$\int_1^\infty \frac{e^{\sin{x}}\sin{2x}}{x^a}dx$ convergence I have to check the following: When does integral $\int_1^\infty \frac{e^{\sin{x}}\sin{2x}}{x^a}dx$ converges?
In my opinion, this converges for $a>0$ by Dirichlet's test, otherwise diverges. Am I correct?
Also, when does integral from $\int_0^\infty $ of the same function  absolutely converges? I got that it diverges for every $a$. Am I right?
Thanks in advance.
 A: $$\int_1^\infty \frac{e^{\sin{x}}\sin{2x}}{x^a}dx=2^{a-1}\int_2^\infty \frac{e^{\sin{(t/2)}}\sin{t}}{t^a}dt$$
For investigation of the convergence of the integral, it suffices to consider
$$I(a)=\int_{2\pi}^\infty \frac{e^{\sin{(t/2)}}\sin{t}}{t^a}dt=\int_{2\pi}^{4\pi}+\int_{4\pi}^{6\pi}+ ...$$
Due to periodicity of $\sin t$ we can write
$$I(a)=\int_{0}^{2\pi}\Bigl(\frac{e^{\sin{((t+2\pi)/2)}}}{(t+2\pi)^a}+\frac{e^{\sin{((t+4\pi)/2)}}}{(t+4\pi)^a} + ...\Bigr)\sin t \,dt$$
$$=\int_{0}^{\pi}\sin t \,dt\sum_{n=1}^{\infty}\Bigl(\frac{e^{\sin{(t/2+n\pi)}}}{(t+2\pi n)^a}-\frac{e^{\sin{(t/2+n\pi+\pi/2)}}}{(t+2\pi n+\pi)^a}\Bigr)$$
$$=4\int_{0}^{\pi}\sin (t/2)\cos(t/2) \,d(t/2)\sum_{n=1}^{\infty}\Bigl(\frac{e^{\sin{(t/2+n\pi)}}}{(t+2\pi n)^a}-\frac{e^{\cos{(t/2+n\pi)}}}{(t+2\pi n+\pi)^a}\Bigr)$$
$$=4\int_{0}^{\pi/2}\sin x\cos x \,dx\sum_{n=1}^{\infty}\Bigl(\frac{e^{\sin{(x+n\pi)}}}{(2x+2\pi n)^a}-\frac{e^{\cos{(x+n\pi)}}}{(2x+2\pi n+\pi)^a}\Bigr)$$
Making change in the second term $x\to\pi/2-x$ we can write
$$I(a)=4\int_{0}^{\pi/2}\sin x\cos x \,dx\sum_{n=1}^{\infty}e^{\sin{(x+n\pi)}}\Bigl(\frac{1}{(2x+2\pi n)^a}-\frac{1}{(\pi-2x+2\pi n+\pi)^a}\Bigr)$$
$$I(a)=4\int_{0}^{\pi/2}\sin x\cos x \,dx\sum_{n=1}^{\infty}e^{(-1)^n\sin x}a_n$$
where $a_n=\frac{1}{(2x+2\pi n)^a}-\frac{1}{(\pi-2x+2\pi n+\pi)^a}$ is positive.
Let's evaluate $a_n$
On the one hand,
$$a_n=\frac{(2\pi-2x+2\pi n)^a-(2x+2\pi n)^a}{(2x+2\pi n)^a(2\pi-2x+2\pi n)^a}>\frac{\Bigl(1+\frac{2\pi-4x}{2\pi +2\pi n}\Bigr)^a-1}{(2\pi+2\pi n)^{a}}$$
But for $\alpha<1$ and $a\in(0,1) \,\, (1+\alpha)^a<1+\alpha a\,$ (the function $(1+\alpha)^a$ is convex on the interval $(0,1)$ and equal to $1+\alpha a\,$ at $a=0,1$)
$$a_n>\frac{2\pi-4x}{2\pi(n+1)}\frac{a}{(2\pi(n+1))^a}=(2\pi-4x)\frac{a}{(2\pi)^{a+1}}\frac{1}{n^{a+1}}\bigl(1-O(1/n)\bigr)$$
On the other hand in the same way we can show that
$$a_n<(2\pi-4x)\frac{a}{(2\pi)^{a+1}}\frac{1}{n^{a+1}}$$
It means that the series (and, therefore, $I(a)$) converges at any $a>0$.
For $n\to \infty$
$$I(a)\sim\sum_{n} b_n\,\text {, where}\, b_n\sim \frac{8a}{(2\pi)^{a+1}}\frac{1}{n^{a+1}}\int_{0}^{\pi/2} e^{(-1)^n\sin x}\sin x\cos x(\pi-2x)\,dx$$
A: Your solution is correct. Indeed for $f(x)=e^{\sin x}\sin2x$ we have for any $M>1$
$$
\left|\int_1^M f(x)dx\right|=\left|2\int_1^M  e^{\sin x}\sin x\, d(\sin x)\right|=
\left|\Big[2(\sin x- 1)e^{\sin x}\Big]_1^M\right|\le 4e
$$
and for $a>0$:
$$
g(x)=\frac1{x^a}>0,\quad g'(x)=-\frac a{x^{a+1}}<0.
$$
Thus $\int_1^\infty f(x)g(x) dx$ converges by Dirichlet's test.
For $a\le0$ the integral diverges since $f(x)g(x)$ does not converge to $0$ as $x\to\infty$.
What conncerns the second question
the integral $\int_0^\infty|f(x)g(x)|\,dx$ converges only if $1 <a<2$.
Indeed for convergence at $x\to+\infty$ we need $a>1$, whereas the convergence at $x\to0$ requires $a<2$.
