If \(X \sim G(a, b)\) with PDF
\[ f_X(x) = \frac{b^a}{\Gamma(a)} x^{a-1} e^{-bx}, \quad x > 0, \; a > 0, \; b > 0, \]
then show that for \(a > 1\),
\[ E\left[\frac{1}{X}\right] = \frac{b}{a-1}. \]
We start with the definition of expectation:
\[ E\left[\frac{1}{X}\right] = \int_0^\infty \frac{1}{x} \cdot f_X(x) \, dx \]
Substitute the PDF:
\[ E\left[\frac{1}{X}\right] = \int_0^\infty \frac{1}{x} \cdot \frac{b^a}{\Gamma(a)} x^{a-1} e^{-bx} \, dx \]
Simplify \(\frac{1}{x} \cdot x^{a-1} = x^{a-2}\):
\[ E\left[\frac{1}{X}\right] = \frac{b^a}{\Gamma(a)} \int_0^\infty x^{a-2} e^{-bx} \, dx \]
Let \(u = bx\), so \(du = b \, dx\) and \(x = \frac{u}{b}\).
Then \(x^{a-2} = \left(\frac{u}{b}\right)^{a-2}\) and \(dx = \frac{du}{b}\).
The integral becomes:
\[ \int_0^\infty x^{a-2} e^{-bx} \, dx = \int_0^\infty \left(\frac{u}{b}\right)^{a-2} e^{-u} \cdot \frac{du}{b} \]
\[ = \frac{1}{b^{a-2}} \cdot \frac{1}{b} \int_0^\infty u^{a-2} e^{-u} \, du \]
\[ = \frac{1}{b^{a-1}} \int_0^\infty u^{a-2} e^{-u} \, du \]
The integral \(\int_0^\infty u^{a-2} e^{-u} \, du = \Gamma(a-1)\), valid for \(a > 1\).
Thus:
\[ \int_0^\infty x^{a-2} e^{-bx} \, dx = \frac{\Gamma(a-1)}{b^{a-1}} \]
\[ E\left[\frac{1}{X}\right] = \frac{b^a}{\Gamma(a)} \cdot \frac{\Gamma(a-1)}{b^{a-1}} \]
\[ = \frac{b^{a-(a-1)}}{\Gamma(a)} \cdot \Gamma(a-1) \]
\[ = \frac{b}{\Gamma(a)} \cdot \Gamma(a-1) \]
For \(a > 1\), \(\Gamma(a) = (a-1) \Gamma(a-1)\).
Therefore:
\[ \frac{\Gamma(a-1)}{\Gamma(a)} = \frac{1}{a-1} \]
\[ E\left[\frac{1}{X}\right] = \frac{b}{a-1} \]
\[ \boxed{E\left[\frac{1}{X}\right] = \frac{b}{a-1}, \quad a > 1} \]
This matches the formula given in the problem.