Convergence or divergence for $\int _0 ^\infty e^{\sin x} \frac{\sin 2x}{x^\lambda } dx,\lambda>0$

Question

Consider the following integral:

$$\int _0 ^\infty e^{\sin x} \frac{\sin 2x}{x^\lambda } dx,\lambda>0$$

Here's what I did

First I studied convergence at $\infty$ $$|e^{\sin x} \frac{\sin 2x}{x^\lambda }|\le\frac{e}{x^\lambda} \Rightarrow \lim\limits_{x\to\infty}x^\alpha \frac{e}{x^\lambda}=e\text{, when }\alpha-\lambda=0$$

So, using criterion of convergence by comparing to $\frac{1}{x^\alpha}$ we get that this integral is Absolute Convergent ( so Convergent ) for $\lambda>1$

Then I studied convergence at $0$ $$\lim\limits_{x\to0}x^\alpha e^{\sin{x}}\frac{\sin2x}{x^{\lambda}}= \lim\limits_{x\to0}\frac{\sin2x}{2x}e^{\sin x}2 x^{\alpha-\lambda+1}= 2 \text{, when }\alpha= \lambda-1 $$

So, using criterion of convergence by comparing to $\frac{1}{x^\alpha}$ we get that for $\lambda-1\ge1\Rightarrow\lambda\ge2$ the integral diverges and for $\lambda \lt2$ it convergerges

Conclusion: for $\lambda\in(1,2)$ the integral is convergent.

I'm not sure if any of this is correct so I ask you to correct me or use a different way of solving this exercice. Somebody told me that it can be done with Abel criterion but I'm not very familiar with it Thank you!

Answer 1

Note that we have

$$\int_{2\pi}^\infty\frac{e^{\sin(x)}\sin(2x)}{x^\lambda}~\mathrm dx=\sum_{n=1}^\infty I_n+J_n$$

where

$$I_n=\int_0^{\pi/2}\left[\frac1{(2\pi n+x)^\lambda}-\frac1{((2n+1)\pi-x)^\lambda}\right]e^{\sin(x)}\sin(2x)~\mathrm dx$$

$$J_n=\int_0^{\pi/2}\left[\frac1{((2n+1)\pi+x)^\lambda}-\frac1{(2(n+1)\pi-x)^\lambda}\right]e^{-\sin(x)}\sin(2x)~\mathrm dx$$

is obtained by splitting the interval $[2\pi n,2\pi(n+1)]$ into four equal subintervals.

Note though that

$$0\le\frac1{(t+x)^\lambda}-\frac1{(t+\pi-x)^\lambda}\le\frac1{t^\lambda}-\frac1{(t+\pi)^\lambda}\le\frac{\pi\lambda}{t^{\lambda+1}}$$

for all $x\in[0,\pi/2]$, since it is decreasing in $x$ and attains a maximum value at $x=0$ and a minimum of $0$ at $x=\pi/2$. The second inequality is deduced from the mean value theorem.

We may then conclude that we have

$$0\le\int_{2\pi}^\infty\frac{e^{\sin(x)}\sin(2x)}{x^\lambda}~\mathrm dx\le\sum_{n=1}^\infty\frac{\pi\lambda}{(2\pi n)^{\lambda+1}}\int_0^{\pi/2}\cosh(\sin(x))\sin(2x)~\mathrm dx$$

and that it certainly converges for $\lambda>0$.