- #1

CGandC

- 326

- 34

- Homework Statement
- Prove/Disprove: the integral ## J=\int_{0}^{\infty} \frac{\sin x+\arctan x}{x^{2}} d x ## converges.

- Relevant Equations
- Using integral comparison test in the sense of comparing between integrals, like here https://web.njit.edu/~bg263/Lecture%20notes%20and%20supplements/L19.pdf ( and not in the context of sums )

*My attempt:*Disprove. Note that ## \int_{0}^{\infty} \frac{-1 - \frac{\pi}{2} }{x^2} d x \leq \int_{0}^{\infty} \frac{\sin x+\arctan x}{x^{2}} d x ## and that ## \int_{0}^{\infty} \frac{-1 - \frac{\pi}{2} }{x^2} d x ## diverges, hence by the integral Direct Comparison Test, ## J ## diverges.

Was the above proof specious?

*My professor gave the following proof:*Disprove. In the interval ## (0, \frac{\pi}{2}] ## the integrand is positive and also

## \frac{\sin x+\arctan x}{x^{2}} \geq \frac{ \sin x }{ x^2} ##

And ## \lim _{x \rightarrow 0} \frac{\frac{\sin x}{x^{2}}}{\frac{1}{x}}=\lim _{x \rightarrow 0} \frac{\sin x}{x}=1 ##

Hence by the Limit Comparison Test, ## \frac{ \sin x }{ x^2} ## , ## \frac{1}{x} ## converge/diverge together.

Since ## \int_{0}^{\infty} \frac{1}{x} d x ## diverges, thus ## \frac{ \sin x }{ x^2} ## diverges, hence by the integral Direct Comparison Test, the integral in ## (0, \frac{\pi}{2}] ## over the integrand ## \frac{\sin x+\arctan x}{x^{2}} ## diverges which means ## J ## diverges.