I guess.. fourier transform

[matematicar73]

Hi everybody,

can somebody help me with this question. I know that the Riemann-Lebesgue lemma says that INT_a^b f(t)sin(Rt)dt tends to zero as R tends to infinity for f piecewise in [a,b].

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Show that

. . . . .\(\displaystyle \dfrac{\sin\,(2m\, +\, 1)\,\theta}{\sin\,\theta}\, =\, 1\, +\, 2\, \cos\, 2\theta\, +\, 2\, \cos\, 4\theta\, +\, ...\, +\, 2\, \cos\, 2m\theta\)

for positive integer m, and hence that

. . . . .\(\displaystyle \displaystyle \int_0^{\pi / 2}\, \)\(\displaystyle \dfrac{\sin\, (2m\, +\, 1)\, \theta}{\sin\, \theta}\, d\theta\, =\, \dfrac{\pi}{2}\)

Why is this result not in contradiction to the Riemann-Lebesgue lemma (Theorem 8.1.1 of Dettman, p. 350)?

Apply the Riemann-Lebesgue lemma to

. . . . .\(\displaystyle \displaystyle \int_{\pi /4}^{\pi /2}\, \)\(\displaystyle \begin{array}{c}\sin\, (2m\, +\, 1)\, \theta \\ \sin\, \theta \end{array}\, d \theta\)

and hence obtain a series expansion for \(\displaystyle \pi /4.\)

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I really don't know what this exercise is asking me to do. So any clue of starting it would be a great help.
Thanks

 

[Rasalhague]

I think you may be overthinking this a bit...

For the first part, try representing the expression as a function of m, say, g(m). Then determine an expression for g(m) - g(m-1) and simplify using trigonometric identities. Armed with that and a value for g(0), you can inductively obtain the expansion. The integral will follow directly from that.

hth