if $\phi(x)={1\over x}\int\nolimits_0^x F(t)dt$ and $F(x):=\int_0^x f(t)dt$ ,how does $$\phi'(x)=-{1\over x^2}\int_0^xF(t)dt +{1\over x}F(x)={1\over x^2}\int_0^x t f(t)dt\ ?$$

Please explain step by step since I am confused how to get to the last two equalities.

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Out of curiosity, what have you tried so far? It is easier to help and give advice if you tell us exactly where you are stuck. – Eric Naslund Jul 20 '11 at 17:55
@Dylan Moreland - i am not that advanced... – Victor Jul 20 '11 at 18:43
@Dylan: I am curious, how does the above feature in the proof of Hardy's inequality? – Eric Naslund Jul 20 '11 at 18:53

Here are two hints. Give it a try, I can expand on the details for Hint 2 if needed.

Hint 1:
$$\phi'(x)=\frac{d}{dx} \frac{1}{x} \int_0^x F(t)dt\\ =\left(\frac{d}{dx}\frac{1}{x}\right) \int_0^xF(t)dt+\frac{1}{x} \left(\frac{d}{dx}\int_0^x F(t)dt\right).$$
Then the fundamental theorem of calculus says $$\left(\frac{d}{dx}\int_0^x F(t)dt\right)=F(x).$$ Also, $$\frac{d}{dx}\frac{1}{x}=\frac{-1}{x^2}.$$

Hint 2: For any nice function $f$, we have that $$\int_0^x \int_0^t f(u) du dt= \int_0^x (x-t)f(t)dt.$$ This is a case of Cauchy's formula for repeated integration.

Hope that helps,

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very concise,thanks! – Victor Jul 20 '11 at 18:33

For our purposes, we may assume that $f$ is continuous.

Step 1: $$\phi '(x) = \bigg(\frac{d}{{\,dx}}\frac{1}{x}\bigg)\int_0^x {F(t)\,dt} + \frac{1}{x}\frac{d}{{\,dx}}\int_0^x {F(t)\,dt}.$$

Step 2: $$\phi '(x) = - \frac{1}{{x^2 }}\int_0^x {F(t)\,dt} + \frac{1}{x}F(x).$$

Step 3: $$x^2 \phi '(x) = - \int_0^x {F(t)dt} + xF(x) = xF(x) - \int_0^x {F(t)dt}.$$

Step 4 -- integration by parts: $$\int_0^x {tf(t) \,dt} = tF(t) \big|_0^x - \int_0^x {\bigg(\frac{d}{{\,dt}}t \bigg)F(t)\,dt} = xF(x) - 0F(0) - \int_0^x {1F(t)\,dt} = xF(x) - \int_0^x {F(t)\,dt}.$$

Step 5: It follows that $$x^2 \phi '(x) = \int_0^x {tf(t) \,dt}.$$

Step 6: It follows that $$\phi '(x) = \frac{1}{{x^2 }}\int_0^x {tf(t) \,dt}.$$

EDIT: For the integration by parts (beginning of Step 4), note that $F$ is an antiderivative of $f$, since, by the Fundamental theorem of calculus, $$F'(x) = \frac{d}{{dx}}F(x) = \frac{d}{{dx}}\int_0^x {f(t)dt} = f(x).$$ (Here we used the assumption that $f$ is continuous.)

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