Estimate the value of the integral: I = integral from 100

Estimate the value of the integral: $$ I = \int_{100}^{200} \frac{\sin \pi x}{x} \, dx $$

Key observation: Oscillation and cancellation

The function $\sin \pi x$ has period 2 and oscillates between $-1$ and $1$.

For integer values: $\sin(\pi n) = 0$ for all $n \in \mathbb{Z}$.

Between consecutive integers, $\sin \pi x$ alternates positive and negative regions.

Method 1: Exact calculation using periodicity

Step 1: Observe the pattern

For $x \in [n, n+1]$ where $n$ is an integer:

$$\sin \pi x = \begin{cases} > 0 & \text{if } n \text{ is even} \\ < 0 & \text{if } n \text{ is odd} \end{cases}$$

More precisely, $\sin \pi x = \sin \pi(x-n)$ shifted by sign.

Step 2: Use substitution on each interval

Consider one period $[n, n+2]$:

$$\int_n^{n+2} \frac{\sin \pi x}{x} \, dx = \int_n^{n+1} \frac{\sin \pi x}{x} \, dx + \int_{n+1}^{n+2} \frac{\sin \pi x}{x} \, dx$$

On $[n, n+1]$: $\sin \pi x > 0$ (for even $n$)

On $[n+1, n+2]$: $\sin \pi x < 0$ (for odd $n+1$)

Due to oscillation, there is significant cancellation.

Step 3: Split into integer intervals

$$I = \sum_{k=100}^{199} \int_k^{k+1} \frac{\sin \pi x}{x} \, dx$$

For each interval $[k, k+1]$, let $u = x - k$, so $x = k + u$ and $du = dx$:

$$\int_k^{k+1} \frac{\sin \pi x}{x} \, dx = \int_0^1 \frac{\sin \pi(k+u)}{k+u} \, du$$

Since $\sin \pi(k+u) = \sin(\pi k + \pi u) = \sin(\pi k)\cos(\pi u) + \cos(\pi k)\sin(\pi u)$

and $\sin(\pi k) = 0$:

$$\sin \pi(k+u) = \cos(\pi k) \sin(\pi u) = (-1)^k \sin(\pi u)$$

Therefore:

$$\int_k^{k+1} \frac{\sin \pi x}{x} \, dx = (-1)^k \int_0^1 \frac{\sin(\pi u)}{k+u} \, du$$

Method 2: Recognize alternating series structure

$$I = \sum_{k=100}^{199} (-1)^k \int_0^1 \frac{\sin(\pi u)}{k+u} \, du$$

Since $\frac{1}{k+u}$ is slowly varying on $[0,1]$ for large $k$:

$$\int_0^1 \frac{\sin(\pi u)}{k+u} \, du \approx \frac{1}{k+1/2} \int_0^1 \sin(\pi u) \, du$$

And:

$$\int_0^1 \sin(\pi u) \, du = \left[-\frac{\cos(\pi u)}{\pi}\right]_0^1 = \frac{1}{\pi}[\cos 0 - \cos \pi] = \frac{1}{\pi}[1-(-1)] = \frac{2}{\pi}$$

Therefore:

$$\int_k^{k+1} \frac{\sin \pi x}{x} \, dx \approx (-1)^k \cdot \frac{2}{\pi(k+1/2)}$$

Method 3: Alternating series estimation

$$I \approx \sum_{k=100}^{199} \frac{(-1)^k \cdot 2}{\pi(k+1/2)}$$

This is an alternating series with terms decreasing in magnitude.

Since 100 is even:

$$I \approx \frac{2}{\pi} \left(\frac{1}{100.5} - \frac{1}{101.5} + \frac{1}{102.5} - \frac{1}{103.5} + \cdots\right)$$

For an alternating series $\sum_{k=m}^{n} (-1)^k a_k$ with decreasing $a_k$, the sum is approximately bounded by the first term.

The first few terms:

$$\frac{2}{\pi \cdot 100.5} \approx \frac{2}{3.14159 \cdot 100.5} \approx \frac{2}{315.73} \approx 0.00633$$

$$-\frac{2}{\pi \cdot 101.5} \approx -\frac{2}{318.87} \approx -0.00627$$

The pairs cancel significantly. With 100 terms total (50 pairs), the sum is very small.

Method 4: Cancellation estimate

Pairing consecutive terms:

$$\frac{2}{\pi}\left(\frac{1}{k+1/2} - \frac{1}{k+3/2}\right) = \frac{2}{\pi} \cdot \frac{1}{(k+1/2)(k+3/2)}$$

For $k \approx 150$ (middle of range):

$$\frac{2}{\pi \cdot 150.5 \cdot 151.5} \approx \frac{2}{\pi \cdot 22801} \approx \frac{0.00003}{\pi} \approx 10^{-5}$$

With 50 such pairs:

$$I \approx 50 \times 10^{-5} \approx 0.0005$$

But this is still an overestimate due to further cancellations.

Final estimate

Due to the alternating nature and slow decay of $\frac{1}{x}$:

$$\boxed{I \approx 0 \quad \text{(very small, less than } 0.001\text{)}}$$

More precisely:

$$\boxed{|I| < 0.01}$$

The integral is extremely small due to oscillatory cancellation.

Summary

Estimate:

$$\boxed{I \approx 0}$$

Bounds:

$$\boxed{|I| < 0.01}$$

More precisely: $|I| < 0.001$

Key insights:

• $\sin \pi x$ oscillates with period 2
• Each integer interval contributes an alternating term
• Terms have form $(-1)^k \cdot \frac{2}{\pi(k+1/2)}$
• This forms an alternating series with decreasing terms
• Massive cancellation occurs between positive and negative contributions
• The $\frac{1}{x}$ factor decreases slowly, so cancellation is nearly complete
• The integral is essentially zero (magnitude less than $0.001$)
• Exact value requires summing an alternating series of 100 terms
• Due to symmetry and cancellation, $I \approx 0$