Integration by parts appears to prove the existence of integrals that have discontinuity.

Question

Consider the following integral:

$$\int \lfloor x \rfloor dx$$

I know the integral reduces to:

$$x \lfloor x \rfloor - \frac {\lfloor x \rfloor ^2}2 - \frac {\lfloor x \rfloor}2$$

However, if we use integration by parts:

$$\int \lfloor x \rfloor dx = x\lfloor x \rfloor - \int x (\lfloor x \rfloor)' dx$$

Using simple algebra:

$$\int x (\lfloor x \rfloor)' dx = \frac {\lfloor x \rfloor ^2}2 + \frac {\lfloor x \rfloor}2$$

Yet, this is a contradiction. The indefinite integral is always continuous, yet the final expression is a piecewise constant function with infinite jump discontinuities. Does integration by parts only work for continuous functions or something? This is incredibly confusing.

Answer 1

The problem with your argument is, as stated in the comments, the continuity of $u$ and $v$ are necessary conditions for applying the integration by parts. Take a look at this article:

The theorem can be derived as follows. Suppose $u(x)$ and $v(x)$ are two continuously differentiable functions...

As a side-note, according to the fact that: $$\int_a^b f(x)\delta(x-x_0)dx=\begin{cases}f(x_0)&a<x_0<b\\0&\text{otherwise}\end{cases}$$ and knowing: $$d(\lfloor x \rfloor)=(\lfloor x \rfloor)'dx=\delta(x-\lfloor x \rfloor)dx$$ where $\delta$ is the Dirac delta function, one can write: $$\int x d\lfloor x \rfloor=\int x\delta(x-\lfloor x \rfloor)dx$$ Therefore $$\int_a^b x\delta(x-\lfloor x \rfloor)dx=\sum_{a<\lfloor x\rfloor<b}\lfloor x \rfloor$$