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I want to determine the convergence of the following improper integral $$ I := \int_{0}^{+\infty}\frac{\sin(x)}{x^2-x}dx. $$ What I did is to rewrite the integral as $$ \int_{0}^{+\infty}\frac{\sin(x)}{x^2-x}dx = \int_{0}^{1}\frac{\sin(x)}{x^2-x}dx + \int_{1}^{2}\frac{\sin(x)}{x^2-x}dx + \int_{2}^{+\infty}\frac{\sin(x)}{x^2-x}dx =: I_{1} + I_{2} + I_{3}. $$ It's easy to see that $I_{3}$ converges, $I_{1} = -\infty$, $I_{2} = +\infty$, therefore I can't deduce anything, since $I_{1}$ and $I_{2}$ diverge with different signs. Is there another way of doing it? Thanks in advance!

user_12345
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  • Note that $\int_{1}^{\infty}\frac{\sin(x)}{x}=\frac{\pi}{2}$ – B E I R U T Dec 16 '20 at 18:52
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    Wait, you're already done. If any of the integrals are infinite, it's game over. Why do you say you can't deduce anything? – Ninad Munshi Dec 16 '20 at 18:53
  • $I_{1}$ and $I_{2}$ should diverge with the same sign if we want to conclude that, isn't it? – user_12345 Dec 16 '20 at 19:00
  • @Smm no it doesn't matter – Ninad Munshi Dec 16 '20 at 19:05
  • Wait, you already solved your problem, and concluded convergence. You are already done. – B E I R U T Dec 16 '20 at 19:10
  • @NinadMunshi Why? – user_12345 Dec 16 '20 at 19:37
  • @NinadMunshi Why doesn’t it matter? – user_12345 Dec 16 '20 at 21:10
  • @GAUSS1860 No, in fact the integral doesn’t converge. – user_12345 Dec 16 '20 at 21:11
  • @Smm the burden of explanation is on you. Why did you believe the sign matters? – Ninad Munshi Dec 16 '20 at 21:11
  • @NinadMunshi Well, because $\infty -\infty$ is undefined. The answer is that the integral diverges, yes, but proceeding the way I did, we should get an indetermination. If I’m wrong, please show me why. – user_12345 Dec 16 '20 at 21:23
  • @smm that's if it were on the same interval. For an integral to converge each sub interval must converge independently. – Ninad Munshi Dec 16 '20 at 21:38
  • Ah sorry, I got confused, so basically if you have a function $f \colon [a,b] \longrightarrow \mathbb{R}$, and a $c \in [a,b]$ such that $f$ is not bounded at $[a,c)$ nor $(c,b]$, but $f$ is integrable at every interval $[a, c- \varepsilon)$ and $(c + \varepsilon, b]$ for every $\varepsilon > 0$, $f$ converges by definition iff it does in both intervals (that is, if the 2 improper integrals exist, when $\varepsilon \to 0^{+}$), and it diverges by definition if it does the contrary, right? – user_12345 Dec 16 '20 at 21:49
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    @Smm (1) Do not delete your question when you've received answers. (2) Do not vandalize your post before or after deletion to remove the question you asked in the first place. – amWhy Dec 16 '20 at 22:51

1 Answers1

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Let $$f(x)=\frac{\sin(x)}{x^2-x}$$

$ f $ is not locally integrable at $ (0,+\infty)$, so we cannot write $$\int_0^{+\infty} f $$

If you want

$ I_1=\int_0^{\frac 12}f $ converges because $ \lim_{x\to0^+}f(x)=-1$.

$ I_2=\int_{\frac 12}^1f $ diverges because $$f(x)\sim \frac{\sin(1)}{x-1}\; \;(x\to 1) $$

$ I_3=\int_2^{+\infty} f$ is absolutely convergent since, $$|f(x)|\le \frac{1}{x^{\frac 32}}$$

hamam_Abdallah
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