I was reading The Laplace Transform by Widder and have a problem in understanding the proof of theorem 6.1 in chapter II. Please see the following image
I am unable to understand how equation 2 becomes equation 4 with the help of equation 3. So my question is how to get equation 4 from equation 2.
Since $\alpha$ is integrable, $\displaystyle\beta(t) = \int_0^t \alpha(u) \, du$ is of bounded variation and
$$\int_0^1 t^n d \beta(t) = \int_0^1 t^n \alpha(t) \, dt = 0$$
This follows formally by writing $d\beta(t) = \alpha(t) \, dt$. However, a rigorous proof is needed in general, for $\alpha$ not everywhere continuous, and is given here.
We can now follow the same steps (IBP) leading from (1) to (2) to obtain (4):
$$\int_0^1 t^n \beta(t)\, dt = 0 \quad\quad\quad\quad (n=0,1,2,\ldots) $$
