where am I wrong in this calculation?

Question

In one of the paper I have seen that $$\int_0^{\infty} \ln(1+x)f(x)dx=\int_0^{\infty}(1+x)^{-1}[1-F(x)]dx$$ where $X$ is a random variable with PDF $f(x)$ and CDF $F(x)$. However, my steps does not yield the right hand side. Please point me where is the mistake. After applying the integration by parts I can get $$\ln(1+x)F(x)|_0^{\infty}-\int_0^{\infty}\frac{1}{1+x}F(x)dx$$ but this result is not equal to the right hand side of my first equation please correct me. Thanks in advance.

Answer 1

Just an old trick in proving the layer-cake representation of expectation:

$$\begin{align*} \int_0^{\infty} \ln(1+x)f(x)dx &= \int_0^{\infty} \int_0^x (1+u)^{-1}duf(x)dx \\ &= \int_0^{\infty} (1+u)^{-1}\int_u^{\infty} f(x)dxdu \\ &= \int_0^{\infty} (1+u)^{-1}[1 - F(u)]du \end{align*}$$

Answer 2

Hint: When you integrate by parts, you can add any constant to the antiderivative that is introduced. Often, nothing is added, but sometimes a specific constant may be interesting.

Usually:

$$\int Uv \mathrm dx=UV-\int uV\mathrm dx$$

More generally:

$$\int Uv \mathrm dx=U(V+a)-\int u(V+a)\mathrm dx$$

Here, to make the limit at $\infty$ converge in $[\ln(1+x)(F(x)+a)]_0^\infty$, you have to set $a=-1$.