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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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    @Dylan: I am cu$r$ious, how does the above featu$r$e in the p$r$oof of Ha$r$dy's inequality?2011-07-20

2 Answers 2

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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!2011-07-20
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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.)