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I'd like to know how to prove or disprove that

$$\int\limits_0^\infty {\frac{{\sin \left( {2 \omega x} \right)}}{{\sin x}}\frac{{dx}}{{1 + {x^2}}} = \frac{\pi }{{{e^2} - 1}}\frac{{{e^{2 \omega}} - 1}}{{{e^{2\omega - 1}}}}} $$

I always try to solve this problems with differential equations but this one yields

$$\int\limits_0^\infty {\frac{{\sin \left( {2wt} \right)}}{{\sin t}}dt} = I\left( w \right) - \frac{{I''\left( w \right)}}{4}$$

...and the integral of the LHS is not defined.

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Do you implicitly assume $\omega$ to be an integer ? –  Sasha Mar 1 '12 at 2:37
    
@Sasha Maybe I should've used $n$ instead of $\omega$. Yes. –  Pedro Tamaroff Mar 1 '12 at 2:43

1 Answer 1

up vote 5 down vote accepted

Let $\omega = n \in \mathbb{N}$ $$ \frac{\sin(2 \omega x)}{\sin(x)} = U_{2n-1}\left(\cos(x)\right) $$ Then $$ \mathcal{I}(\omega) = \int_0^\infty \frac{\sin(2 \omega x)}{\sin(x)} \cdot \frac{\mathrm{d} x}{1+x^2} = \int_0^\infty U_{2n-1}\left(\cos(x)\right) \frac{\mathrm{d} x}{1+x^2} $$ By Chebyshev polynomial of the second kind for odd index can be written as a sum of polynomials of the first kind: $$ U_{2n-1}(z) = 2 \sum_{k=1}^{n} T_{2k-1}(z) $$ Thus, using $T_n(\cos x) = \cos(n x)$ and parity $$ \mathcal{I}(n) = 2 \sum_{k=1}^n \int_0^\infty T_{2k-1}(\cos x) \frac{\mathrm{d} x}{1+x^2} = \sum_{k=1}^n \int_{-\infty}^\infty \cos((2k-1) x) \frac{\mathrm{d} x}{1+x^2} =\\ \sum_{k=1}^n \pi \mathrm{e}^{-(2k-1)} = \frac{\pi}{\mathrm{e}^2-1} \frac{\mathrm{e}^{2n}-1}{\mathrm{e}^{2n-1}} $$


Added One does not really need to use Chebyshev polynomials above. Instead we can use $$ \sum_{k=1}^n 2 \sin(x) \cos((2k-1)x) = \sum_{k=1}^n \left( \sin(2 k x) - \sin(2(k-1) x) \right) \stackrel{\text{telescope}}{=} \\ \sin(2nx) - \sin(0 x) = \sin(2n x) $$

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Thanks for that. However, I'm not familiar with Chebyshev polynomials. Under what topic are they studied? And in what level of studies? They don't seem to complex, yet I have never studied them. –  Pedro Tamaroff Mar 1 '12 at 3:30
    
In that case, you can directly prove that $\sin(2 n x) = 2 \sin(x) \left( \sum_{k=1}^n \cos((2k-1)x) \right)$. Do you see how ? –  Sasha Mar 1 '12 at 5:42
    
Hm, not really. –  Pedro Tamaroff Mar 1 '12 at 5:49
    
@PeterT.off I have edited the post to add the proof. –  Sasha Mar 1 '12 at 6:01
    
Great! I knew that could be done, just didn't know what identity I had to use. –  Pedro Tamaroff Mar 1 '12 at 6:02

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