Prove that if f(x) is Riemann integrable on [a, b], then the integral from a to b of f(x)cos(px) dx tends to 0 as p tends to infinity.

Step-by-step solution

1. 1

   Choose the right approximating class. A Riemann integrable $f$ is bounded and can be squeezed between step functions, so for every $\varepsilon>0$ there is a step function $\varphi$ with $\int_a^b\left|f(x)-\varphi(x)\right|dx<\varepsilon .$ Here $\varphi(x)=c_k$ on each $(x_{k-1},x_k)$ for a partition $a=x_0<x_1<\cdots<x_n=b$. Step functions are the right class because $\int\cos(px)$ over an interval can be written down exactly.

2. 2

   Settle the step-function case exactly. On each piece, $\int_{x_{k-1}}^{x_k}\cos(px)\,dx=\frac{\sin(px_k)-\sin(px_{k-1})}{p},$ so $\int_a^b\varphi(x)\cos(px)\,dx=\frac1p\sum_{k=1}^{n}c_k\left[\sin(px_k)-\sin(px_{k-1})\right].$

3. 3

   Bound the step case by 1/p. Each bracket has modulus at most $2$, so $\left|\int_a^b\varphi(x)\cos(px)\,dx\right|\le\frac{2}{p}\sum_{k=1}^{n}|c_k| .$ The partition is fixed once $\varepsilon$ is chosen, so the sum is a constant and the whole bound tends to $0$ as $p\to\infty$. This is where the oscillation does the work: neighbouring cancellation inside each subinterval costs a factor $1/p$.

4. 4

   Split the general integral into main term and error. Write $\int_a^b f\cos(px)\,dx=\int_a^b \varphi\cos(px)\,dx+\int_a^b (f-\varphi)\cos(px)\,dx .$ Since $|\cos(px)|\le 1$, the second piece obeys $\left|\int_a^b (f-\varphi)\cos(px)\,dx\right|\le\int_a^b|f-\varphi|\,dx<\varepsilon$ uniformly in $p$.

5. 5

   Combine the two estimates. By the previous step there is $P$ such that $\left|\int_a^b\varphi\cos(px)\,dx\right|<\varepsilon$ for all $p>P$. Then $\left|\int_a^b f\cos(px)\,dx\right|<2\varepsilon$ for all $p>P$. Since $\varepsilon>0$ was arbitrary, the limit is $0$. The same argument with $\sin(px)$ gives the sine version, and together they are the Riemann-Lebesgue lemma.