Integrale improprio : Verifica
Posted: Wednesday 5 September 2012, 12:30
Calcolare il seguente integrale improrio:
[tex]\displaystyle\int_1^{+\infty}\frac{\ln 7x}{(x+1)^2}\,\,dx[/tex]
anzitutto osserviamo che nell'intervallo [tex][1,+\infty)[/tex] la funzione integranda risulta positiva e presenta singolarità in [tex]+\infty;[/tex] considerando il comportamento asintotico della funzione integranda, si osserva che, quando [tex]x\to +\infty,[/tex]
[tex]\displaystyle \frac{\ln 7x}{(x+1)^2}\sim \frac{\ln 7x}{ x ^2}=\frac{\ln 7 }{ x ^2}+\frac{\ln x}{ x ^2} \to \mbox{converge.}[/tex]
Per calcolare il valore dell'area, calcoliamo anzitutto l'insieme delle primitive della funzione integranda:
[tex]\displaystyle\int \frac{\ln 7x}{(x+1)^2}\,\,dx=\int \frac{\ln 7 }{ (x+1) ^2}+\frac{\ln x}{ (x+1) ^2}\,\,dx[/tex] [tex]\displaystyle=\int \frac{\ln 7 }{ (x+1) ^2}\,\,dx+\int \frac{\ln x}{ (x+1) ^2}\,\,dx[/tex] [tex]\displaystyle =-\frac{\ln 7}{x+1}-\int \ln x\,\,d\Big(-\frac{1}{x+1}\Big)[/tex]
[tex]\displaystyle =-\frac{\ln 7}{x+1}+\int \ln x\,\,d\Big(\frac{1}{x+1}\Big)\stackrel{\bf(P)}{=}[/tex] [tex]\displaystyle-\frac{\ln 7}{x+1}+\Big[\frac{\ln x}{x+1}-\int \frac{1}{x+1}\,\,d\Big( \ln x\Big)\Big][/tex] [tex]\displaystyle=-\frac{\ln 7}{x+1}+\Big[\frac{\ln x}{x+1}-\int \frac{1}{x+1}\cdot\frac{1}{x}\,\,dx\Big][/tex]
[tex]\displaystyle=-\frac{\ln 7}{x+1}+\Big[\frac{\ln x}{x+1}-\int \frac{1}{x(x+1)} \,\,dx\Big][/tex] [tex]\displaystyle=-\frac{\ln 7}{x+1}+\Big[\frac{\ln x}{x+1}-\int \frac{A}{x }+\frac{B}{x+1 } \,\,dx\Big][/tex] [tex]\displaystyle=-\frac{\ln 7}{x+1}+\Big[\frac{\ln x}{x+1}-\int \frac{A}{x }\,\,dx-\int \frac{B}{x+1 } \,\,dx\Big][/tex]
[tex]\displaystyle=-\frac{\ln 7}{x+1}+\Big[\frac{\ln x}{x+1}- A\ln| x|- B\ln|x+1|\Big],[/tex] [tex]\quad\mbox{ed essendo} \quad A=1, B=-1[/tex]
[tex]\displaystyle=-\frac{\ln 7}{x+1}+\Big[\frac{\ln x}{x+1}- \ln| x|+\ln|x+1|\Big][/tex] [tex]\displaystyle=\Big[-\frac{\ln 7}{x+1}+\frac{\ln x}{x+1} + \ln\Big| \frac{x+1}{x}\Big|\Big]_1^{+\infty}[/tex]
[tex]\displaystyle= \lim_{k \to +\infty} \Big(-\frac{\ln 7}{k+1}+\frac{\ln k}{k+1} + \ln \frac{k+1}{k}\Big)[/tex] [tex]\displaystyle-\Big(-\frac{\ln 7}{1+1}+\frac{\ln 1}{1+1} + \ln\Big| \frac{1+1}{1}\Big|\Big)=0-\Big(-\frac{\ln 7}{2}+\ln 2\Big)=[/tex] [tex]\displaystyle\frac{\ln 7}{2}-\ln 2[/tex]
[tex]\displaystyle\int_1^{+\infty}\frac{\ln 7x}{(x+1)^2}\,\,dx[/tex]
anzitutto osserviamo che nell'intervallo [tex][1,+\infty)[/tex] la funzione integranda risulta positiva e presenta singolarità in [tex]+\infty;[/tex] considerando il comportamento asintotico della funzione integranda, si osserva che, quando [tex]x\to +\infty,[/tex]
[tex]\displaystyle \frac{\ln 7x}{(x+1)^2}\sim \frac{\ln 7x}{ x ^2}=\frac{\ln 7 }{ x ^2}+\frac{\ln x}{ x ^2} \to \mbox{converge.}[/tex]
Per calcolare il valore dell'area, calcoliamo anzitutto l'insieme delle primitive della funzione integranda:
[tex]\displaystyle\int \frac{\ln 7x}{(x+1)^2}\,\,dx=\int \frac{\ln 7 }{ (x+1) ^2}+\frac{\ln x}{ (x+1) ^2}\,\,dx[/tex] [tex]\displaystyle=\int \frac{\ln 7 }{ (x+1) ^2}\,\,dx+\int \frac{\ln x}{ (x+1) ^2}\,\,dx[/tex] [tex]\displaystyle =-\frac{\ln 7}{x+1}-\int \ln x\,\,d\Big(-\frac{1}{x+1}\Big)[/tex]
[tex]\displaystyle =-\frac{\ln 7}{x+1}+\int \ln x\,\,d\Big(\frac{1}{x+1}\Big)\stackrel{\bf(P)}{=}[/tex] [tex]\displaystyle-\frac{\ln 7}{x+1}+\Big[\frac{\ln x}{x+1}-\int \frac{1}{x+1}\,\,d\Big( \ln x\Big)\Big][/tex] [tex]\displaystyle=-\frac{\ln 7}{x+1}+\Big[\frac{\ln x}{x+1}-\int \frac{1}{x+1}\cdot\frac{1}{x}\,\,dx\Big][/tex]
[tex]\displaystyle=-\frac{\ln 7}{x+1}+\Big[\frac{\ln x}{x+1}-\int \frac{1}{x(x+1)} \,\,dx\Big][/tex] [tex]\displaystyle=-\frac{\ln 7}{x+1}+\Big[\frac{\ln x}{x+1}-\int \frac{A}{x }+\frac{B}{x+1 } \,\,dx\Big][/tex] [tex]\displaystyle=-\frac{\ln 7}{x+1}+\Big[\frac{\ln x}{x+1}-\int \frac{A}{x }\,\,dx-\int \frac{B}{x+1 } \,\,dx\Big][/tex]
[tex]\displaystyle=-\frac{\ln 7}{x+1}+\Big[\frac{\ln x}{x+1}- A\ln| x|- B\ln|x+1|\Big],[/tex] [tex]\quad\mbox{ed essendo} \quad A=1, B=-1[/tex]
[tex]\displaystyle=-\frac{\ln 7}{x+1}+\Big[\frac{\ln x}{x+1}- \ln| x|+\ln|x+1|\Big][/tex] [tex]\displaystyle=\Big[-\frac{\ln 7}{x+1}+\frac{\ln x}{x+1} + \ln\Big| \frac{x+1}{x}\Big|\Big]_1^{+\infty}[/tex]
[tex]\displaystyle= \lim_{k \to +\infty} \Big(-\frac{\ln 7}{k+1}+\frac{\ln k}{k+1} + \ln \frac{k+1}{k}\Big)[/tex] [tex]\displaystyle-\Big(-\frac{\ln 7}{1+1}+\frac{\ln 1}{1+1} + \ln\Big| \frac{1+1}{1}\Big|\Big)=0-\Big(-\frac{\ln 7}{2}+\ln 2\Big)=[/tex] [tex]\displaystyle\frac{\ln 7}{2}-\ln 2[/tex]