Why is $\int_0^1e^{e^x}dx$ equal to $\int_1^e{e^uu^{-1}}du$?

Question

Why is this equality $\displaystyle\int_0^1e^{e^x}dx=\int_1^e{e^uu^{-1}}du$?

I don't see the change of variable that was used to pass from one integral to the other?

Could someone please explain?

And how to solve the latter?

Thank you

Answer 1

$u\mapsto e^x$. This change of variable can be seen pretty easily from the bounds of integration and the integrand itself. $$\begin{bmatrix}u\\\mathrm du\end{bmatrix}\mapsto\begin{bmatrix}e^x\\e^x\mathrm dx\end{bmatrix}$$ Notice that $\mathrm du=e^x\mathrm dx$ is equivalent to saying $\mathrm du/u=\mathrm dx$. $$\int_{0}^{1}e^{e^x}\mathrm dx=\int_{1}^{e}e^u\cdot\mathrm du/u$$ This is a special integral and has a name, it is the exponential integral denoted $\mathrm {Ei}(x)$. The definite integral can be approximated, and doing so gives $6.316563839027679$.

Answer 2

The change of variables is $u = e^x$. Notice that $du = e^x dx = u dx $

Therefore,

$$ \int\limits_{u(0)}^{u(1)} e^{u} \cdot \frac{1}{u} du = \int\limits_1^e e^u u^{-1} du $$