I am facing problem in $\int (\sin(x)×\sin(2x)×\sin(3x)×...×\sin(nx))dx$ for $n\geq 4$ .

Question

Q)I am facing problem in $$\int [\sin(x)×\sin(2x)×\sin(3x)×....×\sin(nx)]dx$$ for $n\geq 4$. I want a shortcut method to this Integral.

My Approach:

I can evaluate $\int [\sin(x)×\sin(2x)]dx$ and $\int[\sin(x)×\sin(2x)×\sin(3x)]dx$ very easily.

Let me show how to integrate these two integrals.

For the integral $\int[\sin(x)×\sin(2x)]dx$ :

Let $$I=\int[\sin(x)×\sin(2x)]dx$$

$$\implies I=\frac{1}{2}\int[2×\sin(x)×\sin(2x)]dx$$

$$\implies I=\frac{1}{2}[\int\cos(x)dx-\int\cos(3x)dx]$$

Now after this step it is easy to integrate.

Similarly to integrate $\int[\sin(x)×\sin(2x)×\sin(3x)]dx$ we have to write $\sin(x)×\sin(3x)$ as $\frac{1}{2}[\cos(2x)-\cos(4x)]$. Therefore finally the integral will become $$\frac{1}{4}(\int[2×\cos(2x)×\sin(2x)]dx-\int[2×\cos(4x)×\sin(2x)]dx)$$

Now after this step it is easy to integrate.

But how to integrate $\int [\sin(x)×\sin(2x)×\sin(3x)×....×\sin(nx)]dx$ for $n\geq 4$. My process will be very much lengthy for $n\geq 4$.

Please help me out with any shortcut process to this Integral.

Answer 1

Let $c \in \{0,1\}^n$ denote a binary string, $sgn(c): = (-1)^{\text{#ones in the string}}$ and $s(c) := \sum_{k=1}^{n} (-1)^{c_k}k$.

Now, observe that $$P = \prod_{m=1}^{n}{\sin{mx}} = \frac{1}{2^ni^n}\underbrace{\prod_{m=1}^n {e^{ixm}-e^{-ixm}}}_{A}$$

Let $C$ be the set of all $c$, such that $c_n = 0$. Then by choosing the first term of the factor if $c_m = 0$ or the second if $c_m=1$ we can write $P$ as the sum:

$$ A = \begin{cases} \sum_{c\in C} sgn(c) (e^{ixs(c)}+e^{-ixs(c)}) &\text{ if $n$ is even} \\ \sum_{c\in C} sgn(c) (e^{ixs(c)}-e^{-ixs(c)}) &\text{ if $n$ is odd} \end{cases} $$

This simplifies to $$ A = \begin{cases} 2 \sum_{c\in C} sgn(c) \cdot \cos(s(c)x) &\text{ if $n$ is even} \\ 2i \sum_{c\in C} sgn(c) \cdot \sin(s(c)x) &\text{ if $n$ is odd} \end{cases} $$

Substituting $A$ into $P$ gives $$ P = \begin{cases} \frac{(-1)^k}{2^{n-1}} \sum_{c\in C} sgn(c) \cdot \cos(s(c)x) &\text{ if $n = 2k$} \\ \frac{(-1)^k}{2^{n-1}} \sum_{c\in C} sgn(c) \cdot \sin(s(c)x) &\text{ if $n = 2k+1$} \end{cases} $$

Integrating this expresion, given extra care when s(c) = 0 (if $n$ is even then then the term integrates to $sgn(c) x$, if $n$ is odd the term vanishes) $$\boxed{ \int \prod_{m=1}^{n}{\sin{mx}} \cdot dx = \begin{cases} \frac{(-1)^k}{2^{n-1}} \sum_{c\in C} a(c) &\text{if $n = 2k$} \\ \frac{(-1)^{k+1}}{2^{n-1}} \sum_{c\in C} b(c) &\text{if $n = 2k+1$} \\ \end{cases}} $$ where $$ \boxed{a(c) = \begin{cases} \frac {sgn(c)}{s(c)} \cdot \sin(s(c)x) &\text{if $s(c) \neq 0$} \\ sgn(c) \cdot x &\text{if $s(c) = 0$} \\ \end{cases} \\ b(c) = \begin{cases} \frac {sgn(c)}{s(c)} \cdot \cos(s(c)x) &\text{if $s(c) \neq 0$} \\ 0 &\text{if $s(c) = 0$} \\ \end{cases}} $$

For example when $n=4$ the binary strings $c$ are \begin{array}{|c|c|} \hline c & 0000 & 0001 & 0010 & 0011 & 0100 & 0101 & 0110 & 0111 \\ \hline sgn(c) & 1 & -1 & -1 & 1 & -1 & 1 & 1 & -1 \\ \hline s(c) & 10 & 8 & 6 & 4 & 4 & 2 & 0 & -2 \\ \hline \end{array} and $k$=2 so the integral is equal to $$ \frac{1}{8} \left(1/10 \sin(10x) - 1/8 \sin(8x) - 1/6 \sin(6x) +\\ 1/4 \sin(4x) - 1/4 \sin(4x) + 1/2 \sin(2x) +\\ x + 1/2 \sin(-2x)\right) = \\\boxed{\frac{1}{8} \left(1/10 \sin(10x) - 1/8 \sin(8x) - 1/6 \sin(6x) + x\right)} $$

Answer 2

Zetko gave a very nice answer. Another way is through partitions.

Let $l(\lambda)$ denote the length of a partition $\lambda\vdash k$ of $k$. Define $$s_n(k)=\sum_{\lambda\vdash k\\\lambda\in[1,n]\\\lambda_i\neq \lambda_j}(-1)^{l(\lambda)},$$ a sum over the partitions of $k$ whose components are distinct and not exceeding $n$.

$n$ is odd: $$\frac{(-1)^{\frac{n+1}2}}{2^{n-2}}\sum_{k=0}^{\lfloor\frac{n(n+1)}4\rfloor}\frac{s_{n}(k)}{n(n+1)-4k}\\\cos\left(\frac{n(n+1)}2-2k\right)+C$$ $n$ is even: $$\frac{(-1)^{\frac n2}}{2^{n-2}}\sum_{k=0}^{\lfloor\frac{n(n+1)}4\rfloor}\frac{s_{n}(k)}{n(n+1)-4k}\\\sin\left(\frac{n(n+1)}2-2k\right)+C$$ where if $n(n+1)-4k=0,$ then that term inside the summation will be $\frac12s_n(\tfrac{n(n+1)}4)x.$

Proof follows after we write the integrand as $$\frac{(-1)^n}{2^ni^n}t^{-\frac{n(n+1)}4}\sum_{k=0}^{\frac{n(n+1)}2}s_n(k)t^k$$ where $t=e^{2ix}.$