Advanced Applications of Integration

Problem Statement

In high school, we often view integration merely as "finding the area under a curve." However, integration is a universal tool. It is used extensively to solve differential equations in mechanics, model multi-variable systems in vector calculus, prove mathematical inequalities, and process digital signals via convolution in electrical engineering.

Let's explore how results involving integration are used in proofs and inequalities, specifically focusing on bounding areas to prove algebraic relationships.

Consider the curve $y = \frac{1}{x}$ for $x > 0$.

(a) By considering the area under the curve $y = \frac{1}{x}$ between $x = n$ and $x = n+1$ (where $n$ is a positive integer), and comparing it to the area of an inscribed rectangle, prove the inequality: $\frac{1}{n+1} < \ln\left(1 + \frac{1}{n}\right)$

(b) Using the result from (a), or otherwise, deduce that the sequence defined by $S_k = \sum_{n=1}^{k} \frac{1}{n}$ (the harmonic series) diverges as $k \to \infty$.

Hints

• Part (a): Sketch the graph of $y = 1/x$. Draw the region between $x=n$ and $x=n+1$. Calculate the exact area under the curve using integration. Then, draw a rectangle within this interval with width $1$ and height equal to the function's value at the right endpoint ($x=n+1$). Since the curve is decreasing, the rectangle's area is strictly less than the integral's area.
• Part (b): Rearrange the inequality from (a) so you can sum both sides from $n=1$ to $k$. You will end up with $S_{k+1} - 1$ on one side and a sum of logarithms on the other. Use log laws to collapse the sum of logs into a single logarithm, and take the limit as $k \to \infty$.

Solutions

Part (a): Proving the Inequality via Integration

1. Consider the interval $[n, n+1]$. The exact area under the curve $y = \frac{1}{x}$ is: $\text{Area} = \int_n^{n+1} \frac{1}{x} \, dx$ $\text{Area} = [\ln x]_n^{n+1} = \ln(n+1) - \ln(n) = \ln\left(\frac{n+1}{n}\right) = \ln\left(1 + \frac{1}{n}\right)$
2. Now consider an inscribed rectangle on the same interval. Since $y = \frac{1}{x}$ is a decreasing function, the minimum height on the interval $[n, n+1]$ occurs at the right endpoint, $x = n+1$.
   • Height of rectangle = $\frac{1}{n+1}$
   • Width of rectangle = $(n+1) - n = 1$
3. The area of this inscribed rectangle is: $\text{Rectangle Area} = 1 \times \frac{1}{n+1} = \frac{1}{n+1}$
4. Because the curve is strictly decreasing and concave up, the area of the inscribed rectangle is strictly less than the area under the curve. $\text{Rectangle Area} < \text{Area under curve}$
5. Substituting our calculated values: $\frac{1}{n+1} < \ln\left(1 + \frac{1}{n}\right)$ This proves the required inequality.

Part (b): Divergence of the Harmonic Series

1. Take the inequality from (a) and sum both sides from $n=1$ to $k$: $\sum_{n=1}^{k} \frac{1}{n+1} < \sum_{n=1}^{k} \ln\left(\frac{n+1}{n}\right)$
2. Expand the left side: $\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{k+1}$ Notice this is $S_{k+1} - 1$.
3. Expand the right side (this is a telescoping sum): $\left( \ln\left(\frac{2}{1}\right) \right) + \left( \ln\left(\frac{3}{2}\right) \right) + \dots + \left( \ln\left(\frac{k+1}{k}\right) \right)$
4. Use the logarithm law $\ln A + \ln B = \ln(AB)$: $\ln \left( \frac{2}{1} \times \frac{3}{2} \times \frac{4}{3} \times \dots \times \frac{k+1}{k} \right)$
5. Everything cancels out diagonally, leaving only the final numerator: $\ln(k+1)$
6. Substitute the sums back into the inequality: $S_{k+1} - 1 < \ln(k+1)$ (Wait, the inequality is the wrong way around to prove divergence. Let's use the circumscribed rectangle instead!)
7. Correction: To prove divergence, we need $S_k > \text{something that diverges}$. Let's use the rectangle whose height is the left endpoint, $x=n$. This circumscribed rectangle has area $\frac{1}{n}$.
8. Area of circumscribed rectangle > Area under curve: $\frac{1}{n} > \ln\left(\frac{n+1}{n}\right)$
9. Sum both sides from $n=1$ to $k$: $\sum_{n=1}^{k} \frac{1}{n} > \sum_{n=1}^{k} \ln\left(\frac{n+1}{n}\right)$
10. The left side is exactly $S_k$. The right side is the telescoping sum $\ln(k+1)$ as shown in step 5. $S_k > \ln(k+1)$
11. As $k \to \infty$, $\ln(k+1) \to \infty$.
12. Since $S_k$ is strictly greater than a value that approaches infinity, $S_k$ must also approach infinity. Thus, the harmonic series diverges.

Takeaways

• Integration as an Estimator: Integration isn't just for exact calculations; bounding integrals with rectangles (Riemann sums) is a powerful technique for proving algebraic inequalities.
• The Harmonic Series: This is a classic paradox in mathematics. The terms get infinitely small ($\lim_{n \to \infty} \frac{1}{n} = 0$), yet their sum is infinite. The comparison to the integral of $1/x$ (which is $\ln x$) is the most elegant proof of this.
• Telescoping Sums: Recognising when a sum collapses (like the sum of logarithms) is a vital skill for HSC Extension 2.

Further Readings

• HSC Integrals: https://vumaths.com/booklets/hsc-integrals/
• HSC Proofs: https://vumaths.com/booklets/hsc-proofs/
• HSC Inequalities: https://vumaths.com/booklets/hsc-inequalities/

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If you want to see how these abstract integration concepts apply to mechanics, engineering, and beyond, dive into Vu's Maths Hub. I regularly post solutions merging proofs and calculus on my YouTube channel. Follow me on LinkedIn or subscribe to my Substack for a deeper look at the mathematics driving the real world.