Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Help me please with this one.

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

Thanks!

share|improve this question
    
Use substitution $t=x^4$ –  Norbert Feb 3 '12 at 14:14

2 Answers 2

up vote 2 down vote accepted

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.