Dos ejercicios de integración definida en los que interviene la función piso

Deseamos calcular el valor de la integral definida $$\displaystyle \int_{1}^{3}\,\dfrac{x}{\left \lfloor x \right \rfloor}\,dx$$

Recordemos la definición de la función piso: $$\left \lfloor x \right \rfloor := \text{máximo}(\{\ell \le x: \ell \in \mathbb{Z}\}$$

Entonces, $$\displaystyle \int_{1}^{3}\,\dfrac{x}{\left \lfloor x \right \rfloor}\,dx=\int_{1}^{2}\,\dfrac{x}{1}\,dx+\int_{2}^{3}\,\dfrac{x}{2}\,dx=\int_{1}^{2}\,x\,dx+\dfrac{1}{2}\,\int_{2}^{3}\,x\,dx=\left[ \dfrac{1}{2}\,x^2 \right]_{1}^{2}+\dfrac{1}{2}\cdot \left[ \dfrac{1}{2}\,x^2 \right]_{2}^{3}=$$ $$\displaystyle = \dfrac{1}{2} \cdot \left[ x^2 \right]_{1}^{2}+\dfrac{1}{2}\cdot \dfrac{1}{2}\cdot \left[ x^2 \right]_{2}^{3}=\dfrac{1}{2} \cdot \left( 2^2-1^2 \right)+\dfrac{1}{4}\cdot \left( 3^2-2^2 \right)=\dfrac{1}{2} \cdot 3 +\dfrac{1}{4}\cdot 5 = \dfrac{3}{2}+\dfrac{5}{4}=\dfrac{6}{4}+\dfrac{5}{4}=\dfrac{11}{4}$$

-oOo-

Ahora voy a calcular el valor de la integral definida $$\displaystyle \int_{1}^{3}\,\dfrac{\left \lfloor x \right \rfloor}{x}\,dx$$ que no es la misma que la primera que hemos resuelto, ya que $\dfrac{x}{\left \lfloor x \right \rfloor} \neq \dfrac{\left \lfloor x \right \rfloor}{x}$

$$\displaystyle \int_{1}^{3}\,\dfrac{\left \lfloor x \right \rfloor}{x}\,dx=\int_{1}^{2}\,\dfrac{1}{x}\,dx+\int_{2}^{3}\,\dfrac{2}{x}\,dx=\left[\ln(x)\right]_{1}^{2}+2\,\left[\ln(x)\right]_{2}^{3}=\left(\ln(2)-\ln(1)\right)+2\cdot \left( \ln(3)-\ln(2)\right)=$$ $$=\ln(2)-0+2\,\ln(3)-2\,\ln(2)=2\,\ln(3)-\ln(2)=\ln(3^2)-\ln(2)=\ln\left(3^2/2\right)=\ln\left(9/2\right)$$

-oOo-

Y para acabar este artículo, expondré el cálculo de la integral definida $$\displaystyle \int_{-1}^{1}\,\dfrac{|x|}{\left \lfloor x \right \rfloor+2}\,dx$$

Teniendo en cuenta las definiciones de las funciones valor absoluto y piso: $$\displaystyle \int_{-1}^{1}\,\dfrac{|x|}{\left \lfloor x \right \rfloor+2}\,dx=\int_{-1}^{0}\,\dfrac{-x}{-1+2}\,dx+\int_{0}^{1}\,\dfrac{x}{0+2}\,dx=\int_{-1}^{0}\,\dfrac{-x}{1}\,dx+\int_{0}^{1}\,\dfrac{x}{2}\,dx=$$ $$\displaystyle =-\int_{-1}^{0}\,x\,dx+\dfrac{1}{2}\,\int_{0}^{1}\,x\,dx=-\left[\dfrac{1}{2}\,x^2\right]_{-1}^{0}+\dfrac{1}{2}\cdot \left[\dfrac{1}{2}\,x^2\right]_{0}^{1}=-\dfrac{1}{2}\cdot \left[x^2\right]_{-1}^{0}+\dfrac{1}{4}\cdot \left[x^2\right]_{0}^{1}=$$ $$=-\dfrac{1}{2}\cdot \left(0^2-(-1)^2\right)+\dfrac{1}{4}\cdot \left(1^2-0^2\right)=-\dfrac{1}{2}\cdot (-1)+\dfrac{1}{4}=\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{2}{4}+\dfrac{1}{4}=\dfrac{3}{4}$$

$\diamond$