Integrale di $$$\frac{1}{- k^{2} + r^{2}}$$$ rispetto a $$$r$$$

Trova $$$\int \frac{1}{- k^{2} + r^{2}}\, dr$$$.

Esegui la decomposizione in fratti semplici:

$${\color{red}{\int{\frac{1}{- k^{2} + r^{2}} d r}}} = {\color{red}{\int{\left(- \frac{1}{2 \left(r + \left|{k}\right|\right) \left|{k}\right|} + \frac{1}{2 \left(r - \left|{k}\right|\right) \left|{k}\right|}\right)d r}}}$$

Integra termine per termine:

$${\color{red}{\int{\left(- \frac{1}{2 \left(r + \left|{k}\right|\right) \left|{k}\right|} + \frac{1}{2 \left(r - \left|{k}\right|\right) \left|{k}\right|}\right)d r}}} = {\color{red}{\left(\int{\frac{1}{2 \left(r - \left|{k}\right|\right) \left|{k}\right|} d r} - \int{\frac{1}{2 \left(r + \left|{k}\right|\right) \left|{k}\right|} d r}\right)}}$$

Applica la regola del fattore costante $$$\int c f{\left(r \right)}\, dr = c \int f{\left(r \right)}\, dr$$$ con $$$c=\frac{1}{2 \left|{k}\right|}$$$ e $$$f{\left(r \right)} = \frac{1}{- k + r}$$$:

$$- \int{\frac{1}{2 \left(r + \left|{k}\right|\right) \left|{k}\right|} d r} + {\color{red}{\int{\frac{1}{2 \left(r - \left|{k}\right|\right) \left|{k}\right|} d r}}} = - \int{\frac{1}{2 \left(r + \left|{k}\right|\right) \left|{k}\right|} d r} + {\color{red}{\left(\frac{\int{\frac{1}{- k + r} d r}}{2 \left|{k}\right|}\right)}}$$

Sia $$$u=- k + r$$$.

Quindi $$$du=\left(- k + r\right)^{\prime }dr = 1 dr$$$ (i passaggi si possono vedere »), e si ha che $$$dr = du$$$.

Quindi,

$$- \int{\frac{1}{2 \left(r + \left|{k}\right|\right) \left|{k}\right|} d r} + \frac{{\color{red}{\int{\frac{1}{- k + r} d r}}}}{2 \left|{k}\right|} = - \int{\frac{1}{2 \left(r + \left|{k}\right|\right) \left|{k}\right|} d r} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2 \left|{k}\right|}$$

L'integrale di $$$\frac{1}{u}$$$ è $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$- \int{\frac{1}{2 \left(r + \left|{k}\right|\right) \left|{k}\right|} d r} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2 \left|{k}\right|} = - \int{\frac{1}{2 \left(r + \left|{k}\right|\right) \left|{k}\right|} d r} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2 \left|{k}\right|}$$

Ricordiamo che $$$u=- k + r$$$:

$$\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2 \left|{k}\right|} - \int{\frac{1}{2 \left(r + \left|{k}\right|\right) \left|{k}\right|} d r} = \frac{\ln{\left(\left|{{\color{red}{\left(- k + r\right)}}}\right| \right)}}{2 \left|{k}\right|} - \int{\frac{1}{2 \left(r + \left|{k}\right|\right) \left|{k}\right|} d r}$$

Applica la regola del fattore costante $$$\int c f{\left(r \right)}\, dr = c \int f{\left(r \right)}\, dr$$$ con $$$c=\frac{1}{2 \left|{k}\right|}$$$ e $$$f{\left(r \right)} = \frac{1}{k + r}$$$:

$$\frac{\ln{\left(\left|{k - r}\right| \right)}}{2 \left|{k}\right|} - {\color{red}{\int{\frac{1}{2 \left(r + \left|{k}\right|\right) \left|{k}\right|} d r}}} = \frac{\ln{\left(\left|{k - r}\right| \right)}}{2 \left|{k}\right|} - {\color{red}{\left(\frac{\int{\frac{1}{k + r} d r}}{2 \left|{k}\right|}\right)}}$$

Sia $$$u=k + r$$$.

Quindi $$$du=\left(k + r\right)^{\prime }dr = 1 dr$$$ (i passaggi si possono vedere »), e si ha che $$$dr = du$$$.

Quindi,

$$\frac{\ln{\left(\left|{k - r}\right| \right)}}{2 \left|{k}\right|} - \frac{{\color{red}{\int{\frac{1}{k + r} d r}}}}{2 \left|{k}\right|} = \frac{\ln{\left(\left|{k - r}\right| \right)}}{2 \left|{k}\right|} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2 \left|{k}\right|}$$

L'integrale di $$$\frac{1}{u}$$$ è $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$\frac{\ln{\left(\left|{k - r}\right| \right)}}{2 \left|{k}\right|} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2 \left|{k}\right|} = \frac{\ln{\left(\left|{k - r}\right| \right)}}{2 \left|{k}\right|} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2 \left|{k}\right|}$$

Ricordiamo che $$$u=k + r$$$:

$$\frac{\ln{\left(\left|{k - r}\right| \right)}}{2 \left|{k}\right|} - \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2 \left|{k}\right|} = \frac{\ln{\left(\left|{k - r}\right| \right)}}{2 \left|{k}\right|} - \frac{\ln{\left(\left|{{\color{red}{\left(k + r\right)}}}\right| \right)}}{2 \left|{k}\right|}$$

Pertanto,

$$\int{\frac{1}{- k^{2} + r^{2}} d r} = \frac{\ln{\left(\left|{k - r}\right| \right)}}{2 \left|{k}\right|} - \frac{\ln{\left(\left|{k + r}\right| \right)}}{2 \left|{k}\right|}$$

Semplifica:

$$\int{\frac{1}{- k^{2} + r^{2}} d r} = \frac{\ln{\left(\left|{k - r}\right| \right)} - \ln{\left(\left|{k + r}\right| \right)}}{2 \left|{k}\right|}$$

Aggiungi la costante di integrazione:

$$\int{\frac{1}{- k^{2} + r^{2}} d r} = \frac{\ln{\left(\left|{k - r}\right| \right)} - \ln{\left(\left|{k + r}\right| \right)}}{2 \left|{k}\right|}+C$$

$$$\int \frac{1}{- k^{2} + r^{2}}\, dr = \frac{\ln\left(\left|{k - r}\right|\right) - \ln\left(\left|{k + r}\right|\right)}{2 \left|{k}\right|} + C$$$A