Integration by parts in expectation of a random variable Prove the integration by parts property of expectation of a random variable, that is, for a random variable $X$ with cumulative distribution function $F_X$ and probability density function $f_X$,
\begin{align*}
E[X] = \int_{-\infty}^{\infty}x f_X(x)\, dx = \int_{0}^{\infty}(1 - F_X(x)) \,dx - \int_{-\infty}^{0}x F_X(x)dx
\end{align*}
My attempt:
\begin{align*}
E[X] &= \int_{-\infty}^{\infty}x f_X(x)\, dx  \\
&=  \bigg[xF(x)\bigg]_{-\infty}^{\infty} - \int_{-\infty}^{\infty}F_X(x)\, dx
\end{align*}
But I get stuck here itself and it leads nowhere. So how do I use integration by parts to prove the result?
 A: The original statement is simply false! The correct statement is
$$\mathbb{E}[X]=\int_0^{\infty}\left[1-F_X(x)\right]dx-\int_{-\infty}^0 F_X(x)dx=I_1-I_2$$
this can be easy proved integrating by parts
$$I_1=\underbrace{\left[x(1-F_X(x) )  \right]_0^{\infty}}_{=0}+\int_0^{\infty}xf_X(x)dx$$
$$I_2=\underbrace{\left[xF_X(x)   \right]_{-\infty}^0}_{=0}-\int_{-\infty}^0xf_X(x)dx$$
thus
$$\mathbb{E}[X]=I_1-I_2=\int_{-\infty}^{\infty}xf_X(x)dx$$
The expressions =0 in the above $I_1,I_2$ can be easily proved applying de l'Hôpital to the resulting limit

Geometrically, this means that the expectation is the difference between Purple and Yellow Areas

A: As right hand side guides you, First split negative and positive parts. Then try to prove it.
An intuitive proof (Maybe you didn't believe it is a proof!):
Consider a vertical element in graph of $f(x)$, and prove it occurs in left hand side $(xf(x))$ as many as in right hand side$((1-F(x))$ or $(xF(x)))$.
A: Maybe it's too late for the asker but useful for the later readers.
As Ali Ashja' said, first separate the integral in two: in positive values and negative values: $\int_0^{\infty}xf(x)dx+\int_{-\infty}^0xf(x)dx$.
Then apply integration by parts with $u=x$ and $dv=f(x)dx$. You have to choose a primitive of $f(x)$ for $v$, surely you think $v=F(x)$ is an option, that's true but in this case it's better to choose $v=F(x)-1$ and you will see a more similar expression to the final result.
