# Question on Riemann integration

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• May 30th 2011, 03:32 AM
worc3247
Question on Riemann integration
Suppose that f is Riemann integrable on $\displaystyle (-\pi,\pi )$. Prove that:
$\displaystyle \lim_{n \to \infty} \int_{-\pi}^{\pi} \! f(x)cos(nx)dx=0$.
I can't see how to start this question, could someone help?
Thanks.
• May 30th 2011, 03:55 AM
Also sprach Zarathustra
Quote:

Originally Posted by worc3247
Suppose that f is Riemann integrable on $\displaystyle (-\pi,\pi )$. Prove that:
$\displaystyle \lim_{n \to \infty} \int_{-\pi}^{\pi} \! f(x)cos(nx)dx=0$.
I can't see how to start this question, could someone help?
Thanks.

Hint:

Integration by parts.

$\displaystyle \cos(nx)=(\frac{sin(nx)}{n})'$
• May 30th 2011, 04:00 AM
girdav
Show the result when $\displaystyle f$ is a simple function (a function which only takes a finite number of values).
• May 30th 2011, 05:14 AM
Opalg
Quote:

Originally Posted by worc3247
Suppose that f is Riemann integrable on $\displaystyle (-\pi,\pi )$. Prove that:
$\displaystyle \lim_{n \to \infty} \int_{-\pi}^{\pi} \! f(x)cos(nx)dx=0$.
I can't see how to start this question, could someone help?
Thanks.

This result is known as the Riemann–Lebesgue lemma. If the function f is differentiable, then Also sprach Zarathustra's hint provides a quick and easy proof. If f is continuous then the R–L lemma is an easy consequence of Fejér's theorem. But if all you know about f is that it is Riemann integrable, then you have to follow girdav's advice and approximate f by a simple function.

The R–L lemma is easy to prove for the characteristic function of an interval (by explicitly calculating the integral). The result then follows for a simple function (which is just a finite linear combination of characteristic functions). Finally, to approximate a Riemann integrable function by a simple function, take a Riemann sum corresponding to a suitably fine dissection of the interval.