the present discounted value of equation $(5)$ is given by equation $(6)$ $$\tag 6 \frac{x_0^\theta}{r-(\theta{\alpha}+\frac{1}{2}\theta(\theta-1)\sigma^2)}$$

As far as I understand, deriving this result involves multiplying equation $(5)$ by $\exp[-rt]$ and integrating with respect to time between $t=0$ and $t=\infty$. Doing this, we arrive at the integral given by equation $(7)$ $$\tag 7 \int_{0}^{\infty}\frac{x_0^\theta}{\exp[t(r-(\theta{\alpha}+\frac{1}{2}\theta(\theta-1)\sigma^2))]}dt$$ However, I do not understand how $(5)$ follows from $(7)$, if this is indeed the correct approach. Any assistance in understanding this problem would be appreciated.

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The anti-derivative of (7) is $$\left[\frac{- x_o^\theta}{r - (\theta \alpha + \frac{1}{2} \theta (\theta - 1) \sigma^2)} exp^{-t (r - (\theta \alpha + \frac{1}{2} \theta (\theta - 1) \sigma^2)}\right]^\infty_0 $$ Now evaluate as $t \rightarrow \infty$ and $t \rightarrow 0$ $$\lim_{t \rightarrow \infty} \left[\frac{- x_o^\theta}{r - (\theta \alpha + \frac{1}{2} \theta (\theta - 1) \sigma^2)} exp^{-t(r - (\theta \alpha + \frac{1}{2} \theta (\theta - 1) \sigma^2)}\right] + \lim_{t \rightarrow 0} \left[\frac{x_o^\theta}{r - (\theta \alpha + \frac{1}{2} \theta (\theta - 1) \sigma^2)} exp^{-t(r - (\theta \alpha + \frac{1}{2} \theta (\theta - 1) \sigma^2)}\right]$$ $$0 + \frac{ x_o^\theta}{r - (\theta \alpha + \frac{1}{2} \theta (\theta - 1) \sigma^2)}$$ Which is equal to equation (6).

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