Is there a way to compute the intergral of a bijection by using its symetry?
Meaning is there a way to proove ( maybe by a substitution) that
$$\int_0^2 e^x dx = \int_1^{l} \ln(x)dx$$
(i m not sur about the bounds of the integral... that's why i ve written $l$, I can't find the upper bound)
and then to generalize this idea for any bijection. Especially I was thinking about $ \frac{1}{x} $ which we can't compute the integral on $\mathbb{R}^+$ but nevertheless we have :
$$\int_0^1 \frac{1}{x} dx = \int_1^{\infty} \frac{1}{x} dx$$
thank you :)
Perhaps the following result is what you are after.
If $f$ is a bijective function with continuous derivative on the interval $[a,b]$, then $$\int_a^b f(x) \, dx + \int_{f(a)}^{f(b)} f^{-1} (x) \, dx = b f(b) - a f(a).\tag1$$ Here $f^{-1}$ denotes the inverse of $f$.
For the case of your first example, setting $f(x) = \ln x$ with $a = 1, b = e^2$, as $f^{-1} (x) = e^x$ application of (1) gives $$\int_1^{e^2} \ln x \, dx + \int_0^2 e^x \, dx = e^2 \cdot \ln (e^2) - 1 \cdot \ln (1),$$ or $$\int_1^{e^2} \ln x \, dx = 2e^2 - \int_0^2 e^x \, dx.$$