$\int_0^\infty 3^{-x}x^4cos(2x)dx$

I succeeded to prove that this integral is conditionally convergent with Dirichlet's test. I don't know how to prove/disprove absolutely convergent..

Thanks !

  • $\begingroup$ What do you mean by " uniform convergence" of the integral ???? $\endgroup$ – Fred Aug 28 '17 at 9:08
  • $\begingroup$ You right, fixed it. Thanks ! $\endgroup$ – Jill Aug 28 '17 at 9:20

You have that $$\lim_{x\to \infty }x^23^{-x}x^4\cos(2x)=0,$$ and thus $$3^{-x}x^4\cos(2x)=\mathcal O\left(\frac{1}{x^2}\right),$$ at the neighborhood of $+\infty $. Therefore it's absolutely integrable on $[1,+\infty )$. The integrability on $[0,1]$ is obvious. The claim follow.

  • $\begingroup$ First of all, Thanks ! Can you please give a more detailed explenation ? How did you get to O(1\x^2) ? $\endgroup$ – Jill Aug 28 '17 at 9:48
  • $\begingroup$ Because $$0=\lim_{x\to \infty }x^2 3^{-x}x^4\cos(2x)=\lim_{x\to \infty }\frac{3^{-x}x^4\cos(2x)}{\frac{1}{x^2}}$$ and thus $3^{-x}x^4\cos(2x)=o(1/x^2)$ which is in particular a $\mathcal O(1/x^2)$. @Jill $\endgroup$ – Surb Aug 28 '17 at 10:07

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