Find the values of a>0 for which the improper integral $\int_{0}^{\infty}\frac{\sin x}{x^{a}} $ converges . Do I have to expand integrand using series expansion??
Find the values of a>0 for which the improper integral $\int_{0}^{\infty}\frac{\sin x}{x^{a}} $ converges .
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improper-integrals
2 Answers
2
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.