Proof that an integral is convergent

Question

Prove that $$\int_0^∞ \frac{\sin(2x)}{x+2}\;dx$$ is a convergent integral.

Should I prove this by trying to show that it is bounded and monotone or should I use another proof? Any help please?

Answer 1

Following Surb's suggestion, your integral is $$\left[-\frac12\frac{\cos 2x}{x+2}\right]_0^\infty-\frac12\int_0^\infty\frac{\cos 2x dx}{(x+2)^2}.$$The surface term is $\frac14$, while the integral is bounded between $\pm\int_0^\infty\frac{dx}{(x+2)^2}=\pm\frac12$.

Answer 2

Let $n\in\Bbb Z_{\ge 0}$. Then $$\begin{align}\int_{n\pi}^{(n+1)\pi}\frac{\sin(2t)}{t+2}dt&=\left|\int_{n\pi}^{(n+0.5)\pi}\frac{\sin(2t)}{t+2}dt\right|-\left|\int_{(n+0.5)\pi}^{(n+1)\pi}\frac{\sin(2t)}{t+2}dt\right|\\ &\le\frac1{n\pi+2}-\frac{\left|\int_{(n+0.5)\pi}^{(n+1)\pi}\sin(2t)dt\right|}{(n+1)\pi+2}\\ &=\frac1{n\pi+2}-\frac1{(n+1)\pi+2} \end{align}$$ Now, $$\int_{0}^{\infty}\frac{\sin(2t)}{t+2}dt\le\sum_{n=0}^\infty\left(\frac1{n\pi+2}-\frac1{(n+1)\pi+2}\right)=\frac12$$

Answer 3

$$\int_0^\infty\frac{\sin(2x)}{x+2}dx\le\int_0^\infty\frac{\sin(2x)}{x}dx$$ and this integral can be solved by letting: $$I(a)=\int_0^\infty\frac{\sin(ax)}{x}dx$$ and using Feynman's trick.