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Inverted Gamma distribution is: $$ \frac{1}{\Gamma(a)b^a} \left( \frac{1}{y} \right)^{a+1} e^{-1/by} $$ So, $$ \mathbb{E}Y = \frac{1}{\Gamma(a)b^a} \int_{0}^{\infty} \left( \frac{1}{y} \right)^{a} e^{-1/by} \ \ \ (1) $$ I think, we need to use the definition of the Gamma function: $$ \Gamma(z) = \int_0^\infty t^{z-1} e^{-t} dt \ \ \ (2) $$ In (1), how can I use Gamma function (how can I deal with $b$)?

If there isn't $b$, I think I can use $$ \Gamma(a+1) = \int_0^\infty \left( \frac{1}{y} \right)^{a} e^{-1/y} dt \ \ \ (3) $$

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    what if you substitute $t = 1/by$? – 2017-01-25
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    @user159517 I got $\frac{1}{\Gamma(a) b^a} \int_0^\infty (tb)^a e^{-t}$, but I couldn't get $1/(a-1)b$ (answer) from here. – 2017-01-25

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We have $$\mathbb{E}[Y] = \frac{1}{\Gamma(a)b^a}\int_{0}^{\infty} \left(\frac{1}{y}\right)^{a} e^{-1/by} dy$$ Now substitute $t = 1/by$, giving $dt = - dy/by^{2}$ to get

$$\mathbb{E}[Y] = \frac{1}{\Gamma(a)b^a}\int_{0}^{\infty} b\left(tb\right)^{a-2} e^{-t} dt = \frac{1}{b} \frac{\Gamma(a-1)}{\Gamma(a)} = \frac{1}{(a-1)b}$$

Note that in your comment, you did not substitute correctly.

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    Thanks. Following your advice, I am still not sure how to deal with `-` in $dt = -dy/by^2$. What I got after integration by parts is: $(1/\Gamma(a) b^a) \int_0^\infty (bt)^a e^{-t} \cdot \left(- \frac{1}{b} \cdot \frac{1}{t^2} \right) dt$ – 2017-01-26
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    @user51966 You have to keep the limits in mind too. You should get $(1/\Gamma(a) b^a) \int_{\infty}^{0} -b(bt)^{a-2} e^{-t} dt $ after substitution, so changing the order gives the right integral. – 2017-01-26