This problem comes from Brian Fisher and Biljana Jolevska-Tuneska's 2010 paper 'On the logarithmic integral'. As it is freely available I will refer to it rather than copy large chunks here that will undoubtably make things more unclear.
Given $(f * g)(x)=\int_{-\infty}^{\infty}f(t)g(x-t)dt$
they show that,
using where PV is the Cauchy principle value,
Now whilst step 1 is not clear how it relates to $(f * g)(x)=\int_{-\infty}^{\infty}f(t)g(x-t)dt$, though it might be something to do with $(f * g)(x)=[\int_{-\infty}^{\infty}f(x)dx][\int_{-\infty}^{\infty}g(t)dt]$ as shown here, I do not understand then how they get from 1 to 2 then to 3? Step 4 is fairly obvious.
First, the definitions of $\operatorname{li}_+$ and $x_+$: $$ \operatorname{li}_+(x) = \begin{cases} \mathrm{PV} \int_0^x \frac{dt}{\ln t}, & x>0 \\ 0 & x \leq 0 \end{cases} $$
$$ x_+ = \begin{cases} x & x>0 \\ 0 & x\leq 0 \end{cases} $$
I will skip the prefix $\mathrm{PV}$ though.
Now, by definition of convolution, $$ (\operatorname{li}_+ * x_+^r)(x) = (x_+^r * \operatorname{li}_+)(x) = \int_0^\infty (x-t)_+^r \operatorname{li}_+(t) \, dt $$ Here, $(x-t)_+$ vanishes if $x-t<0,$ i.e. if $t>x$. Therefore, $$ \int_0^\infty (x-t)_+^r \operatorname{li}_+(t) \, dt = \int_0^x (x-t)^r \operatorname{li}_+(t) \, dt $$
Then we insert the definition of $\operatorname{li}_+$ and get $$ \int_0^x (x-t)^r \left( \int_0^t \frac{du}{\ln u} \right) \, dt $$
The next step is to make a change of order of integration. The domain of integration is $\{ (t,u) \mid 0 < t < x, \ 0 < u < t \}$ which can also be written $\{ (t,u) \mid 0 < u < x, \ u < t < x \},$ so $$ \int_0^x (x-t)^r \left( \int_0^t \frac{du}{\ln u} \right) \, dt = \int_0^x \frac{1}{\ln u} \left( \int_u^x (x-t)^r \, dt \right) \, du $$
Then we evaluate the inner integral, $$ \int_u^x (x-t)^r \, dt = \left[ \frac{(-1)^r (x-t)^{r+1}}{r+1} \right]_{t=u}^x = \frac{(u-x)^{r+1}}{r+1} $$
By the binomial theorem, $(u-x)^{r+1} = \sum_{i=0}^{r+1} { r+1 \choose i } (-1)^i x^i u^{r+1-i}$. Inserting this we get step 3.