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I was interested in calculating $E\left(\dfrac{1}{1-X}\right)$ where $X\sim$ Gamma ($n,\lambda$), but I wasn't able to solve the associated integral using standard integration techniques.

$$E\left(\dfrac{1}{1-X}\right)=\frac{\lambda^n}{\Gamma(n)}\int_{0}^{\infty}\frac{1}{1-x}x^{n-1}e^{-\lambda x}\,dx.$$

I know $\int_0^\infty x^{n-1}e^{-\lambda x}\,dx=\frac{\Gamma(n)}{\lambda^n}$ but the $\frac{1}{1-x}$ term throws me off.

How can I calculate this expected value?

ZirconPetrichor
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    You have an issue when $X$ is close to $1$ and has a positive density there. What make you think this has an expectation? – Henry Oct 29 '21 at 16:20
  • If I input fixed values of $n$ and $\lambda$ in Wolfram Alpha I get a numerical result, so I think the integral converges in Principal Value. However I wasn't able to get a symbolical expression in terms of $n$ and $\lambda$. – ZirconPetrichor Oct 29 '21 at 16:24
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    Cauchy principal value is not good enough for an expectation: it does not work with the Cauchy distribution and does not work here for $\frac{1}{1-X}$ – Henry Oct 29 '21 at 16:32
  • If I make a change of variables $Y=\frac{1}{1-X}$ I get the following pdf: $f_Y(y)=\frac{\lambda^n}{\Gamma(n)}\frac{(y-1)^{n-1}}{y^{n+1}}e^{\frac{-\lambda (y-1)}{y}},, y\in (-\infty,0)\cup(1,\infty)$. I graphed $y\cdot f_Y(y)$ and I don't see any discontinuities. Do you think the expected value exists for this transformation? If so, would it mean the expected value of $\frac{1}{1-X}$ of a Gamma exists as well? – ZirconPetrichor Oct 29 '21 at 16:42
  • Simply graph the integrand $x^{n-1}e^{-x}/(1-x)$ near $x=1$ to see the problem. The "discontinuity" to examine when you graph the PDF for $Y$ corresponds to $Y=\pm\infty.$ The problem is that the integral of that PDF, when correctly calculated, diverges at those extremes. – whuber Oct 29 '21 at 16:47
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    I see, the integral diverges indeed at infinite. I thought I just needed to derive a general expression since I was able to compute the expected value for fixed $n$ and $k$, but it seems there is a deeper issue in that the expected value might not exist. Thank you both for your help! – ZirconPetrichor Oct 29 '21 at 16:53

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