Improper integral convergence Help me please with this one.
$\displaystyle\int_1^\infty x^2 \cos (x^4) \; dx$ converges on not?
Thanks!
 A: 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.
A: 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.
