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Calculate this:

$$\int_{0}^{\pi }\frac{\sin2017x}{\sin x}\mathrm dx$$ I have calculated this in the answers, but I'm not sure whether it's right or wrong. I'll be glad if you check it. Thank you guys!

Peter338
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  • Could you please comment why did you downvote? I hope there are some valid reasons for this. – Peter338 Apr 01 '18 at 08:21
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    No need to have valid reasons to downvote: anyone can do practically whatever he wants here. I don't like it, but that's the way it is. I've no idea who did downvote this, but perhaps not writing down your try in the body of the question instead of an "answer" did annoy someone...again, I don't know why. – DonAntonio Apr 01 '18 at 08:23
  • I have tried to use addition formula for $\sin$, $\frac{\sin 2017x}{\sin x}=\frac{(\sin 2016x)(\cos x)}{\sin x}+\cos 2016x$. The later part is easy but I tried to use substitution to the former part but I fail. – Tony Ma Apr 01 '18 at 08:25
  • numerical approximation: 5.23513535678 by Desmos – Tony Ma Apr 01 '18 at 08:29
  • https://math.stackexchange.com/questions/478121/can-we-simplify-int-0-pi-left-frac-sin-nx-sin-x-rightmdx – Tony Ma Apr 01 '18 at 08:43
  • Replace $2017$ by any odd integer, the value remains the same. Use reccurence. – FDP Apr 01 '18 at 08:55
  • https://math.stackexchange.com/questions/263705/compute-int-0-pi-2-frac-sin-2013x-sin-x-dx-space – lab bhattacharjee Apr 01 '18 at 10:17

1 Answers1

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So at first I used $\sin{x} = \frac{e^{ix}-e^{-ix}}{2i}$,

then $\int^{\pi}_{0} \frac{\sin{2017x}}{\sin{x}}dx=\int^{\pi}_{0} \frac{e^{2017ix}-e^{-2017ix}}{e^{ix}-e^{-ix}}dx=\int^{\pi}_{0} \sum_{k=0}^{2016}e^{(2016-2k)ix}dx=$

$=\int_{0}^{\pi}dx+\int^{\pi}_{0} \sum_{k=0, k≠1008}^{2016}e^{(2016-2k)ix}dx=$

$=\pi+\sum_{k=0,k≠1008}^{2016}\frac{e^{(1008-k)i2\pi}}{(2016-2k)i}-\sum_{k=0,k≠1008}^{2016}\frac{e^{(1008-k)i2*0}}{(2016-2k)i}=$

$=\pi + \sum_{k=0,k≠1008}^{2016}\frac{1}{(2016-2k)i}-\sum_{k=0,k≠1008}^{2016}\frac{1}{(2016-2k)i}=\pi$

Peter338
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