I'm trying to prove for $$f:(a,b]\to\mathbb{R}$$ be a continuous and strictly decreasing function, with $$\displaystyle \lim_{x\to a^{+}} f(x)=\infty$$, that: $$\int_a^b (-1)^{\left \lfloor f(x) \right \rfloor}\mathrm dx=(-1)^{\left \lceil f(b) \right \rceil -1}b +2\sum_{n=\left \lceil f(b) \right \rceil}^\infty (-1)^{n}f^{-1}(n)$$ where $$\left \lfloor m \right \rfloor$$ and $$\left \lceil m \right \rceil$$ are floor and ceiling function of $$m$$, respectively. Intuitively, I start with the substitution $$f(x)=y$$ and then:

$$\int_a^b (-1)^{\left \lfloor f(x) \right \rfloor}\mathrm dx=\int_{f(a)}^{f(b)} (-1)^{\left \lfloor y \right \rfloor}\left[f^{-1}(y)\right]'\mathrm dy= -\int_{f(b)}^{\infty} (-1)^{\left \lfloor y \right \rfloor}\left[f^{-1}(y)\right]'\mathrm dy$$

Here, I'm having trouble developing the last integral and what I know is that I need to add integrals over some interval which for me is still a mystery. Thanks for some clarification.

• Don't shout!!!! Commented Jul 31, 2021 at 5:48

Since the floor function is not continuous (let alone differentiable), it seems more promising to start with letting $$x_n=f^{-1}(n)$$ for all naturals $$n\ge n_0:=\lceil f(b)\rceil$$. Then $$\{x_n\}_n$$ is a sequence in $$(a,b]$$ and strictly decreasing to $$a$$. As the integrand is piecewise constant, we have $$\int_{x_{n_0}}^b(-1)^{\lfloor f(x)\rfloor}\,\mathrm dx = \int_{x_{n_0}}^b(-1)^{\lfloor f(b)\rfloor}\,\mathrm dx =(b-x_{n_0})(-1)^{\lfloor f(b)\rfloor} =(-1)^{n_0}(x_{n_0}-b)$$ and for $$n\ge n_0$$, $$\int_{x_{n+1}}^{x_{n}}(-1)^{\lfloor f(x)\rfloor}\,\mathrm dx=\int_{x_{n+1}}^{x_{n}}(-1)^{n}\,\mathrm dx=(-1)^n(x_{n}-x_{n+1}).$$ By summing and telescoping, \begin{align}\int_{x_{N+1}}^{b}&=(-1)^{n_0}(x_{n_0}-b)+\sum_{n=n_0}^N(-1)^n(x_{n}-x_{n+1})\\ &=(-1)^{n_0+1}b+(-1)^{n_0}x_{n_0}+\sum_{n=n_0}^N(-1)^nx_{n}+\sum_{n=n_0+1}^{N+1}(-1)^nx_{n}\\ &=(-1)^{n_0+1}b+2\sum_{n=n_0}^N(-1)^nx_{n}+(-1)^{N+1}x_{N} \end{align} Let's assume $$a=0$$ for the moment. Then if we take the limit as $$N\to\infty$$ of the right hand side above, we arrive at $$(-1)^{n_0+1}b+2\sum_{n=n_0}^\infty(-1)^nx_{n}$$ where the series converges by the Leibniz criterion, while at the same time the left hand side converges to the improper integral from $$a$$ to $$b$$, as desired.
In the general case when $$a\ne 0$$, the series does not converge, while clearly the improper integral does. We can adjust for that: Define $$g\colon (0,b-a]\to \Bbb R$$, $$g(x)=f(x+a)$$, do the above with $$g$$. We can then express the result in terms of $$f$$ and note that the original claim has to be adjusted to have $$(-1)^n(f^{-1}(n)-a)$$ instead of $$(-1)^nf^{-1}(n)$$ as series summand.