![\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](latexrender/pictures/150cecc0186fe38918ddc92e15364886.png "\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")
La funzione integranda è definita e positiva
infatti: 
![\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}](latexrender/pictures/b5eeaea25f7eecc81d58bb9969c5b407.png "\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}")
dunque la funzione ha come punto critico, oltre
anche il punto 
, Considerando il comportamento asintotico, si ha
![\displaystyle e^{-\sqrt[3]{|x|}} \stackrel{\pm\infty}{\longrightarrow} 0,\qquad](latexrender/pictures/388fc1dc2b4d31fdef2e513d1deae649.png "\displaystyle e^{-\sqrt[3]{|x|}} \stackrel{\pm\infty}{\longrightarrow} 0,\qquad")
allora:
![\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|}](latexrender/pictures/357fc0b15810eb319ff533bff9bc4f50.png "\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|}")
![\displaystyle\le\frac{\frac{ \pi}{2} e^{-\sqrt[3]{|x|}} \ln |x | }{ \sqrt {|x|}\left|\frac{\pi}{2}\right|}=](latexrender/pictures/c633473bc3bff2d42ab31ab9ad6f9c70.png "\displaystyle\le\frac{\frac{ \pi}{2} e^{-\sqrt[3]{|x|}} \ln |x | }{ \sqrt {|x|}\left|\frac{\pi}{2}\right|}=")
![\displaystyle \frac{ \ln |x | }{e^{ \sqrt[3]{|x|}}}\cdot\frac{1}{ \sqrt {|x|} }=](latexrender/pictures/150241120e68f6a088110eb7e25ed783.png "\displaystyle \frac{ \ln |x | }{e^{ \sqrt[3]{|x|}}}\cdot\frac{1}{ \sqrt {|x|} }=")
![\displaystyle \frac{ 1}{e^{ \sqrt[3]{|x|}}\cdot \ln^{-1} |x |\cdot|x|^{1/2} }\to \mbox{converge}](latexrender/pictures/4597c52d16ec5a0d0b39c5fb014100c2.png "\displaystyle \frac{ 1}{e^{ \sqrt[3]{|x|}}\cdot \ln^{-1} |x |\cdot|x|^{1/2} }\to \mbox{converge}")
e dunque per
l'integrale dato converge.
anche in questo caso possiamo usare il confronto asintotico: osservando che:![\displaystyle e^{-\sqrt[3]{|x|}} \stackrel{0}{\longrightarrow} 1](latexrender/pictures/9d157e368139530f61450a0c36db373d.png "\displaystyle e^{-\sqrt[3]{|x|}} \stackrel{0}{\longrightarrow} 1")
, 
3 
allora:
![\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](latexrender/pictures/2b2193340bdf77392ef52faaa9379e07.png "\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")
e dunque per
l'integrale dato convergeSi conclude quindi che
![\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}](latexrender/pictures/799ba5c6d682552cbbdb29bf17c16dea.png "\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}")