Evaluate $\int_a^b\frac{1}{ x^2}dx$ using limit of a sum definition

definite integralssummation

From the definition of a definite integral as the limit of a sum,
evaluate $$\int_a^b\frac{1}{ x^2}dx$$

Step 1

To simplify working with $x^2$, divide the interval $[a,b]$ using variable length intervals: $$[a,b]=\bigcup_{j=1}^n\bigg[a+\frac{\sqrt {j-1}(b-a)}{\sqrt n}, a+\frac{\sqrt j(b-a)}{\sqrt n}\bigg]$$

Step 2

Now rewriting the integral to remove the lower limit ($0\leq a\leq b$) : $$\int_a^b\frac1{x^2}dx=\int_0^b\frac{1}{x^2}dx-\int_0^a\frac{1}{x^2}dx$$

Step 3

Dividing $[0,b]$ as described in Step 1:$$[0,b]=\bigcup_{j=1}^n\bigg[\frac{\sqrt {j-1}(b)}{\sqrt n}, \frac{\sqrt j(b)}{\sqrt n}\bigg]$$

Step 4

Using limit of a sum definition on $\int_0^b\frac{1}{x^2}dx$ :

(result can then be used for $\int_0^a\frac{1}{x^2}dx$)

$$\begin{align}
\int_0^b\frac1{x^2} &= \lim_{n\to\infty}\sum_{j=1}^n\bigg(\frac{\sqrt j(b)}{\sqrt n}\bigg)^{-2}\times \bigg[\frac{\sqrt j(b)}{\sqrt n}-\frac{\sqrt {j-1}(b)}{\sqrt n} \bigg] \\
&=\frac1{b}\lim_{n\to\infty}\sqrt n \times \sum_{j=1}^n\frac1{j}\big[\sqrt j – \sqrt {j-1}\big]
\end{align}$$

After this point, I can not seem to find a nice form of $\sum_{j=1}^n\frac1{j}\big[\sqrt j – \sqrt {j-1}\big]$ to work with. Any help would be appreciated.

Best Answer

We obtain for $0<a\leq b$: \begin{align*} \color{blue}{\int_a^b\frac{1}{x^2}\,dx}&=\lim_{n\to\infty}\sum_{j=1}^nf\left(a+j\frac{b-a}{n}\right)\frac{b-a}{n}\\ &=\lim_{n\to\infty}\sum_{j=1}^n\frac{1}{\left(a+j\frac{b-a}{n}\right)^2}\cdot\frac{b-a}{n}\\ &\,\,\color{blue}{=\lim_{n\to\infty}\frac{n}{b-a}\sum_{j=1}^n\frac{1}{\left(\frac{an}{b-a}+j\right)^2}}\tag{1} \end{align*}

We calcluate the limit (1) by squeezing it with lower and upper bounds which can be easily calculated using telescoping.

We consider the inequality chain \begin{align*} \frac{1}{\left(\frac{an}{b-a}+j\right)\left(\frac{an}{b-a}+j+1\right)} &\leq \frac{1}{\left(\frac{an}{b-a}+j\right)^2}\leq \frac{1}{\left(\frac{an}{b-a}+j-1\right)\left(\frac{an}{b-a}+j\right)}\\ \frac{1}{\frac{an}{b-a}+j}-\frac{1}{\frac{an}{b-a}+j+1} &\leq \frac{1}{\left(\frac{an}{b-a}+j\right)^2}\leq \frac{1}{\frac{an}{b-a}+j-1}-\frac{1}{\frac{an}{b-a}+j}\tag{2} \end{align*}

The left-most and right-most part admit telescoping which makes summation and taking the limit an easy job.

We start with the left-most part of (2) and obtain

\begin{align*} \color{blue}{\lim_{n\to\infty}}&\color{blue}{\frac{n}{b-a}\sum_{j=1}^n\left(\frac{1}{\frac{an}{b-a}+j}-\frac{1}{\frac{an}{b-a}+j+1}\right)}\\ &=\lim_{n\to\infty}\frac{n}{b-a}\left(\frac{1}{\frac{an}{b-a}+1}-\frac{1}{\frac{an}{b-a}+n+1}\right)\\ &=\lim_{n\to\infty}\frac{n}{b-a}\left(\frac{b-a}{an+b-a}-\frac{b-a}{bn+b-a}\right)\\ &=\lim_{n\to\infty}\left(\frac{n}{an+b-a}-\frac{n}{bn+b-a}\right)\\ &\,\,\color{blue}{=\frac{1}{a}-\frac{1}{b}}\tag{3} \end{align*}

We continue with the right-most part of (2)

\begin{align*} \color{blue}{\lim_{n\to\infty}}&\color{blue}{\frac{n}{b-a}\left(\frac{1}{\frac{an}{b-a}}-\frac{1}{\frac{an}{b-a}+n}\right)}\\ &=\lim_{n\to\infty}\frac{n}{b-a}\left(\frac{b-a}{an}-\frac{b-a}{bn}\right)\\ &\,\,\color{blue}{=\frac{1}{a}-\frac{1}{b}}\tag{4} \end{align*}

We finally conclude since (1) is squeezed by (3) and (4)

\begin{align*} \color{blue}{\int_a^b\frac{1}{x^2}\,dx=\frac{1}{a}-\frac{1}{b}} \end{align*}

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