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For every $odd$ $n \geq 1$ the answer should be $\pi/2$

For every $even$ $n \geq 2$ the possible answers are :

$A )$ $ 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots + (-1)^{n/2+1}\frac{1}{n-1} $

$B )$ $ 3(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots + (-1)^{n/2+1}\frac{1}{n-1} )$

$C )$ $ 2(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots + (-1)^{n/2+1}\frac{1}{n-1} )$

$D )$ $ \frac{1}{2}(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots + (-1)^{n+1}\frac{1}{n-1} )$

$E )$ $ 1-\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\cdots - \frac{1}{n-1} $

Does the recurrence $ (n-1)(I_{n}-I_{n-2})=2sin(n-1)x $ help?

Bernard
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SADBOYS
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2 Answers2

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For $n$ even, say $2m$, we can write

$$\begin{align} \int_0^{\pi/2}\frac{\sin(2mx)}{\sin(x)}\,dx&=\text{Im}\int_0^{\pi/2}\frac{e^{i2mx}}{\sin(x)}\,dx\\\\ &=2\text{Re}\int_0^{\pi/2}\frac{e^{i2mx}}{e^{ix}-e^{-ix}}\,dx\tag1 \end{align}$$

Then, letting $z=e^{ix}$ and using long division reveals that

$$\begin{align} \text{Re}\left(\frac{e^{i2mx}}{e^{ix}-e^{-ix}}\right)&=\text{Re}\left(\frac{z^{2m+1}}{z^2-1}\right)\\\\ &=\text{Re}\left(\sum_{k=1}^{m}z^{2k-1}+\frac{1}{z-1/z}\right)\\\\ &=\sum_{k=1}^{m} \cos((2k-1)x)\tag 2 \end{align}$$

Then, substituting $(2)$ into $(1)$ we find that

$$\int_0^{\pi/2}\frac{\sin(2mx)}{\sin(x)}\,dx=2\sum_{k=1}^{m} \frac{(-1)^{k-1}}{2k-1}$$

If $n=2m+1$, then we see that

$$\begin{align} \int_0^{\pi/2}\frac{\sin((2m+1)x)}{\sin(x)}\,dx&=\text{Im}\int_0^{\pi/2}\frac{e^{i(2m+1)x}}{\sin(x)}\,dx\\\\ &=2\text{Re}\int_0^{\pi/2}\frac{e^{i(2m+1)x}}{e^{ix}-e^{-ix}}\,dx\tag1 \end{align}$$

Then, letting $z=e^{ix}$ and using long division reveals that

$$\begin{align} \text{Re}\left(\frac{e^{i(2m+1)x}}{e^{ix}-e^{-ix}}\right)&=\text{Re}\left(\frac{z^{2m+2}}{z^2-1}\right)\\\\ &=\text{Re}\left(\sum_{k=0}^{m}z^{2k}+\frac{1/z}{z-1/z}\right)\\\\ &=\frac12+\sum_{k=1}^{m} \cos(2kx)\tag 2 \end{align}$$

Then, substituting $(2)$ into $(1)$ we find that

$$\int_0^{\pi/2}\frac{\sin(2mx)}{\sin(x)}\,dx=\frac{\pi}{2}$$

Mark Viola
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  • Thanks much for your time/work, I'm in my last year of highschool and it's pretty hard to understand, but one day I will come back to this^^ – SADBOYS Mar 04 '18 at 12:14
  • You're welcome. My pleasure. All I did was express $\frac{\sin(nx)}{\sin)x)}$ as a finite sum of cosine functions and the integrated. What was the challenging part? – Mark Viola Mar 04 '18 at 16:20
4

first we note that $I_0=0$ and $I_1 = \frac{\pi}2$.

now, using the trig identity $$ \sin A - \sin B = 2\sin \frac{A-B}2 \cos \frac{A+B}2 $$ we obtain $$ I_{n+2} = I_n + \frac2{n+1} \sin \frac{(n+1)\pi}2 $$ if $N$ is odd, therefore, we have $I_N =I_{N-2}=\dots = I_1 = \frac{\pi}2$.

for even values of $N$, say $N=2M$, we have $$ I_{2M}= I_0+2\bigg(\frac11\sin \frac{\pi}2 + \frac13\sin \frac{3\pi}2+\dots +\frac1{2M-1}\sin \frac{(2M-1)\pi}2 \bigg) = 2\sum_{k=1}^{M} \frac{(-1)^{k-1}}{2k-1} $$

David Holden
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