Can you integrate by parts with one integral inside another? From the definition of the Laplace transform:
$$\mathcal{L}[f(t)]\equiv \int_{t=0}^{\infty}f(t)e^{-st}\mathrm{d}t$$
where $s \in \mathbb{R^+}$. 
$$\mathcal{L}\left[\int_{u=0}^{t}f(u)\mathrm{d}u\right]=\int_{t=0}^{\infty}e^{-st}\mathrm{d}t\int_{u=0}^{t}f(u)\mathrm{d}u$$
$$=\int_{t=0}^{\infty}e^{-st}\mathrm{d}t\int_{u=0}^{t} f(u)\mathrm{d}u=\int_{u=0}^{t} f(u) \mathrm{d}u\int_{t=0}^{\infty}e^{-st}\mathrm{d}t$$
$$\left[-\cfrac{1}{s}e^{-st} \int_{u=0}^{t} f(u)\mathrm{d}u \right]_{\color{red}{0}}^{\color{red}{\infty}}+\int_{\color{red}{0}}^{\color{red}{\infty}}\cfrac{1}{s}e^{-st}f(\color{blue}{t})\mathrm{d}\color{blue}{t}$$
$$=\cfrac{1}{s}\mathcal{L}\left[\color{#180}{f}\right]$$
Assuming that on page 486 in this book it was done by parts I have $\mathbf{3}$ questions about the calculation above:
$\mathbf{\color{red}{1)}}$ For the $2$ sets of limits marked red above for which variable ($u$ or $t$) do they belong?
$\mathbf{\color{blue}{2)}}$ For the variables marked blue, shouldn't the $t$'s be $u$'s?
$\mathbf{\color{#180}{3)}}$ For the part marked green should this be $f(t)$ instead of $t$?
Please explain your answers.
Thank you.
 A: $$\mathcal{L}\left[\int_{u=0}^{t}f(u)\mathrm{d}t\right]$$ should be $$\mathcal{L}\left[\int_{u=0}^{t}f(u)\mathrm{d}u\right]$$ denote the argument to $\mathcal{L}$ a $g(t)= \int_{u=0}^{t}f(u)\mathrm{d}u$ to get
$$\mathcal{L}[g(t)]=\int_{t=0}^{\infty}g(t)e^{-st}\mathrm{d}t = \int_{t=0}^{\infty} \left(\int_{u=0}^{t}f(u)\mathrm{d}u\right) e^{-st}\mathrm{d}t  =
\int_{t=0}^{\infty} \int_{u=0}^{t}f(u)  e^{-st}\,\mathrm{d}u\,\mathrm{d}t
\ne \int_{t=0}^{\infty}e^{-st}\mathrm{d}t\int_{u=0}^{t}f(u)\mathrm{d}u = \left(\int_{t=0}^{\infty}e^{-st}\mathrm{d}t\right)\cdot g(t)$$
Note also that you can't swap the integrals since $\mathrm{\color{purple}{the\space limit\space of\space inner\space one\space is\space an\space integration\space variable\space of\space the\space outer\space one.}}$ 
This is pretty much put your first two question less relevant, but to give a general answer:


*

*always look at the differential $\mathrm{d}X$ so if $\frac{\partial F}{\partial x}=f$ you will have $$\int_a^b f(X,y,z,...) \mathrm{d}X =
    F(X,y,z,...)\big|_{X=a}^{X=b}$$

*$f(t)={d\over dt}\int_0^t f(u)\,du$ by fundamental theorem of calculus

*$\mathcal{L}$ is an operator which take functions and return another functions, similar to functions that take numbers and return another numbers. So, the correct notations are $\mathcal{L}f$ or when it is not clear then $\mathcal{L}(f)$ (you can use any kind of brackets in general as long as it is become clear), and therefore $\mathcal{L}f(t)\equiv \left\{\mathcal{L}f\right\}(t)$. In some sense the notation $\mathcal{L}[f(t)]$ is not correct, because if you evaluate it, say at $t=1$, you get a number $f(1)$ which is not exactly what you what you want to supply to the operator $\mathcal{L}$

A: 1) $t$
2) No: The $f(t)$ there should be thought of as ${d\over dt}\int_0^t f(u)\,du$, in accordance with the integration by parts.
3) No, $\mathcal L[f]$ is correct.
