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Suppose $f$ is integrable on $[0,b]$. Let $g(x)=\displaystyle\int^b_x\frac{f(t)}t\;dt,0.

Now I want to prove that $g$ is integrable on $[0,b]$. But I don't know how to show that.

Can anyone give me some hints?

Thank you very much.

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    If you now understand how to solve the problem, you should write up an answer. It may seem odd to answer your own question, but site policy actually encourages doing that. Then you can accept your answer (and answers to some of your other questions, too!).2011-12-09

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Fix $\delta>0$. Then we have, using Fubini's theorem (as @Zarrax suggests) \begin{align*} \int_{\delta}^b|g(x)|dx&\leq\int_{\delta}^b\int_x^b\left|\frac{f(t)}t\right|dtdx\\ &=\int_{{\delta\leq x\leq t\leq b}}\frac{|f(t)|}tdtdx\\ &=\int_{\delta}^b\int_{\delta}^t\frac{|f(t)|}tdxdt\\ &=\int_{\delta}^b\frac{t-\delta}t|f(t)|dt\\ &\leq \int_{\delta}^b|f(t)|dt\\ &\leq \int_0^b|f(t)|dt,\\ \end{align*} and we are done since the map $\delta\mapsto \int_{\delta}^b|g(x)|dx$ is decreasing.