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Let's suppose

$$E\left(\frac{1}{x}\right)=\int_{0}^{\infty}{M_X(-t)dt}$$

Could you please help me to find $$E\left(\frac{1}{x}\right)$$ where $$X \sim\textrm{Gamma}(\alpha, \beta)$$ and $$E(X) = \alpha\beta~?$$

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This is straightforward integration:

Note that since $\mathbb{E}(X) = \alpha\beta$ then the $\beta$ is a scale parameter. Thus

$$ \begin{aligned} M_X(t) =& (1-\beta t)^{-\alpha}\\ \therefore\mathbb{E}\left(\frac{1}{X}\right) =& \int_0^\infty M_X(-t)dt = \int_0^\infty (1+\beta t)^{-\alpha}dt\\ =&\frac{(1+\beta t)^{-\alpha + 1}}{-\beta(\alpha-1)}\Bigg|_0^\infty = \frac{1}{\beta(\alpha-1)} \end{aligned} $$

Also note that the expectation could be computed directly:

$$ \begin{aligned} \mathbb{E}\left(\frac{1}{X}\right) &= \int_0^\infty\frac{1}{x} \frac{1}{\Gamma(\alpha)\beta^\alpha}x^{\alpha-1}e^{-\frac{x}{\beta}}dx\\ &=\frac{1}{\Gamma(\alpha)\beta^\alpha}\int_0^\infty x^{\alpha-1-1}e^{-\frac{x}{\beta}}dx \\ &=\frac{\Gamma(\alpha - 1)\beta^{\alpha - 1}}{\Gamma(\alpha)\beta^\alpha} = \frac{\Gamma(\alpha - 1)\beta^{\alpha - 1}}{(\alpha-1)\Gamma(\alpha - 1)\beta^\alpha} = \frac{1}{\beta(\alpha-1)} \end{aligned} $$

Lastly you can note that since $X\sim\Gamma(\alpha, 1/\beta)$ then $Y=1/X\sim\Gamma^{-1}(\alpha,1/\beta)$ and therefore $$\mathbb{E}\left(\frac{1}{X}\right) = \mathbb{E}(Y) = \frac{1}{\beta(\alpha-1)}$$

You can easily look at wikipedia for the moments of the inverse gamma distribution

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