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.
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.
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.