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$$\int_0^\pi\sin(2t)e^{-in2t} \, dt$$

wolfram alpha say the answer is

$$\frac{1-e^{-2 i n π}}{2-2 n^2}$$

although using the integral trig identity

$$\int \sin(bt)e^{at}\,dt=e^{at}\dfrac{1}{a^2+b^2}\left(a \sin(bt)-b \cos(bt)\right)$$

and the integral across the period of sine and cosine is 0.

I want to express this as a Fourier Series.

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  • $\begingroup$ You can compute the integral by writing $\sin x = \frac{1}{2i} (e^{ix} - e^{-ix})$. If you're trying to express $\sin (2t)$ as a Fourier series, though, you don't need to go through any computation. $\endgroup$ – anomaly Jun 22 '14 at 22:05
  • $\begingroup$ well actually there are more terms to what I am trying to calculate, but I can solve their series separately I assume. What I am trying to do is solve for the fourier coefficient in exponential form. $\endgroup$ – RohanLatinSindhi Jun 22 '14 at 22:12
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The trig identity is wrong

$$ \int \sin(ax) \exp(bx) dx = \frac{1}{a^2+b^2} \exp(bx) \Big( b \sin(ax) - a \cos(bx) \Big) $$

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  • $\begingroup$ I'm pretty sure that your equation might be wrong. I just looked it up on wolfram. $\endgroup$ – RohanLatinSindhi Jun 22 '14 at 22:04
  • $\begingroup$ @RohanLatinSindhi It is indeed wrong. The $a\cos{(bx)}$ term at the end should be amended to $a\cos{(ax)}$. $\endgroup$ – David H Jun 22 '14 at 22:07
  • $\begingroup$ Yes indeed - the $b$ should be an $a$... But it is the $\exp(bx)$ that is missing... $\endgroup$ – johannesvalks Jun 22 '14 at 22:21
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Hint: A slightly easier trig identity to derive* is,

$$\int_{0}^{2\pi}\sin{(x)}\exp{(bx)}\,\mathrm{d}x=\frac{1-e^{2\pi b}}{1+b^2}.$$

Your integral can of course be put in this form via the substitution $x=2t$:

$$\int_{0}^{\pi}\sin{(2t)}\,e^{-2int}\,\mathrm{d}t=\frac12\int_{0}^{2\pi}\sin{(x)}\,e^{-inx}\mathrm{d}x.$$


*Derivation of trig integral identity:

$$\begin{align} \int_{0}^{2\pi}e^{bx}\sin{x}\,\mathrm{d}x&=-\frac{1}{b}\int_{0}^{2\pi}e^{bx}\cos{x}\,\mathrm{d}x\\ &=-\frac{e^{bx}\cos{(x)}}{b^2}\bigg{|}_{0}^{2\pi}-\frac{1}{b^2}\int_{0}^{2\pi}e^{bx}\sin{x}\,\mathrm{d}x\\ \left(1+\frac{1}{b^2}\right)\int_{0}^{2\pi}e^{bx}\sin{x}\,\mathrm{d}x&=-\frac{e^{2\pi b}}{b^2}-(-\frac{1}{b^2})\\ \left(\frac{b^2+1}{b^2}\right)\int_{0}^{2\pi}e^{bx}\sin{x}\,\mathrm{d}x&=\frac{1-e^{2\pi b}}{b^2}\\ \int_{0}^{2\pi}e^{bx}\sin{x}\,\mathrm{d}x&=\frac{1-e^{2\pi b}}{1+b^2} \end{align}$$

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  • $\begingroup$ thanks. I can find the others similar to this using wolfram. It would be nice to find a source though. oh well can't have your cake and eat pizza. $\endgroup$ – RohanLatinSindhi Jun 22 '14 at 22:41

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