Determine convergence of integral

Question

I have the general question: What is to check if an (improper) integral is convergent, semi-convergent or divergent?

That is: How can I see this?

For example, if I've the following integral:

$$ \int\limits_{0}^{\infty} \frac{1 + \arctan(\sin(x))}{x^3+2}\,\text{dx}$$

Thank you very much for your help.

Answer 1

Your integral is convergent if and only if $$\int_{0}^t \frac{1+\mathrm{arctan}(\sin(x))}{x^3 +2} \mathrm{dx}$$ admits a limit when $t \rightarrow +\infty$.

You have $$\int_{0}^t \left| \frac{1+\mathrm{arctan}(\sin(x))}{x^3 +2} \right| \mathrm{dx} \leq \left( 1+\frac{\pi}{2}\right)\int_{0}^t \frac{1}{x^3 +2} \mathrm{dx} $$

which is a convergent integral. So your integral is absolutely convergent, and therefore convergent.

Answer 2

The above answer is almost 90 percent. The remaining 10 percent is this answer . The max value of $\arctan(x)$ is $\frac{\pi}{2}$ as the range of arctan is from $[-\frac{\pi}{2},\frac{\pi}{2}]$. Thus the upper bound of I is $\int_0^{\infty} \frac{1+\frac{\pi}{2}}{x^3+2}dx$ which is convergent (quite easy to see.) The general tactics include 1)seeing whether a part of function like here arctan (most of the trigonometric and inverse would have such max value) has some maximum finite value. 2) checking whether the function is odd so that integral is $0$. 3) checking the boundedness of a function like $e^{-x}\leq 1,x>0$.4) comparing the integral with known convergent integrals like Riemann Zeta, Gamma functions,Beta functions. The list may go on as each integral may require different technique.