Mean with CDF approach

Question

Since $$E[X]=\int_{0}^{\infty}\Pr[X\geq t]\,dt;~X\geq 0$$ I tried following in similar way with random variable $x$ which follows exponential distribution $e^{-x}$ for $x\geq 0$, and $a\geq 0,d\geq 1$: $$E\left[\ln[1+(x+a)^{-d}]\right]=\int\Pr[\ln[1+(x+a)^{-d}]\geq t]\,dt=\int e^{-\left((e^t-1)^{-\frac{1}{d}}-a\right)}\,dt$$ However, I am not sure about the integration limits. Can someone please guide me?

Answer 1

$$E[Y]=\int_{0}^{\infty}\Pr[Y\geq t]\,dt;~Y\geq 0$$

Let $Y = \ln [1+(X+a)^{-d}]$

Assuming $a$, $d$ positive.

\begin{align}E[Y]&=\int_{0}^{\infty}\Pr[Y\geq t]\,dt\\ &= \int_0^\infty \Pr[ [(X+a)^{-d}]\geq \exp(t)-1]\, dt \\ &= \int_0^\infty \Pr[ [(X+a)^d]\leq \frac1{\exp(t)-1}]\, dt \\ &= \int_0^\infty \Pr\left[ X\leq \left(\frac1{\exp(t)-1}\right)^\frac{1}{d}-a\right]\, dt \\ &= \int_0^\infty \Pr\left[ X\leq \max\left(\left(\frac1{\exp(t)-1}\right)^\frac{1}{d}-a,0\right)\right]\, dt \\\end{align}