How do I approach showing that integral $$\int^{\infty}_0\sin^3(x^2+2x)\,\mathrm dx$$ converges absolutely/conditionally? I have calculated via software that $\int^{\infty}_0\sin^3(x^2+2x)\,\mathrm dx$ is indeed convergent, but how is it done exactly?

What I tried:

Let $U = x^2+2x, N\in \mathbb{N}, N>1$, then

$$\int^{2\pi N}_0|\sin^3(u)|\,\mathrm dx=\sum^{N-1}_{n=0}\int^{2\pi (n+1)}_{2\pi n}|\sin^3(u)|\,\mathrm du\leq \cdots$$

I am not sure what I even did is correct. I would appreciate a hint :)


1 Answer 1


By making the substitution $u=(x+1)^2$, we find that $\sqrt{u}=x+1$ and $\frac{du}{2\sqrt{u}}=dx$. Then $$\int^{\infty}_0\sin^3(x^2+2x)\,dx=\int_1^{\infty}\frac{\sin^3(u-1)}{2\sqrt{u}}\,du.$$ Now, by recalling that $4\sin^3(t)=3\sin(t)-\sin(3t)$, show that the last integral is convergent by using Dirichlet's test for convergence of improper integrals

Can you take it from here?


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