# Convergence/divergence of the improper integral $\intop_{1}^{\infty}\sin\left(x^{p}\right)dx$

I have to determine if the improper integral $$\intop_{1}^{\infty}\sin\left(x^{p}\right)dx$$

convergent/divergent for any $$0

Here's what Ive done :

We can substitute $$x^{p}=y$$ and then we'll get

$$\intop_{1}^{\infty}\sin\left(x^{p}\right)dx=\intop_{1}^{\infty}y^{\frac{1-p}{p}}\sin\left(y\right)$$

Thus, for $$p>1$$ the integral will converge by Dirichlet's theorem.

My intuition is that for $$p<1$$ the integral diverges.

I'll write it again, let $$\frac{1-p}{p}=\alpha$$

How do we prove that $$\intop_{1}^{\infty}x^{\alpha}\sin\left(x\right)dx$$ diverge for $$\alpha > 0$$ ?

I tried to show that $$x^{\alpha}\sin\left(x\right)$$ will not follow Cauchy's condition but it got complicated.

$$\int_{\pi}^{n\pi} x^{\alpha}\sin(x)dx=\sum_{k=1}^{n-1}\int_{k\pi}^{(k+1)\pi}x^{\alpha}\sin(x)dx=\sum_{k=1}^{n-1}(-1)^k\int_{0}^{\pi}(u+k\pi)^{\alpha}\sin(u)du$$ The integral $$\int_{0}^{\pi}(u+k\pi)^{\alpha}\sin(u)du$$ does not converge to $$0$$ as $$k\rightarrow +\infty$$ because, using $$\sin(x)\geqslant\frac{2}{\pi}x$$ for $$x\in[0,\pi/2]$$, $$\int_{0}^{\pi}(u+k\pi)^{\alpha}\sin(u)du\geqslant\int_0^{\pi/2}(u+k\pi)^{\alpha}\sin(u)du\geqslant\frac{2}{\pi}\int_0^{\pi/2}(u+k\pi)^{\alpha}udu\geqslant\frac{2}{\pi}\int_0^{\pi/2}u^{\alpha+1}du\geqslant\frac{ (\pi/2)^{\alpha+1}}{\alpha+2}$$ Thus the integral $$\int_1^{\infty}x^{\alpha}\sin(x)dx$$ diverges for $$\alpha>0$$.