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Suppose $f(x)$ has a support $x\in(0,\infty)$. By integration by part, $$\begin{align*} \int xf'(x) dx & = xf(x)|^{\infty}_{0} -\int f(x) dx\\ & = xf(x)|^{\infty}_{0} -1 \end{align*}$$

My question is how to solve $xf(x)|^{\infty}_{0}$. I would assume that this will equal zero...

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    "has a support $x \in (0,\infty)$" makes no sense. Do you mean that its support is compactly contained in $(0,\infty)$, i.e. there exists $0$f$ is contained in $[a,b]$? If so then yes, the first term is just zero. The same is true if $a=0$. – 2017-01-21
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    @lan "has a support $x\in(0,\infty)$ means that $f(x)$ is only defined on that region and 0 otherwise. – 2017-01-21
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    So you mean its support is contained in $(0,\infty)$. – 2017-01-21

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$xf(x)|^\infty _0$ means you take limit $\lim_{x \to \infty}xf(x)- \lim_{y \to 0}y f(y)$. Both terms have to go to 0 because the positive function $f$ must integrate to 1. One can show that if any one of the two limits are non-zero then $f$ integrates to infinity

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    The limit at $+\infty$ is not trivial, it is possible for $xf$ to not go to zero and $f$ can still be integrable. – 2017-01-21
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    Yes, that part does require the existence of expected value. – 2017-01-21
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    That doesn't fix it either: $xf$ can be integrable without going to zero at infinity. – 2017-01-21
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    Ur right, something else is needed, maybe something like $f(x)$ is monotonically decreasing when $x>N$ for some large $N$. – 2017-01-21
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    @jimh Suppose that $f(x)$ is monotonically decreasing for some large $N$, then how should I prove that $x'f(x)|^\infty_0$ goes to 0? – 2017-01-21
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    Try to show that integrability of $f$ implies for $N$ large enough, if $x>N$ then $f(x)\leq 1/x$. Then try contrapositive of the limit not being $0$. – 2017-01-21
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    wow.. thanks!!! – 2017-01-21
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    Again that doesn't work by itself. Consider $f(x)=\begin{cases} x & x \in [0,1/2] \\ 1-x & x \in [1/2,1] \\ 0 & \text{otherwise} \end{cases}$ and $g(x)=\sum_{n=1}^\infty f(2^n (x-n))$. Then $g$ is integrable but does not vanish at infinity. Monotonicity removes examples like this one. So does uniform continuity (but not continuity: the $g$ I wrote is continuous). – 2017-01-22
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    @Ian Yea my last comment was meant to use monotonicity. Although uniform continuity removes your example, it seems not enough to prove what user1292919 want. – 2017-01-22
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    @jimh A uniformly continuous integrable function does vanish at infinity. – 2017-01-22
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    @Ian yes but multiplied with x? – 2017-01-22
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    Right, this should be referring to $xf$, not $f$ itself. – 2017-01-22
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    @Ian I see, you also want to assume the expectation exists. That also works. – 2017-01-22