How to prove $ \int^{\pi}_{-\pi} \sin(ax)\sin(bx) = 0 $

Question

Question:

I've been working on a problem involving inner product spaces with inner product given by: $$ \langle f, g\rangle = \int^{\pi}_{-\pi} f(x)g(x)dx \hspace{0.5cm} \text{for every } f,g \in \mathcal{C}[-\pi, \pi] $$ As part of the problem, I've come to a point where the following inner product $$ \int^{\pi}_{-\pi} \sin(ax)\sin(bx) = 0 \hspace{0.5cm} \text{ if } a \not = b \text{ and } a, b \in \mathbb{Z^{+}}$$ must be true for the purposes of the problem.

Approach:

First I tried to simply evaluate the integral using integration by parts: $$ -\frac{\sin(ax)\cos(bx)}{b}\bigg|^{\pi}_{-\pi} + \frac{a}{b}\int^{\pi}_{-\pi}\cos(ax)\cos(bx)dx $$ evaluating to $$ 0 + \frac{a}{b}\int^{\pi}_{-\pi}\cos(ax)\cos(bx)dx $$ I took this another step arriving at: $$ \frac{a}{b} \cdot \frac{\cos(ax)\sin(bx)}{b} + \frac{a^2}{b^2} \cdot \int^{\pi}_{-\pi} \sin(ax)\sin(bx)dx $$ $$ 0 + \frac{a^2}{b^2} \cdot \int^{\pi}_{-\pi} \sin(ax)\sin(bx)dx $$ At this point, it seems that the method I am using to evaluate the integral will lead to the integral repeating itself infinitely with a larger coefficient each time but still evaluating to zero each time. Is this enough to conclude that the integral evaluates to zero?

Answer 1

We have \begin{align} \int^{\pi}_{-\pi} \sin(ax)\sin(bx) \mathbb{d}x&=2\int^{\pi}_{0} \sin(ax)\sin(bx) \mathbb{d}x \end{align} Because the function is even. \begin{align} &=2 \left( \int^{\pi/2}_{0} \sin(ax)\sin(bx)\mathbb{d}x + \int^{\pi}_{\pi/2} \sin(ax)\sin(bx) \mathbb{d}x\right)\\ &=2 \left( \int^{\pi/2}_{0} \sin(a(\pi/2-x))\sin(b(\pi/2-x))\mathbb{d}x + \int^{\pi}_{\pi/2} \sin(ax)\sin(bx) \mathbb{d}x\right)\\ &\pi/2-x=u, \frac{du}{dx} = -1, \text{ bounds are now $\pi$ and $\pi/2$}\\ &=2 \left( -\int^{\pi}_{\pi/2} \sin(au)\sin(bu)\mathbb{d}u + \int^{\pi}_{\pi/2} \sin(ax)\sin(bx) \mathbb{d}x\right)\\ &=0 \end{align}

Answer 2

Alternatively, set $z=e^{i\theta}$ so that you get: \begin{align} \int^\pi_{-\pi} \sin(ax)\sin(bx) dx &=\int_{|z|=1} \frac{z^a-z^{-a}}{2i} \frac{z^b-z^{-b}}{2i} \frac{dz}{iz}\\ & = \frac{i}{4}\int_{|z|=1} z^{a+b-1} -z^{a-b-1} -z^{b-a-1}+z^{-a-b-1}dz \end{align} Since $b,a \in \mathbb{Z}^+$ and $a\neq b$ we have no $z$ with power $-1$. So we get: \begin{align} \int^\pi_{-\pi} \sin(ax)\sin(bx) dx = 0 \end{align} By the residue theorem.

Answer 3

$\cos x - \cos y =$

$-2\sin((x+y)/2) \sin((x -y)/2);$

$x,y \in \mathbb{R}.$

Set:

$az:= (x+y)/2$, $bz:= (x -y)/2,$ :

$(1/2)[-\cos (a+b)z +\cos(a-b)z]$

$=\sin (az) \sin (bz)$.

Note:

$a-b, a+b \in \mathbb{Z}$.

Since $ \cos$ is even, we can choose

$n=a+b, |a-b|$, i.e.

$ n \in \mathbb{Z+}$, and get:

$\int_{0}^{π} cos(nz)dz = 0.$

$\rightarrow:$

$\int_{-π}^π [(-1/2)\cos (a+b)z +$

$(1/2)\cos(a-b)z]dz =0$

$\int_{-π}^π \sin(az) \sin(bz) dz $.