Prove that $\int_x^{+\infty} \frac{e^{\sin^2(t)}\sin(t)}{t}dt = O(x^{-1})$ as $x \to +\infty$ Prove that $\int_x^{+\infty} \frac{e^{\sin^2(t)}\sin(t)}{t}dt = O(x^{-1})$ as $x \to +\infty$
First of all seems like I need to prove that integral tends to something using this:
$$\int_x^{+\infty} \frac{e^{\sin^2(t)}\sin(t)}{t}dt = \int_1^{+\infty}f(t)g(t)dt$$
For some $f(t)$ and $g(t)$.
And use one of these tests: Abel's and Dirichlet's
But what can give me an asymptotic estimate for the integral?
 A: Let $N = \left\lceil {x/\pi } \right\rceil$. Then
\begin{align*}
\int_x^{ + \infty } {\frac{{e^{\sin ^2 t} \sin t}}{t}dt} & = \int_x^{\pi N} {\frac{{e^{\sin ^2 t} \sin t}}{t}dt}  + \sum\limits_{k = N}^\infty  {\int_{\pi k}^{\pi (k + 1)} {\frac{{e^{\sin ^2 t} \sin t}}{t}dt} } 
\\ &
 = \int_x^{\pi N} {\frac{{e^{\sin ^2 t} \sin t}}{t}dt}  + \sum\limits_{k = N}^\infty  {( - 1)^k \int_0^1 {\frac{{e^{\sin ^2 (\pi s)} \sin (\pi s)}}{{k + s}}ds} } .
\end{align*}
First, note that
$$
\left|\int_x^{\pi N} {\frac{{e^{\sin ^2 t} \sin t}}{t}dt} \right| \le \frac{1}{x}\int_x^{\pi N} {e^{\sin ^2 t} |\sin t|dt} \le \frac{1}{x}\int_{\pi(N-1)}^{\pi N} {e^{\sin ^2 t} |\sin t|dt} = \mathcal{O}\!\left( {\frac{1}{x}} \right).
$$
Second, observe that
$$
a_k  = \int_0^1 {\frac{{e^{\sin ^2 (\pi s)} \sin (\pi s)}}{{k + s}}ds} 
$$
is a monotonically decreasing sequence of positive numbers. Thus, by the alternating series test,
\begin{align*}
&\left| {\sum\limits_{k = N}^\infty  {( - 1)^k \int_0^1 {\frac{{e^{\sin ^2 (\pi s)} \sin (\pi s)}}{{k + s}}ds} } } \right| \le \int_0^1 {\frac{{e^{\sin ^2 (\pi s)} \sin (\pi s)}}{{N + s}}ds} \\ & \le \frac{1}{N}\int_0^1 {e^{\sin ^2 (\pi s)} \sin (\pi s)ds} \le \frac{\pi }{x}\int_0^1 {e^{\sin ^2 (\pi s)} \sin (\pi s)ds}  = \mathcal{O}\!\left( {\frac{1}{x}} \right).
\end{align*}
