# How to check if this improper integral converges or diverges ?

$$I=\displaystyle \int_{0}^{\infty} \dfrac{\mathrm{d}x}{x^{2}+\sin x}$$ 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

• This integral (clearly) diverges at 0. – Did Apr 5 '17 at 11:41
• yeah that's pretty evident but how i got limit as 1 when applied comparison test on it ? @Did – zeno-san Apr 5 '17 at 11:48
• @Lelouch.D.Light Can you fill in the steps before "and got that the integral converges"? My guess is that you forgot that $|m\int_0^\infty dx/x$ diverges. – Stella Biderman Apr 5 '17 at 12:07
• "how i got limit as 1 when applied comparison test on it " Before I can answer this, you will have to be much clearer... – Did Apr 5 '17 at 12:14
• yeah i got that by doing $$\lim_{x \rightarrow \infty}\dfrac{f(x)}{g(x)}=\lim_{x \rightarrow \infty}\dfrac{\dfrac{1}{x}}{\dfrac{1}{x+(\sin x/x)}}=1$$ pls correct me if i'm wrong @Did – zeno-san Apr 5 '17 at 12:35

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.

• oh okay thanks :) i need to look at the theorem again – zeno-san Apr 5 '17 at 12:50
• @Lelouch.D.Light You also might have analyzed the wrong function. Your function $g(x)$ isn't the function in the stated integral. Your analysis is correct for $g(x)$. – Stella Biderman Apr 5 '17 at 12:52

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

• This doesn't seem to answer the question, which is about the two proofs provided in the OP and the fact that they seem to differ. – Stella Biderman Apr 5 '17 at 12:47
• @StellaBiderman I don't agree with you at all: the OP's question is about the convergence/divergence of the integral (read the title). The mistakes he made seem to be secondary. – DonAntonio Apr 5 '17 at 13:12
• Fair. ${}{}{}{}{}{}{}$ – Stella Biderman Apr 5 '17 at 13:19