# Oscillating integral converges to zero?

An integral like, say, $$\int_0^1 \cos[ nf(x)]~dx$$ with some function $f$ which is well behaved, and maybe almost everywhere non-zero, should be very small for large $n$ since the positive and negative contributions should about cancel each other whenever the oscillation frequency is high enough.

Is there a way to formalise this?

For $f(x)=x$, one could compute $$\int_0^1 \cos nx ~dx=\int_0^1 \frac{d}{dx}\left(\frac{\sin nx}{n}\right) dx=\frac{\sin n}{n}$$ but how could one argue when there is no expression for the antiderivative available? For certain cases a substitution $y=f(x)$ and some integration by parts might help, but it seems that the restrictions on $f$ imposed by that are stronger than necessary.

• What happens when $f(x) = \frac{1}{1-x}$? Apr 18, 2016 at 20:30
• @frogeyedpeas $f$ is well behaved. Apr 18, 2016 at 20:34
• @frogeyedpeas Also for this $f$ the convergence holds true.. Apr 18, 2016 at 20:38
• @Fabian I know about Riemann-Lebesuge, but how does it help? Apr 18, 2016 at 20:38
• Basically, I am asking which conditions one has to impose on $f$. Something like nonzero, continuously differentiable with non-zero derivative is sufficient, but almost certainly not necessary. Apr 18, 2016 at 20:49

If we assume that $f:(0,1)\to\mathbb{R}$ has a continuous and positive derivative, its inverse function $g(x)$ is an increasing $C^1$ function and: $$c_n = \int_{0}^{1}\cos(nf(x))\,dx = \int_{f(0)}^{f(1)} g'(y) \cos(ny)\,dx$$ converges to zero as $n\to +\infty$ by the Riemann-Lebesgue lemma (series version), since $g'(y)$, as a continuous (or just positive) derivative, is obviously an integrable function over $(f(0),f(1))$, and its integral equals one. If we know something more about the regularity of $g(x)$ ($g\in C^2,C^3$, $C^\infty$ or $C^\omega$) and the size of $f(1)-f(0)$ we may also estimate how fast $c_n$ converges to zero by using integration by parts.
I fear we cannot drop the assumptions on the monotonicity of $f(x)$, since the following nightmare holds: there are some differentiable functions on $(a,b)$ whose derivative is not Riemann-integrable.
However, the above argument should be enough for concrete applications. For instance, $\int_{0}^{1}\cos(ne^x)\,dx$ behaves like $\frac{\pi}{2n}$ (despite some odd-looking oscillations) while $\int_{0}^{1}\cos(n\log x)\,dx$ is very regular, it is exactly equal to $\frac{1}{1+n^2}$. On the other hand, $\int_{0}^{1}\cos(n\sin(\pi x))\,dx = J_0(n)$ decays just like $\frac{1}{\sqrt{n}}$.
• This is basically what I wrote in the question, isn't it? I don't understand why you think monotonicity is a natural assumption. For $f(x)=(x-1/2)^2$ the integral roughly behaves like $n^{-1/2}$. I do understand that there might be nasty choices for $f$ where integrability is unclear, which is the reason I asked about well behaved functions. Can you think of a non-zero example where the convergence does not hold? Apr 19, 2016 at 7:10