$e^{-x^4\sin^2x}\in L_2(R)$ proving $\int_{-\infty}^{\infty} e^{-2x^4\sin^2x}dx < \infty$ I am trying to prove that $e^{-x^4\sin^2x}\in L_2(R)$. It means that  $\int_{-\infty}^{\infty} e^{-2x^4\sin^2x}dx &lt; \infty$. So, i tried to use some usual methods, but $\sin(x)$ has no limit of infinity and I can't pick up an integrable upper bound to use Weierstrass theorem. Please give me some hints.
We have
\begin{align*}
&amp; \int_{ - \infty }^{ + \infty } {e^{ - 2x^4 \sin ^2 x} dx}  = 2\int_0^{ + \infty } {e^{ - 2x^4 \sin ^2 x} dx} 
\\ &amp;
 = 2\int_0^{\pi /2} {e^{ - 2x^4 \sin ^2 x} dx}  + 2\sum\limits_{n = 1}^\infty  {\int_{\left( {n - \frac{1}{2}} \right)\pi }^{\left( {n + \frac{1}{2}} \right)\pi } {e^{ - 2x^4 \sin ^2 x} dx} } 
\\ &amp;
 = 2\int_0^{\pi /2} {e^{ - 2x^4 \sin ^2 x} dx}  + 2\sum\limits_{n = 1}^\infty  {\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {e^{ - 2(x + \pi n)^4 \sin ^2 x} dx} } 
\\ &amp;
 &lt; 2\int_0^{\pi /2} {e^{ - 2x^4 \sin ^2 x} dx}  + 2\sum\limits_{n = 1}^\infty  {\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {e^{ - 2(x + \pi n)^4 (2x/\pi )^2 } dx} } 
\\ &amp;
 &lt; 2\int_0^{\pi /2} {e^{ - 2x^4 \sin ^2 x} dx}  + 2\sum\limits_{n = 1}^\infty  {\int_{-\frac{\pi }{2}}^{ \frac{\pi }{2} } {e^{ - 8\pi ^2 \left( {n - \frac{1}{2}} \right)^4 x^2 } dx} } 
\\ &amp;
 &lt; 2\int_0^{\pi /2} {e^{ - 2x^4 \sin ^2 x} dx}  + 2\sum\limits_{n = 1}^\infty  {\int_{-\infty}^{ + \infty } {e^{ - 8\pi ^2 \left( {n - \frac{1}{2}} \right)^4 x^2 } dx} } 
\\ &amp;
 &lt; \pi  + 2\sqrt {\frac{2}{\pi }} \sum\limits_{n = 1}^\infty  {\frac{1}{{(2n - 1)^2 }}}  = \pi  + \frac{{\pi ^{3/2} }}{{2\sqrt 2 }}&lt;+\infty.
\end{align*}
The function $f(x)=2x^4\sin^2x$ is even, so $\int_{-\infty}^{\infty}\exp(-f(x))=2\int_{0}^{\infty}\exp(-f(x))$. Now you can easily bound the integral by noting that $2x^4\sin^2x &gt; 2x^2$ in $(1,\infty)$, so everything is finite.