Suppose $f$ is integrable on $[0,b]$, and
$$g(x) = \int^b_x\frac{f(t)}{t}\;dt\quad\text{for}\quad0 < x \leq b.$$
Prove that $g$ is integrable on $[0,b]$ and
$$\int^b_0 g(x)\;dx=\int_0^bf(t)\;dt$$
Okay, so i'm going need some insight on this. It seems strange to me that equality should hold in both of these circumstances.
Hint: If you were back in calculus, you would write the integral of $g$ as the interated integral of $f$ and then fearlessly switch the order of integration. You can do the same here, but why?