1
$\begingroup$

Let some function $f \in C[a, b]$.

Find $\lim \limits_{n \to \infty} \int \limits_a^b \cos(nx) \cdot f(x) dx$.

What is the generic way solve $\lim \limits_{n \to \infty} \int \limits_a^b g(nx)\cdot f(x) dx$, where $g(nx)$ is some trigonometric function?

It seems it can be tackled using the concept of improper integral.

$\endgroup$
8
  • $\begingroup$ What is $C[a,b]$ could you explain that for me? $\endgroup$ Commented Apr 11, 2017 at 22:44
  • $\begingroup$ @marshalcraft It means that $f$ is continous on $[a, b]$. $\endgroup$ Commented Apr 11, 2017 at 22:48
  • $\begingroup$ @marshalcraft More generally $C[a, b]$ is a set of all functions that are continuous on $[a, b]$. You may also encounter $C^1[a,b]$, then if $f \in C^1[a, b] \Rightarrow\ f$ is differentiable on $[a, b]$ and its derivative is continuous. There is also a $C^2[a, b], \dotso , C^n[a, b]$. Note that $C[a, b] \supsetneq C^1[a, b] \supsetneq C^2[a, b ] \supsetneq \dotso$. $\endgroup$ Commented Apr 11, 2017 at 23:06
  • $\begingroup$ Third line: Where is $n$ in the integral? $\endgroup$
    – zhw.
    Commented Apr 11, 2017 at 23:29
  • 1
    $\begingroup$ The limit involving $\cos nx$ in the integral is $0$ via Riemann Lebesgue Lemma and it holds for all Lebesgue integrable functions. $\endgroup$
    – Paramanand Singh
    Commented Apr 12, 2017 at 4:51

2 Answers 2

2
$\begingroup$

I don't know if this helps but $$\lim_{n \to \infty} \int_{a}^{b} f(x) g(x) dx =$$ $$ \lim_{n \to \infty } \left[ f(x) \int_a^b g(x) dx-\int_a^b \left(\int g(x) dx \right) f'(x) dx\right]$$

If $g(x)$ reduces to a constant after $n$ derivatives then repeated application of integration by parts reducing it to integration of $f(x)$. Making it just a repeated integration of the trig function. If the $f(x) \in C^{\infty}$ like a polynomial for examle.

$\endgroup$
1
$\begingroup$

As Paramanand Singh noted, $$ \int_a^b \cos(nx) f(x) dx $$ is a kind of Fourier coefficient / transform, and vanishes for $n\to \infty$. The proof is pretty short with dominated convergence theorem.

We extend $f$ onto $\mathbb R$ by setting it to $0$ outside $[a,b]$ for notational convenience. Then, substituting $x$ with $x + \frac{\pi}n$ yields \begin{align*} 2\biggl| \int \cos(nx) f(x) dx \biggr| &= \biggl| \int \cos(nx) f(x) dx + \int \underbrace{\cos(nx + \pi)}_{=-\cos(nx)} f(x + \tfrac{\pi}n) dx \biggr| \\ &\le \int |\cos(nx)| |f(x) - f(x + \tfrac{\pi}n)| dx \\ &\le \int |f(x) - f(x + \tfrac{\pi}n)| dx \to 0, \end{align*} as $f$ is integrable and continuous on $\mathbb R \setminus \{a,b\}$.

Notes: Instead of using dominated convergence, you can also extend $f$ on the left with $f(a)$ and on the right with $f(b)$ and exploit that $f$ is uniformly continuous and bounded.

$\endgroup$
6
  • $\begingroup$ Sorry for late response. Why do you avoid $\lim$ here: $2\biggl| \int \cos(nx) f(x) dx \biggr| = \biggl| \int \cos(nx) f(x) dx + \int \underbrace{\cos(nx + \pi)}_{=-\cos(nx)} f(x + \tfrac{\pi}n) dx \biggr| \\$? I think, since you're are pointing out that $x \to x + \frac{\pi}n$ it has to be under the $\lim$ sign. $\endgroup$ Commented Apr 15, 2017 at 17:43
  • $\begingroup$ I'm not sure if the following is correct, but: $\lim_{n \to \infty }2 |\int cos(nx)f(x)dx| = \lim_{n \to \infty } |\int cos(nx)f(x)dx| + \lim_{n \to \infty}|\int cos(nx)f(x)dx| = \lim_{n \to \infty } |\int cos(nx)f(x)dx| + \lim_{n \to \infty}|\int cos(nx + \pi)f(x + \frac{\pi}{n})dx|$ ? $\endgroup$ Commented Apr 15, 2017 at 17:44
  • $\begingroup$ Basically my question is why do you can just write $\int cos(nx + \pi)f(x + \frac{\pi}{n})dx$ instead of $\int cos(nx)f(x)dx$ without $\lim$ sign? $\endgroup$ Commented Apr 15, 2017 at 17:47
  • $\begingroup$ @user23316192 I substitute $x$ by $x+\pi/n$. $\endgroup$
    – user251257
    Commented Apr 15, 2017 at 17:52
  • $\begingroup$ But as $x \to x + \frac{\pi}{n} \Rightarrow \int f(x) \to \int f(x + \frac{\pi}{n})$ (I omitted $\cos$ here for convenience), but not $\int f(x) = \int f(x + \frac{\pi}{n})$... $\endgroup$ Commented Apr 15, 2017 at 17:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .