why $\int_{1}^{\infty} x^{p} \cos \left(x^{2}\right) d x$ converge for $p<1$

Question

why the integral $\int_{1}^{\infty} x^{p} \cos \left(x^{2}\right) d x$ is converge if $p<1$ ?

Here is what I did: $$x^2=t$$ $$dx=2xdt=2\sqrt tdt$$ $$\int_{1}^{\infty} t^{\frac {p}{2} + \frac {1}{2}} \cos \left(t\right) d x$$

now If I substitude $p=0$ I get an diverge integral

Answer 1

I think you made a mistake.

$$x^2=t$$ $$So, \space \frac {dt}{dx}=2x \space or \space dx=\frac{1}{2x}dt=\frac{1}{2}t^{-\frac{1}{2}}dt$$ $$So, \int_{1}^{\infty} x^{p} \cos \left(x^{2}\right) d x = \frac{1}{2}\int_{1}^{\infty} t^{\frac {p}{2} - \frac {1}{2}} \cos \left(t\right)dt$$ This converges for p < 1.