I am trying to understand how the following function is integrated by parts (F(x) is the cdf of f(x)):
$$ \frac {\int_t^{+\infty}[1-F(x)]dx} {\int_{-\infty}^t F(x) dx} $$
After integrating by parts this becomes(I am familiar with the integrating by parts procedure but don't quite see it is applied to arrive at the equation below):
$$ \frac {\int_t^{+\infty}(x-t) f(x)dx} {\int_{-\infty}^t (t-x) f(x) dx} $$
And some more transformation (also not sure what is happening here):
$$ \frac {E[(x-t)^+]} {E[(t-x)^+]} $$
And one more (also unclear to me):
$$ \frac {E(x)-t} {E[(t-x)^+]}+1 $$
Here is a direct argument. Let $X$ be a random variable with cdf $F$ s.t. $\mathsf{E}|X|<\infty$. Then, using the tail sum formula for expectations (you can find the IBP argument there), $$ \int_{t}^{\infty}(1-F(x))\,dx=\int_0^{\infty}(1-F(x+t))\,dx=\mathsf{E}[X-t]^{+}, $$ where $[X-t]^{+}=\max\{X-t,0\}$, and $$ \int_{-\infty}^t F(x)\,dx=\int_{-\infty}^0 F(x+t)\,dx=\int_0^{\infty} F(t-x)\,dx=\mathsf{E}[t-X]^{+}. $$ Thus, \begin{align} \frac{\int_{t}^{\infty}(1-F(x))\,dx}{\int_{-\infty}^t F(x)\,dx}&=\frac{\mathsf{E}[X-t]^{+}}{\mathsf{E}[t-X]^{+}} \\ &=\frac{\mathsf{E}[X-t]+\mathsf{E}[t-X]^{+}}{\mathsf{E}[t-X]^{+}} \\ &=\frac{\mathsf{E}X-t}{\mathsf{E}[t-X]^{+}}+1 \end{align} because $a=\max\{a,0\}+\min\{a,0\}=\max\{a,0\}-\max\{-a,0\}$.