I need to show that $\int_{0}^{\infty} \cos(x^4+x+1) dx$ converges. I showed that $\int_{0}^{\infty} \cos(x^4) dx$ converges but I don't know how to continue. I can't say that $\cos(x^4) \sim \cos(x^4+x+1)$ to conclude as $\cos(x^4)$ keeps changing sign. If I write $\cos(x^4+x+1) = \cos(x^4)\cos(x+1)-\sin(x^4)\sin(x+1)$ it's the same problem, as the integral of $|\cos(x^4)|$ diverges.
Any help would be greatly appreciated
I've often used the technique of introducing a factor that allows me to do integration by parts. Setting $dv=\cos(x^4+x+1)\,dx$ is intractable, but if we introduce $4x^3+1$ then $dv=(4x^3+1)\cos(x^4+x+1)\,dx$ works great: $$ \int _0^{\infty} \underbrace{\frac{1}{4x^3+1}}_{u}\cdot\underbrace{(4x^3+1)\cos(x^4+x+1)\,dx}_{dv} $$ $$=\left.\frac{\sin(x^4+x+1)}{4x^3+1}\right|_0^{\infty}+ \int _0^{\infty} \frac{12x^2}{(4x^3+1)^2}\cdot\sin(x^4+x+1)\,dx $$The boundary term evaluates to $-\sin(1)$ and the new integral is easily seen to be absolutely convergent by considering $\int_{0}^{\infty} \frac{12x^2}{(4x^3+1)^2}\,dx=\int_{1}^{\infty} \frac{1}{z^2}\,dz=1$.