Suppose $f$ is an integral defined on $[0,b]$, and

Question

Suppose $f$ is integral enough on $[0,b],$ and $g(x)=\int_x^b\frac{f(t)}{t} \, dt$ for $0<x\leq b.$

Prove that $g$ is integrable on $[0,b]$ and $\int_0^b g(x) \, dx=\int_0^b f(t) \, dt$

I know we are trying to prove $\limsup_{n\to 0}\int_n^b g < \infty$, but I tried to look at convergence theorems and I’m at lost, any help or hint? Thanks.

Answer 1

Change order of integration. Write $g(x)=\int_0^b\int_x^b\frac{f(t)}{t}dtdx=\int_0^b\frac{f(t)}{t}\int_0^tdxdt=\int_0^bf(t)dt$