integrale improrio e valori assoluti

[Noisemaker]

[tex]\mbox{Si discuta la convergenza dell^{\prime}}[/tex][tex]\mbox{integrale improprio}[/tex]

[tex]\displaystyle \int_{-\infty}^{+\infty} \left(\frac{ e^{-\sqrt[3]{|x|}}\arctan e^{x^2}\ln\left(1+|x||\sin|x||^2\right)}{ \sqrt {|x|}\left|\arctan x^{\frac{1}{3}}\right|}\right)\,\,dx[/tex]

La funzione integranda è definita e positiva [tex]\forall x\not=0,[/tex] infatti:


[tex]\displaystyle \begin{cases} 1+|x||\sin|x||^2>0&\forall x\in\mathbb{R} \\ \sqrt {|x|}\not=0 &x\not=0 \\ \arctan x^{\frac{1}{3}} \not=0 &x\not=0
\end{cases}\,\,\to\,\, x\not=0[/tex]

[tex]\displaystyle\begin{cases} e^{-\sqrt[3]{|x|}}>0&\forall x\in\mathbb{R} \\ \arctan e^{x^2}>0 &\forall x\in\mathbb{R} \\ \ln\left(1+|x||\sin|x||^2\right)>0 &\forall x\in\mathbb{R}\\ \sqrt {|x|}>0&\forall x\in\mathbb{R}\\
\left|\arctan x^{\frac{1}{3}}\right|>0&\forall x\in\mathbb{R}
\end{cases}[/tex]


dunque la funzione ha come punto critico, oltre [tex]\pm\infty,[/tex] anche il punto [tex]0[/tex], Considerando il comportamento asintotico, si ha

[tex]x\to\pm\infty:,\,\,\,\mbox{ osservando che:}[/tex]


[tex]\displaystyle e^{-\sqrt[3]{|x|}} \stackrel{\pm\infty}{\longrightarrow} 0,\qquad[/tex]

[tex]\displaystyle\arctan e^{x^2} \stackrel{\pm\infty}{\longrightarrow} \frac{\pi}{2},[/tex]

[tex]\displaystyle\ln\left(1+|x||\sin|x||^2\right) \le\ln\left(1+ |x | \right) \stackrel{\pm\infty}{\longrightarrow} \ln |x |,[/tex]

[tex]\displaystyle\sqrt {|x|} \stackrel{\pm\infty}{\longrightarrow} +\infty,[/tex]

[tex]\displaystyle \left|\arctan x^{\frac{1}{3}}\right| \stackrel{\pm\infty}{\longrightarrow} \frac{\pi}{2}[/tex]

allora:

[tex]\displaystyle\frac{ e^{-\sqrt[3]{|x|}}\arctan e^{x^2}\ln\left(1+|x||\sin|x||^2\right)}{ \sqrt {|x|}\left|\arctan x^{\frac{1}{3}}\right|}[/tex] [tex]\displaystyle\le\frac{\frac{ \pi}{2} e^{-\sqrt[3]{|x|}} \ln |x | }{ \sqrt {|x|}\left|\frac{\pi}{2}\right|}=[/tex] [tex]\displaystyle \frac{ \ln |x | }{e^{ \sqrt[3]{|x|}}}\cdot\frac{1}{ \sqrt {|x|} }=[/tex] [tex]\displaystyle \frac{ 1}{e^{ \sqrt[3]{|x|}}\cdot \ln^{-1} |x |\cdot|x|^{1/2} }\to \mbox{converge}[/tex]

e dunque per [tex]x\to\pm\infty[/tex] l'integrale dato converge.

[tex]x\to0:[/tex] anche in questo caso possiamo usare il confronto asintotico: osservando che:

[tex]\displaystyle e^{-\sqrt[3]{|x|}} \stackrel{0}{\longrightarrow} 1[/tex],

[tex]\displaystyle \arctan e^{x^2} \stackrel{0}{\longrightarrow} \frac{\pi}{4},[/tex]

[tex]\displaystyle \ln\left(1+|x||\sin|x||^2\right) \sim \ln\left(1+ |x |^3\right) \sim |x |^[/tex]3

[tex]\displaystyle\sqrt {|x|} \stackrel{0}{\longrightarrow} 0,[/tex]

[tex]\displaystyle\left|\arctan x^{\frac{1}{3}}\right| \sim \left| x^{\frac{1}{3}}\right|[/tex]

allora:

[tex]\displaystyle\frac{ e^{-\sqrt[3]{|x|}}\arctan e^{x^2}\ln\left(1+|x||\sin|x||^2\right)}{ \sqrt {|x|}\left|\arctan x^{\frac{1}{3}}\right|}\sim[/tex][tex]\displaystyle\frac{\frac{ \pi}{4} \cdot1 \cdot |x |^3 }{ \sqrt {|x|}\left| x^{\frac{1}{3}}\right|}\sim[/tex] [tex]\displaystyle\frac{ |x |^3 }{ \left| x^{\frac{1}{3}}\right|}=[/tex] [tex]\displaystyle\frac{ 1 }{ \left| x^{-\frac{8}{3}}\right|}\to \mbox {converge}[/tex]

e dunque per [tex]x\to0[/tex]l'integrale dato converge

Si conclude quindi che

[tex]\displaystyle \int_{-\infty}^{+\infty} \left(\frac{ e^{-\sqrt[3]{|x|}}\arctan e^{x^2}\ln\left(1+|x||\sin|x||^2\right)}{ \sqrt {|x|}\left|\arctan x^{\frac{1}{3}}\right|}\right)\,\,dx \to \text{converge}[/tex]