Find the values of $a>0$ for which the improper integral
$$\int_{0}^{\infty}\frac{\sin x}{x^{a}}dx$$
converges.
Will the Taylor series expansion of $\sin$ be a better method than testing the integral?
Asked
Active
Viewed 126 times
-3
-
Not necessarily. – uniquesolution Mar 18 '19 at 13:50
-
Tell me another way !! – sejy Mar 18 '19 at 13:55
-
2@sejy you shouldn't use this tone against people that you want help from for free – Henry Lee Mar 18 '19 at 14:07
-
https://math.stackexchange.com/questions/793595/convergence-i-int-0-infty-frac-sin-xxsdx – StubbornAtom Mar 18 '19 at 14:29
2 Answers
1
We know that $-1\le\sin(x)\le 1$ and so we can approximate the integral with: $$\int_0^\infty\frac{\sin(x)}{x^a}dx\le\int_0^\infty\frac{1}{x^a}$$ and this integral is convergent for $a>1$ so this narrows our region down to somewhere in $0<a<1$
Henry Lee
- 12,554
-
2
-
1@ Henry Lee You should better consider absolute values in your inequalitiy: $| \int_0^\infty \frac{\sin(x)}{x^\alpha},dx | \le \int_0^\infty |\frac{\sin(x)}{x^\alpha}|,dx\le \int_0^\infty \frac{1}{x^\alpha},dx$ – Dr. Wolfgang Hintze Mar 18 '19 at 14:50
-
1For general $\alpha$ the integral defines the function $\cos \left(\frac{\pi \alpha}{2}\right) \Gamma (1-\alpha)$ by analytic continuation. – Dr. Wolfgang Hintze Mar 18 '19 at 14:53
1
For $a \ge 2$ the singularity at $0$ is non-integrable. For $1 < a < 2$ it's absolutely integrable both at $0$ and at $\infty$. For $0 < a \le 1$ the series $\sum_{n=1}^\infty \int_{n\pi}^{(n+1)\pi} \frac{\sin(x)}{x^a}\; dx$ is an alternating series.
Robert Israel
- 470,583