Can I use Lebesgue dominated convergence? Calculate the following:
$$\lim_{n \rightarrow \infty} \int_{0}^{\infty} \exp(−nx(\sin(x))^2)\,dx$$
my idea was using $f_n(x) = \exp(−nx(\sin(x))^2) < 1$ but $1$ is not integrable under $(0,\infty)$. I also think it is possible that dominated convergence theorem cannot be used for this particular problem. Can someone give me a hint on how to solve this? Thank you.
 A: Let $f_k(x) = \exp(-kx \sin^2 x)$. We show that $f_k$ is not integrable on $[0,+\infty)$.
We have
$$
\begin{align*}
\int_{\pi/2}^{+\infty} f_k(x) \, dx &= \sum_{n \geq 1} \int_{n\pi-\pi/2}^{n\pi +\pi/2} \exp(-kx \sin^2 x) \, dx \\
&= \sum_{n \geq 1} \int_{-\pi/2}^{\pi/2} \exp[-k(x + n\pi)\sin^2 x] \, dx \\
&=\int_{-\pi/2}^{\pi/2} \left( \sum_{n \geq 1} \exp[-k(x + n\pi)\sin^2 x] \right) \, dx \\
&=\int_{-\pi/2}^{\pi/2} \exp(-kx \sin^2 x) \sum_{n \geq 1} \exp(-k\pi \sin^2 x)^n  \, dx \\
&= \int_{-\pi/2}^{\pi/2} \frac{\exp(-k(x+\pi) \sin^2 x)}{1 - \exp(-k\pi \sin^2 x)} \, dx \\
&\geq \exp(-3k\pi/2)\int_{-\pi/2}^{\pi/2} \frac{dx}{1 - \exp(-k\pi \sin^2 x)}.
\end{align*}
$$
We used the monotone convergence theorem for the third line. The integrand in the last integral is equivalent to $\frac{1}{k\pi x^2}$ as $x \to 0$. Therefore the integral is divergent.
A: Hint: For positive integer $k$, if $k\pi - 1/\sqrt{k} < x < k \pi$ we have
$x\sin(x)^2 < \pi$, so ...
A: First, break the integral into $\pi$ wide pieces.
$$
\begin{align}
&\int_0^\infty\exp(−nx\sin^2(x))\,\mathrm{d}x\\
&=\int_0^{\pi/2}\exp(−nx\sin^2(x))\,\mathrm{d}x
+\sum_{k=1}^\infty\int_{-\pi/2+k\pi}^{\pi/2+k\pi}\exp(−nx\sin^2(x))\,\mathrm{d}x\\
&=\int_0^{\pi/2}\exp(−nx\sin^2(x))\,\mathrm{d}x
+\sum_{k=1}^\infty\int_{-\pi/2}^{\pi/2}\exp(−n(x+k\pi)\sin^2(x))\,\mathrm{d}x\tag{1}
\end{align}
$$
Then estimate each piece.
$$
\begin{align}
\int_{-\pi/2}^{\pi/2}\exp(−n(x+k\pi)\sin^2(x))\,\mathrm{d}x
&\ge\int_{-\pi/2}^{\pi/2}\exp\left(−n\left(k+\tfrac12\right)\pi x^2\right)\,\mathrm{d}x\\
&=\frac1{\small\sqrt{n\left(k+\tfrac12\right)}}\int_{-{\large\frac\pi2}{\small\sqrt{n\left(k+\tfrac12\right)}}}^{{\large\frac\pi2}{\small\sqrt{n\left(k+\tfrac12\right)}}}\exp\left(−\pi x^2\right)\,\mathrm{d}x\\
&\sim\frac1{\small\sqrt{n\left(k+\tfrac12\right)}}\tag{2}
\end{align}
$$
Since
$$
\sum_{k=1}^\infty\frac1{\small\sqrt{k+\tfrac12}}\tag{3}
$$
diverges, $(1)$ and $(2)$ show that
$$
\int_0^\infty\exp(−nx\sin^2(x))\,\mathrm{d}x\tag{4}
$$
also diverges.
Therefore, since the smallest function which dominates the sequence is
$$
\exp(−x\sin^2(x))\tag{5}
$$
and its integral diverges, Dominated Convergence does not apply.
