Prove that: integral from 0 to (pi/2) (sin (2k+1)

Prove that: $$ \int_0^{\frac{\pi}{2}} \frac{\sin (2k+1) \theta}{\sin \theta} \, d\theta = \frac{\pi}{2}, \quad (k=0,1,2,\ldots) $$

We will prove this by mathematical induction.

Step 1: Base case ($k = 0$)

For $k = 0$:

$$I_0 = \int_0^{\frac{\pi}{2}} \frac{\sin \theta}{\sin \theta} \, d\theta = \int_0^{\frac{\pi}{2}} 1 \, d\theta = \left[\theta\right]_0^{\pi/2} = \frac{\pi}{2}$$

The base case holds. ✓

Step 2: Establish a recurrence relation

Recurrence Relation

$$I_{k+1} - I_k = \int_0^{\frac{\pi}{2}} \frac{\sin(2k+3)\theta - \sin(2k+1)\theta}{\sin\theta} \, d\theta = 0$$

Proof:

Using the sine difference formula:

$$\sin(2k+3)\theta - \sin(2k+1)\theta = 2\cos\left(\frac{(2k+3)\theta + (2k+1)\theta}{2}\right)\sin\left(\frac{(2k+3)\theta - (2k+1)\theta}{2}\right)$$

$$= 2\cos((2k+2)\theta)\sin\theta$$

Therefore:

$$I_{k+1} - I_k = \int_0^{\frac{\pi}{2}} \frac{2\cos((2k+2)\theta)\sin\theta}{\sin\theta} \, d\theta = 2\int_0^{\frac{\pi}{2}} \cos(2(k+1)\theta) \, d\theta$$

$$= 2\left[\frac{\sin(2(k+1)\theta)}{2(k+1)}\right]_0^{\pi/2} = \frac{1}{k+1}\left[\sin((k+1)\pi)\right]_0^{\pi/2}$$

Since $\sin((k+1)\pi) = 0$ for any integer $k+1$:

$$I_{k+1} - I_k = \frac{1}{k+1}[\sin((k+1)\pi) - \sin(0)] = \frac{1}{k+1}[0 - 0] = 0$$

Therefore: $I_{k+1} = I_k$ for all $k \geq 0$.

Step 3: Proof by induction

Base case: We showed $I_0 = \frac{\pi}{2}$ ✓

Inductive hypothesis: Assume $I_k = \frac{\pi}{2}$

Inductive step: From the recurrence relation:

$$I_{k+1} = I_k = \frac{\pi}{2}$$

By mathematical induction:

$$I_k = \frac{\pi}{2} \quad \text{for all } k = 0, 1, 2, \ldots$$

Sine Integral Identity

For any non-negative integer $k$:

$$\int_0^{\frac{\pi}{2}} \frac{\sin (2k+1) \theta}{\sin \theta} \, d\theta = \frac{\pi}{2}$$

$$\boxed{\int_0^{\frac{\pi}{2}} \frac{\sin (2k+1) \theta}{\sin \theta} \, d\theta = \frac{\pi}{2}, \quad k = 0,1,2,\ldots}$$

Alternative Proof: Using Dirichlet Kernel

The integrand can be written using the Dirichlet kernel.

Recall: The Dirichlet kernel is:

$$D_n(\theta) = \frac{\sin((n+1/2)\theta)}{\sin(\theta/2)} = 1 + 2\sum_{j=1}^{n}\cos(j\theta)$$

For our case with $n = k$ and argument $2\theta$:

$$\frac{\sin((2k+1)\theta)}{\sin\theta} = D_k(2\theta) \text{ (after appropriate scaling)}$$

The integral of the Dirichlet kernel over $[0, \pi]$ is known to be $\pi$.

Alternative Proof: Direct expansion

We can also use the identity:

$$\frac{\sin(2k+1)\theta}{\sin\theta} = 1 + 2\sum_{j=1}^{k}\cos(2j\theta)$$

Proof of this identity:

Using the formula for sum of cosines:

$$\sum_{j=0}^{n} \cos(j\alpha) = \frac{\sin((n+1/2)\alpha)}{2\sin(\alpha/2)}$$

With appropriate substitution, we get:

$$\frac{\sin(2k+1)\theta}{\sin\theta} = 1 + 2\cos(2\theta) + 2\cos(4\theta) + \cdots + 2\cos(2k\theta)$$

Integrating term by term:

$$\int_0^{\pi/2} \frac{\sin(2k+1)\theta}{\sin\theta} \, d\theta = \int_0^{\pi/2} 1 \, d\theta + 2\sum_{j=1}^{k}\int_0^{\pi/2}\cos(2j\theta) \, d\theta$$

$$= \frac{\pi}{2} + 2\sum_{j=1}^{k}\left[\frac{\sin(2j\theta)}{2j}\right]_0^{\pi/2}$$

$$= \frac{\pi}{2} + 2\sum_{j=1}^{k}\frac{\sin(j\pi)}{2j} = \frac{\pi}{2} + 2\sum_{j=1}^{k}\frac{0}{2j} = \frac{\pi}{2}$$

Since $\sin(j\pi) = 0$ for all integers $j$. ✓

Verification for small values

$k = 0$:

$$\int_0^{\pi/2} \frac{\sin\theta}{\sin\theta} \, d\theta = \int_0^{\pi/2} 1 \, d\theta = \frac{\pi}{2}$$

$k = 1$:

$$\int_0^{\pi/2} \frac{\sin 3\theta}{\sin\theta} \, d\theta = \int_0^{\pi/2} (1 + 2\cos 2\theta) \, d\theta = \frac{\pi}{2} + [\sin 2\theta]_0^{\pi/2} = \frac{\pi}{2} + 0 = \frac{\pi}{2}$$

$k = 2$:

$$\int_0^{\pi/2} \frac{\sin 5\theta}{\sin\theta} \, d\theta = \int_0^{\pi/2} (1 + 2\cos 2\theta + 2\cos 4\theta) \, d\theta$$

$$= \frac{\pi}{2} + [\sin 2\theta]_0^{\pi/2} + \left[\frac{\sin 4\theta}{2}\right]_0^{\pi/2} = \frac{\pi}{2} + 0 + 0 = \frac{\pi}{2}$$

Summary

1. Base case: Direct integration shows $I_0 = \frac{\pi}{2}$
2. Recurrence: Proved $I_{k+1} = I_k$ using sine difference formula
3. Induction: Therefore $I_k = \frac{\pi}{2}$ for all $k \geq 0$
4. Alternative: Direct expansion using cosine sum formula confirms result

Key insights:

• The integral is independent of $k$ (always equals $\frac{\pi}{2}$)
• Only odd multiples $(2k+1)$ appear in the formula
• The recurrence relation provides an elegant proof
• This result is related to the Dirichlet kernel in Fourier analysis
• The key is that $\sin(n\pi) = 0$ for all integers $n$