Integrar a Lebesgue

1. Integre a função y= I(x) (y é igual ao inteiro máximo em valor absoluto, contido em x) definida no campo dos números inteiros, a Lebesgue

Olha, pelo que eu entendi ele quer a integral da parte inteira. Se for isso,

A integral de y pode ser dada por:

[latex]S(x) = \int_{a}^{x}\left \lfloor x \right \rfloor \mathrm{d} x[/latex]

Perceba que, sendo n ∈ ℤ, se:

[latex]n \leq x < n+1 \Leftrightarrow \left \lfloor x \right \rfloor = n[/latex]

[latex]\int \left \lfloor x \right \rfloor \mathrm{d} x = \int n \cdot \mathrm{d} x = n\cdot x [/latex]

Com isso, dividindo a integral em regiões com limites inteiros temos:

[latex]S(x) = \int_{a}^{\left \lfloor a \right \rfloor + 1}\left \lfloor x \right \rfloor \mathrm{d} x \; + \int_{\left \lfloor a \right \rfloor + 1}^{\left \lfloor a \right \rfloor + 2}\left \lfloor x \right \rfloor \mathrm{d} x \; + ... + \int_{\left \lfloor x \right \rfloor - 1}^{\left \lfloor x \right \rfloor}\left \lfloor x \right \rfloor \mathrm{d} x \; + \int_{\left \lfloor x \right \rfloor}^{x}\left \lfloor x \right \rfloor \mathrm{d} x[/latex]

[latex]S(x) = \int_{a}^{\left \lfloor a \right \rfloor + 1}\left \lfloor x \right \rfloor \mathrm{d} x \; + \left [\sum_{k = \left \lfloor a \right \rfloor + 1}^{\left \lfloor x \right \rfloor - 1} \int_{k}^{k+1} \left \lfloor x \right \rfloor \mathrm{d} x\right ] + \int_{\left \lfloor x \right \rfloor}^{x}\left \lfloor x \right \rfloor \mathrm{d} x[/latex]

[latex]S(x) = \int_{a}^{\left \lfloor a \right \rfloor + 1}a\cdot \mathrm{d} x \; + \left [\sum_{k = \left \lfloor a \right \rfloor + 1}^{\left \lfloor x \right \rfloor - 1} \int_{k}^{k+1} k\cdot \mathrm{d} x\right ] + \int_{\left \lfloor x \right \rfloor}^{x}\left \lfloor x \right \rfloor\cdot \mathrm{d} x[/latex]

[latex]S(x) = a\cdot \int_{a}^{\left \lfloor a \right \rfloor + 1} \mathrm{d} x \; + \left [\sum_{k = \left \lfloor a \right \rfloor + 1}^{\left \lfloor x \right \rfloor - 1} k\cdot \int_{k}^{k+1} \mathrm{d} x\right ] + \left \lfloor x \right \rfloor\cdot \int_{\left \lfloor x \right \rfloor}^{x} \mathrm{d} x[/latex]


[latex]S(x) = a\cdot \left (\left ( \left \lfloor a \right \rfloor + 1 \right ) - a \right ) + \left [\sum_{k = \left \lfloor a \right \rfloor + 1}^{\left \lfloor x \right \rfloor - 1} k\cdot \left ( \left (k + 1 \right ) - k \right )\right ] + \left \lfloor x \right \rfloor\cdot \left ( x - \left \lfloor x \right \rfloor \right )[/latex]

Chamando a * ((int(a) + 1) - a) de c, temos:

[latex]S(x) =\left \lfloor x \right \rfloor\cdot \left ( x - \left \lfloor x \right \rfloor \right ) + \left [\sum_{k = \left \lfloor a \right \rfloor + 1}^{\left \lfloor x \right \rfloor - 1} k\right ] + c[/latex]

[latex]S(x) =\left \lfloor x \right \rfloor\cdot \left ( x - \left \lfloor x \right \rfloor \right ) + \frac{1}{2}\cdot \left ( \left ( \left \lfloor x \right \rfloor - 1 \right )  - \left ( \left \lfloor a \right \rfloor + 1 \right ) + 1\right )\cdot \left ( \left ( \left \lfloor x \right \rfloor - 1 \right )  + \left ( \left \lfloor a \right \rfloor + 1 \right ) \right ) + c[/latex]

[latex]S(x) =\left \lfloor x \right \rfloor\cdot \left ( x - \left \lfloor x \right \rfloor \right ) + \frac{1}{2}\cdot \left ( \left \lfloor x \right \rfloor  - \left \lfloor a \right \rfloor - 1\right )\cdot \left ( \left \lfloor x \right \rfloor + \left \lfloor a \right \rfloor \right ) + c[/latex]

[latex]S(x) =\left \lfloor x \right \rfloor\cdot \left ( x - \left \lfloor x \right \rfloor \right ) + \frac{1}{2}\cdot \left ( \left \lfloor x \right \rfloor^{2}   - \left \lfloor a \right \rfloor^{2} - \left \lfloor x \right \rfloor - \left \lfloor a \right \rfloor\right ) + c[/latex]

[latex]S(x) =\left \lfloor x \right \rfloor\cdot \left ( x - \left \lfloor x \right \rfloor \right ) + \frac{1}{2}\cdot \left \lfloor x \right \rfloor^{2} - \frac{1}{2}\cdot \left \lfloor x \right \rfloor + \frac{1}{2}\cdot \left (- \left \lfloor a \right \rfloor^{2} - \left \lfloor a \right \rfloor\right ) + c[/latex]

Unindo novamente a parte indeterminada, chamando-a de C:

[latex]S(x) =\left \lfloor x \right \rfloor\cdot \left ( x - \left \lfloor x \right \rfloor \right ) + \frac{1}{2}\cdot \left \lfloor x \right \rfloor^{2} - \frac{1}{2}\cdot \left \lfloor x \right \rfloor + C[/latex]

[latex]S(x) =\left \lfloor x \right \rfloor\cdot \left ( x - \frac{1}{2}\cdot\left \lfloor x \right \rfloor - \frac{1}{2} \right ) + C[/latex]