# Check convergence of an improper Integral

Check for convergence and absolut convergence.

$$\int_{0}^{\infty}{\sin^{n}\left(x\right)\over x}\,\mathrm{d}x$$

I know how to do it if $n = 1$ but i dont understand the integration by parts of $\sin^{n}\left(x\right)$, same goes for absolut convergence i cant find something for the direct comparison test of $\sin^{n}\left(x\right)$, I used the harmonic sequence for $n = 1$.

Thanks for Tips and Help

For $n=2m+1$, you can use that $$(\sin(x))^{2m+1}=\sin(x) (1-\cos(x)^2)^m=\sum a_j\sin(x)(\cos(x))^j$$ (the $a_j$ are some constants) and then show that the integrals $\int_0^{+\infty}\frac{\sin(x)\cos(x)^j}{x} dx$ are convergent (use integration by parts, a primitive of $\sin(x)\cos(x)^j$ is $\frac{1}{j+1}(1-\cos(x)^{j+1})$), this show that for $n$ odd, the integral is convergent. For $n$ even, as we have $|\sin(x)|^{n}=(\sin(x))^n$, the integral is divergent, as they are all not absolutely convergent: use that $$\int_{k\pi}^{(k+1)\pi}\frac{|\sin(x)|^n}{x}dx=\int_0^{\pi}\frac{|\sin(u)|^n}{u+k\pi} du \geq \int _0^{\pi}\frac{|\sin(u)|^n}{\pi+k\pi} du=\frac{c}{k+1}$$
• For $n=2m$, you have $(\sin(x))^{2m}=|\sin(x)|^{2m}$. So absolut convergence and convergence are the same. – Kelenner Jun 4 '17 at 9:22