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Need help with this problem. For $p\in \mathbb{R}$ consider the improper integral $$I_p=\int_0^1 t^p \sin t dt$$ Which of the following is statements are true regarding the convergence of $I_p$?

A. $I_p$ is convergent for $p=-1/2$
B. $I_p$ is divergent for $p=-3/2$
C. $I_p$ is convergent for $p=-4/3$
D. $I_p$ is divergent for $p=-1/2$

First of all, I could not understand why is it improper. Secondly the answer is provided as all correct. But could not get explanation for that either. Can you guide me please ? Tried to search in MSE to get any answer for that, but in vain. If anywhere it is showed how to solve, please help me to get the link. I will study my self and try to understand.

KON3
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1 Answers1

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For $-2 < p <0$, using $\sin t \leq t$, we have \begin{align*} I_p & = \int_0^1 t^p \sin t \ dt \\ & \leq \int_0^1 t^{1+ p} \ dt \\ & \leq \frac{1}{p + 2} < \infty. \end{align*} Hence, the integral converges for $-2 < p < 0$. For $p \leq -2$, first note that $\sin t \geq t/2$ for $t \in [0,1]$. Hence, \begin{align*} I_p & = \int_0^1 t^p \sin t \ dt \\ & \geq \frac{1}{2}\int_0^1 t^{(1+ p)} \ dt. \end{align*} Since the integral on RHS diverges for $p \leq -2$, $I_p$ also diverges.

sudeep5221
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  • beautiful. How did you get the clue for $p\leqslant 2, \sin t\geqslant \frac{t}{2}$ ? Can you share the idea ? – KON3 Dec 04 '20 at 15:29
  • I don't know if I would call it a "trick" exactly. It is a pretty standard approach to find a lower bound of integrand. Since behaviour near zero is what is crucial, the idea is to take a polynomial approximation close to 0 that is valid for your interval. – sudeep5221 Dec 04 '20 at 18:34