Finding value of $I=\int^{\pi\over2}_0 \frac{\sin(nx)}{\sin{x}} \ \mathrm{d}x$

Question

Finding value of $$I=\int^{\pi\over2}_0 \frac{\sin(nx)}{\sin{x}} \ \mathrm{d}x.$$

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I got this question in my book. I got if $n$ is even then $I$ is $0$, by using king's rule it turns out to be $I=-I$.

But I couldn't solve it when $n$ is an odd number. I tried by taking $n=1,3$ and $5$ and simply expanding $\sin (3x)$ and $\sin (5x)$ in terms of $\sin x$ and integrating I get $I=\pi$. But can it be proved by in general taking $n$, not assuming any particular odd value of $n$.

Answer 1

Let $$I(n) = \int_0^{0.5\pi} \frac{\sin{(nx)}}{\sin{(x)}}dx$$ Then $$I'(n) = \int_0^{0.5\pi} \frac{\partial}{\partial n}\left(\frac{\sin{(nx)}}{\sin{(x)}}\right)dx=\int_0^{0.5\pi} \frac{1}{{\sin{(x)}}}\frac{\partial}{\partial n}\left({\sin{(nx)}}\right)dx=n\int_0^{0.5\pi} \frac{{\cos{(nx)}}dx}{{\sin{(x)}}}$$ $$ I''(n)=-n^2\int_0^{0.5\pi} \frac{{\sin{(nx)}}dx}{{\sin{(x)}}} \Rightarrow I(n) = -n^2I''(n) $$ Solving the ODE: $$ \frac{d^2I}{{dn}^2}=-I\times n^{-2} \Rightarrow I^{-1}\frac{d^2I}{{dn}^2}=-n^{-2} \Rightarrow \int \int \frac{dI }{I}dI = -\int \int \frac{dn}{n^2}dn $$ So we get: $$ \int C_1 dI + \int \ln{(I)}dI = \int \frac{dn}{n}+\int C_2 dn$$ $$ C_1I + I ln(I) - I = \ln(n)+C_2n + C_3$$ Rearranging for I: $$ {(e^{c_1-1}I)^I}=ne^{c_2n+C_3} $$ But we know that $e^{c_1-1}=c_4$ Let $J = c_4I. $ Then we get: $${J^J}=(ne^{c_2n+C_3})^{c_4}$$ So $$ J = e^{W((ne^{c_2n+C_3})^{c_4})}$$ $$ I(n) = \frac{1}{c_4} e^{W((ne^{c_2n+C_3})^{c_4})}$$ We could find the value of these coefficients by evaluating the original function at different points. This if for all $n$ both even and odd

Answer 2

Start by using the complex definition for $\sin(nx)$, $$\int_{0}^{\frac{\pi}{2}}\frac{e^{inx}-e^{-inx}}{e^{ix}-e^{-ix}}dx$$ You can then expand using geometric series, $$\int_{0}^{\frac{\pi}{2}}\frac{\left(e^{ix}-e^{-ix}\right)\left(e^{\left(n-1\right)ix}+e^{\left(n-2\right)ix}e^{-ix}+...+e^{ix}e^{-\left(n-2\right)ix}+e^{-\left(n-1\right)ix}\right)}{e^{ix}-e^{-ix}}dx$$ Tidy up using summation, $$\int_{0}^{\frac{\pi}{2}}\frac{\left(e^{ix}-e^{-ix}\right)\left(\sum_{k=0}^{n-1}e^{\left(n-1-k\right)ix}e^{-kix}\right)}{e^{ix}-e^{-ix}}dx$$ $$\int_{0}^{\frac{\pi}{2}}\sum_{k=0}^{n-1}e^{\left(n-1-2k\right)ix}dx$$ Case 1: n is odd. Split the sum, $$\int_{0}^{\frac{\pi}{2}}\left(\sum_{k=0}^{\frac{n-1}{2}}e^{\left(n-1-2k\right)ix}+\sum_{k=\frac{n+1}{2}}^{n-1}e^{\left(n-1-2k\right)ix}\right)dx$$ Re-index, $$\int_{0}^{\frac{\pi}{2}}\left(\sum_{k=0}^{\frac{n-1}{2}}e^{2kix}+\sum_{k=1}^{\frac{n-1}{2}}e^{-2kix}\right)dx$$ $$\int_{0}^{\frac{\pi}{2}}\left(\sum_{k=1}^{\frac{n-1}{2}}2\cos\left(2kx\right)+1\right)dx$$ $$\sum_{k=1}^{\frac{n-1}{2}}\frac{2\sin\left(k\pi\right)}{2k}-\sum_{k=1}^{\frac{n-1}{2}}\frac{2\sin\left(0\right)}{2k}+\frac{\pi}{2}=\frac{\pi}{2} \text{ for all n>0 odd}.$$ Case 2: n is even. Split the sum like this, $$\int_{0}^{\frac{\pi}{2}}\sum_{k=0}^{\frac{n-2}{2}}e^{\left(n-1-2k\right)ix}+\sum_{k=\frac{n}{2}}^{n-1}e^{\left(n-1-2k\right)ix}dx$$ Re-index, $$\int_{0}^{\frac{\pi}{2}}\sum_{k=0}^{\frac{n-2}{2}}e^{\left(1+2k\right)ix}+\sum_{k=0}^{\frac{n-2}{2}}e^{-\left(1+2k\right)ix}dx$$ $$\int_{0}^{\frac{\pi}{2}}\sum_{k=0}^{\frac{n-2}{2}}2\cos\left(\left(1+2k\right)x\right)dx$$ $$2\sum_{k=0}^{\frac{n-2}{2}}\frac{\sin\left(\frac{\left(1+2k\right)\pi}{2}\right)}{1+2k}$$ $$2\sum_{k=1}^{\frac{n}{2}}\frac{\left(-1\right)^{k+1}}{2k-1}$$ Not sure is there is a better solution for the even case.

Answer 3

We first deal with the odd one. Noticing $$ \begin{aligned} &2 \sin x[ \cos (2 x)+\cos (4 x)+\ldots+\cos (2 n x)] \\ = & {[\sin (3 x)-\sin (x)]+[\sin (5 x)-\sin (3 x)] } +\ldots +[\sin (2 n+1) x-\sin (2 n-1)] \\ = & \sin (2 n+1) x-\sin x , \end{aligned} $$ we have $$ \int_0^{\frac{\pi}{2}} \frac{\sin (2 n+1) x}{\sin x}dx= \int_0^{\frac{\pi}{2}}2[ \cos (2 x)+\cos (4 x)+\ldots+\cos (2 n x)]+1 d x=\frac{\pi}{2} $$ Similarly for the even, we have $$ \begin{aligned} & 2 \sin x[\cos x+\cos (3 x)+\cdots+\cos (2n -1) x] \\ =&[\sin (2 x)-\sin 0]+[\sin (4 x)-\sin 2 x]+\ldots+ [\sin (2nx)-\sin (2(n-1) x \\ =&\sin (2 n x), \end{aligned} $$ hence $$ \begin{aligned} \int_0^{\frac{\pi}{2}} \frac{\sin (2 n x)}{\sin x}dx & =2 \int_0^{\frac{\pi}{2}}\left[\cos x+\cos (3 x)+\ldots+\cos (2 n-1)x\right] \\ & =2\left[\frac{\sin x}{1}+\frac{\sin (3 x)}{3}+\cdots+\frac{\sin (2 n-1)x}{2 n-1}\right]_0^{\frac{\pi}{2}} \\ & =2\sum_{k=1}^n\frac{(-1)^{k-1}}{2 k-1} \end{aligned} $$

Answer 4

Starting from @Lai and @Archie Brew final answer, if $n=2m$ $$2\sum_{k=1}^{m}\frac{\left(-1\right)^{k+1}}{2k-1}=\frac{\pi }{2}-(-1)^m \Phi \left(-1,1,m+\frac{1}{2}\right)$$ where appears the Lerch transcendent function.

Computing the very first ones, this generates the sequence $$\left\{2,\frac{4}{3},\frac{26}{15},\frac{152}{105},\frac{526}{315} ,\frac{5156}{3465},\frac{73958}{45045},\frac{67952}{45045},\cdots\right\}$$ The denominators form sequence $A025547$ and the numerators are twice the terms of sequence $A007509$ in $OEIS$.

Notice that $$\Phi \left(-1,1,m+\frac{1}{2}\right)=\frac{1}{2} \left(\psi ^{(0)}\left(\frac{2m+3}{4}\right)-\psi ^{(0)}\left(\frac{2m+1}{4}\right)\right)$$

Expanding for large values of $m$

$$2\sum_{k=1}^{m}\frac{\left(-1\right)^{k+1}}{2k-1}=\frac{\pi }{2}-(-1)^m\sum _{n=0}^{\infty } \frac{E_{2 n} } {4^{n} \, m^{2 n+1}}$$