Why is $\int_\varepsilon^{1/e}-\frac{1}{\\\log(t)}dt = \int_1^a \frac{e^{-x}}{x}dx$ Let $\varepsilon = e^{-a} \in [0,e)$, why does the following hold (including $\varepsilon=0$)?
$$
    \int_\varepsilon^{1/e}-\frac{1}{\\\log(t)}dt = 
    \int_1^a \frac{e^{-x}}{x}dx
$$

Do the substitution $x = -\log(t)$. Thus, $dx = -1/t \,dt \implies dt = -e^{-x} dx.$  Notice that $t=\varepsilon \implies u = a$ and  $t=1/e \implies u = 1$.  Hence, you have
$$\int \limits_{\varepsilon}^{1/e} -\frac{dt}{\log(t)} = \int \limits_{a}^{1} -\frac{e^{-x}}{x}\,dx = \int \limits_{1}^{a} \frac{e^{-x}}{x}\,dx.$$
Since $\varepsilon \in (0, e)$, we have $a \in (-1,\infty)$. You shouldn't include $\varepsilon =0$ because the exponential is always positive.
Notice that if $a \in (-1,0]$ the function $\frac{e^{-x}}{x}$ has a singularity at $x =0$ and the integral diverges. Hence, the equality holds if and only if $a \in  (0, \infty)$.
