Does $\displaystyle\int_0^\infty \cos(x^3 -x) \, \mathrm dx$ converge?
What is the standard method of checking convergence of this kind of improper integrals?
[Math] Does $\int\limits_0^\infty \cos(x^3 -x) \, \mathrm dx$ converge
improper-integrals
improper-integrals
Does $\displaystyle\int_0^\infty \cos(x^3 -x) \, \mathrm dx$ converge?
What is the standard method of checking convergence of this kind of improper integrals?
Best Answer
Integration by parts is useful here, and can be tweaked to give good estimates of the integral.
We can start at $x=1$, since the function is well-behaved in the interval $[0,1]$.
Let $u=\frac{1}{3x^2-1}$ and $dv=(3x^2-1)\cos(x^3-x)\,dx$. Then we have $du=-\frac{6x}{(3x^2-1)^2}\,dx$ and we can take $v=\sin(x^3-x)$. Now when we integrate from $1$ to $M$, everything turns out nicely.
For note that $uv\to 0$ as $M\to\infty$, and the integral that is left to do converges absolutely because the integrand has absolute value $\le \frac{6x}{(3x^2-1)^2}$.