How to check if this improper integral converges or diverges ?

Question

$$ I=\displaystyle \int_{0}^{\infty} \dfrac{\mathrm{d}x}{x^{2}+\sin x} $$ I took $$f(x)=\dfrac{1}{x}$$ and $$g(x)=\dfrac{1}{x+ \dfrac{\sin x}{x}}$$ then tried to find out limit of f(x)/g(x) as x tends to infinity with this i got the the integral as convergent but when i went with $$ \dfrac{\sin x}{x} \le 1$$ and got $$ \dfrac{1}{x\left(x+ \dfrac{\sin x}{x}\right)} \ge \dfrac{1}{x(x+1)} \implies \displaystyle \int_{0}^{\infty}\dfrac{1}{x(x+1)} dx =\infty $$ two different results using different method ? please suggest something , thanks

Answer 1

You did the second example correctly, and you did the first example almost correctly as well, but messed it up at the end.

Theorem (Limit Comparison Test): Suppose thatthere are two functions, $f(x)$ and $g(x)$ such that $\lim_{x\to\infty}f(x)/g(x)=c>0$. Then $\int_a^\infty f(x)dx$ converges if and only if $\int_a^\infty g(x)dx$ does.

You correctly computed the limit and found that it is constant. That means that either both functions have convergent integrals or both have divergent integrals. $\int_0^\infty dx/x$ is a divergent integral though, so the correct conclusion to reach with method $1$ is that the integral diverges, not converges.

Answer 2

Observe that in $\;(0,1]\;$ the integrand is a positive function , and

$$\frac{\frac1{x^2+\sin x}}{\frac1x}=\frac x{x^2+\sin x}\xrightarrow[x\to0^+]{}1$$

so by the Limit comparison Theorem , the integral $\;\int\limits_0^1\frac{dx}{x^2+\sin x}\;$ diverges