# How to show that $g(x)=\int^b_x \frac{f(t)}t\;dt,0<x\le b$ is integrable where $f$ is integrable on $[0,b]$

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

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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Did you mean integrable on $[0,b]$? – Henry Shearman Dec 8 '11 at 12:32
I am so sorry for that. I just want to say that $f$ is integrable. – molan Dec 8 '11 at 13:14
First $g$ is continuous on $(0,b]$. Are you talking about the Riemann or Lebesgue integrals? For Lebesgue all that remains is to show $\int_0^b |g(x)|\,dx < \infty$. – GEdgar Dec 8 '11 at 13:33
I mean Lebesgue integrals. – molan Dec 8 '11 at 13:35
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!). – Gerry Myerson Dec 9 '11 at 11:18

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.