Integration of a Periodic Function

Question

Problem:

$f : \mathbb{R} \to \mathbb{R}$ is a continuous and periodic function with period $T>0.$

Prove that:

$$\lim_{n\to +\infty} \int_{a}^{b} f(nx) dx =\frac{b-a}{T} \int_{0}^{T} f(x) dx$$

I tried substituting $nx=t$, but it gave me $\frac{1}{n} \int_{na}^{nb} f(t)dt$, and I don't know what to do. Can anyone give me hints to solve this? Or is there another way to solve this problem?

Answer 1

Hint: notice that, as $n$ grows larger, the interval $n b - n a$ can be split into $k$ subintervals of size $T$, with $k = \text{floor}(\frac{n b -n a}{T})$. Using periodicity, split the integral over this interval as a sum of integrals over $(0,T)$ and a remainder, then take the limit.

Answer 2

Assume $a < b$ and $f$ is bounded.

We have $ b = a + \dfrac{n(b-a)}{T} \dfrac{T}{n} = a + \left\lfloor \dfrac{n(b-a)}{T} \right \rfloor \dfrac{T}{n} + r_n\dfrac{T}{n}$ where $0 \leq r_n < 1.$

So we can write for large $n$ $$\int_{a}^{b} f(nx) dx = \sum_{k=1}^{\lfloor\frac{n(b-a)}{T}\rfloor} \int_{a+(k-1)\frac{T}{n}}^{a+k\frac{T}{n}} f(nx)dx + \int_{a+\lfloor \frac{n(b-a)}{T}\rfloor\frac{T}{n}}^{b} f(nx)dx.$$

Since $f(nx)$ is a periodic function of period $\frac{T}{n}$ we have $$\int_{a+(k-1)\frac{T}{n}}^{a+k\frac{T}{n}}f(nx)dx = \int_{0}^{\frac{T}{n}} f(nx)dx = \dfrac{1}{n} \int_{0}^{T}f(x)dx.$$

If $|f(x)| \leq M$ for all $x \in [0,T]$ and hence all $x$ note that $$\left|\int_{a+\lfloor \frac{n(b-a)}{T}\rfloor\frac{T}{n}}^{b} f(nx)dx\right| \leq M (b - (a+\lfloor \frac{n(b-a)}{T}\rfloor\frac{T}{n})$$ and since $\lfloor \frac{n(b-a)}{T}\rfloor\frac{T}{n} \to b-a$ as $n \to \infty$ this term tends to $0$.

Hence the above sum reduces to $$\left\lfloor \dfrac{ n (b - a) } {T}\right\rfloor \dfrac{1}{n} \int_{0}^{T} f(x)dx + \delta_n $$ where $\delta_n \to 0$.

Since $$\left\lfloor \dfrac{ n (b - a) } {T}\right\rfloor \dfrac{1}{n} \to \frac{b-a}{T}$$ the result follows.