Improper integral convergence

Help me please with this one.

$\displaystyle\int_1^\infty x^2 \cos (x^4) \; dx$ converges on not?

Thanks!

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Use substitution $t=x^4$ –  Norbert Feb 3 '12 at 14:14

My compliments to Davide Giraudo for his answer to this post; which, I'm essentially copying here:

For $b>1$: \eqalign{ \int_1^b x^2\cos x^4\,dx&=\int_1^{b^4} t^{1/2} \cos t \cdot{dt\over 4 t^{3/4}} \cr &={1\over4} \int_1^{b^4} t^{-1/4} \cos t dt \cr &= -{t^{-1/4}\sin t\over4}\biggl|_1^{b^4} -{1\over16} \int_1^{b^4} {\sin t\over t^{5/4}}\,dt\cr &= -{t^{-1/4}\sin t\over4}\biggl|_1^{b^4} -{1\over16} \int_1^{b^4} {\sin t\over t^{5/4}}\,dt }

We have

$$\lim_{b^4\rightarrow\infty}-{t^{-1/4}\sin t\over4}\biggl|_1^{b^4} ={\sin 1\over 4}.$$

And $\lim\limits_{b^4\rightarrow\infty} \int_1^{b^4} {|\sin t|\over t^{5/4}}\,dt$ exists by comparision with the convergent integral $\int_1^\infty {1\over t^{5/4}}\,dt$.

It follows that $\int_1^\infty x^2\cos x^4\,dx$ converges; thus $\int_0^\infty x^2\cos x^4\,dx$ converges.

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By substitution we have $$\int_0^\infty x^2\cos(x^4)\,dx = \int_0^\infty \sqrt{x}\cos(x)(1/4 x^{-3/4})\,dx = {1\over 4} \int_0^\infty x^{-1/4}\cos(x)\,dx.$$ The last integral is improper at 0; however it integrates there so there is no problem; we have $${1\over 4}\int_0^{\pi/2} x^{-1/4}\cos(x)\,dx$$ is finite. We see that the integral $$\int_{\pi/2}^\infty x^{-1/4}\cos(x)\,dx$$ is finite by the alternating series test. The areas of the humps go decrease to zero.