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Let $0 < a < b$. Use Riemann sums to compute

$$ \int_{a}^{b}x^{-2}\ \mathrm{d}x\ . $$

So far, I have gotten to the step where

\begin{align} \int_{a}^{b} x^{-2}\ \mathrm{d}x & = \lim_{n \to \infty}\ \sum_{i = 1}^{n} \left[\left(a + \frac {b - a} {n} \ i\right)^{-2}\left(\frac {b - a} {n}\right)\right] \\[5 mm] & = \lim_{n\to\infty}\left[\frac{b - a}{n}\ \sum_{i = 1}^{n}\left(a + \frac {b - a} {n}\ i\right)^{-2}\ \right]. \end{align}

I assume my equation is correct, but I am not sure how to evaluate the sum.

I also know how to compute

$$ \int_{a}^{b}x^{2} \ \mathrm{d}x\ , $$

but I am not sure how to carry forward the idea (if it is even relevant) to this problem (if it is even relevant).

Riemann sums were just covered in my module and I am still trying to get used to it. I would appreciate some help/guidance on this!.

Ethan Mark
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    A linear partition does not give the simplest sum Try to use the partition $x_k = a (b/a)^{k/n}$. – Winther Oct 05 '20 at 14:33
  • @Winther Hi. Thank you for the suggestion, but may I know how one can come up with that kind of partition? Is there some intuition behind it? Also, geometrically, what kind of partitions would that look like? – Ethan Mark Oct 05 '20 at 14:35
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    This partition is one for which you have an equal spaced points if you use a logarithmic $x$-axis when you plot it (i.e. we have the same number of points between $1$ and $10$ as between $10$ and $100$). It's a matter of try and fail and experience what is a "good" partition to use. And by good I just mean that we get a sum that we can easily evaluate. That being said you will rarely encounter anything other than linear and geometrical in the wild (exercises on riemann sums). – Winther Oct 05 '20 at 14:53
  • @Winther I see. Thank you for the elaboration! So would it be safe to say that I should try the partition you mentioned whenever I encounter similar problems with functions whose graphs look like that of $x^{-2}$? – Ethan Mark Oct 05 '20 at 14:54
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    Yes this partition works very well for any power-law function $f(x) = x^n$ (also positive powers and non-integer powers also). – Winther Oct 05 '20 at 15:12
  • @Winther I am trying out your suggestion to solve the question, but I am not sure what should be in the Riemann sum. Do you mind writing it out as an answer so that I may analyse it more deeply? – Ethan Mark Oct 05 '20 at 16:03
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    See e.g. https://math.stackexchange.com/questions/2432529/non-standard-partition-for-riemann-sums and https://math.stackexchange.com/a/1791418/147873 for more info – Winther Oct 05 '20 at 16:18

1 Answers1

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With a uniform partition, let $X_k = a + \frac{b-a}{n}k$. The Riemann sum is $R_n = \frac{b-a}{n}\sum_{k=1}^n X_k^{-2}$ and it can be squeezed between sums whose limits are easily evaluated, viz.

$$\tag{*}\frac{b-a}{n}\sum_{k=1}^n X_k^{-1}X_{k+1}^{-1}\leqslant R_n \leqslant \frac{b-a}{n}\sum_{k=1}^n X_{k-1}^{-1}X_k^{-1}$$

Note that

$$\frac{b-a}{n}\sum_{k=1}^n X_k^{-1}X_{k+1}^{-1} = \sum_{k=1}^n (X_k^{-1}-X_{k+1}^{-1}) = X_1^{-1} - X_{n+1}^{-1} \\ = \left(a+ \frac{b-a}{n} \right)^{-1}- \left(a+ (b-a)\frac{n+1}{n} \right)^{-1} \\ \underset{n \to \infty} \to(a^{-1} - b^{-1})$$

Similarly, we can show that the sum on the RHS of (*) also converges to $a^{-1} - b^{-1}$, and by the squeeze theorem

$$\lim_{n \to \infty}R_n = a^{-1} - b^{-1} = \int_a^b \frac{dx}{x^2}$$

RRL
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  • Could you elaborate a little more on how you got the $(*)$ inequalities? – Ethan Mark Oct 05 '20 at 15:29
  • $X_{k-1} < X_k < X_{k+1} \implies X_{k+1}^{-1} < X_k^{-1} < X_{k-1}^{-1} \implies X_{k+1}^{-1}X_k^{-1} < X_k^{-2} < X_{k-1}^{-1}X_k^{-1} $ – RRL Oct 05 '20 at 15:32
  • With a little algebra you will see that the sums are telescoping as shown above. – RRL Oct 05 '20 at 15:33
  • Oh. This solution is so ingenious. Thank you :) – Ethan Mark Oct 05 '20 at 15:34
  • @EthanMark: You're welcome. – RRL Oct 05 '20 at 16:03
  • Just one more question. I noticed that your inequalities above are strict. Why is it that when you take summation over the strict inequalities, then equivalence is included? – Ethan Mark Oct 05 '20 at 16:09
  • I was not being super precise, but there is nothing wrong here as $a_n < b_n < c_n$ and $a_n, c_n \to L$ implies $b_n \to L$. Recall that if $a_n < C$ and $a_n \to L$ implies $L \leqslant C$. – RRL Oct 05 '20 at 16:13
  • For example $1 - 1/n <1$ and $1-1/n \to 1$. – RRL Oct 05 '20 at 16:14
  • I see. Also, I understand that we can use method of differences to evaluate the sum, but how did you get the first equal sign after "Note that...", where you multiplied the $\frac {b - a} {n}$ into the sum and the product became a difference? – Ethan Mark Oct 05 '20 at 16:16
  • That is from the algebra I referred to in doing the partial fraction decomposition. – RRL Oct 05 '20 at 16:19
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    $\frac{1}{a + (b-a)k/n} - \frac{1}{a + (b-a)(k+1)/n} = \frac{b-a}{n} \frac{1}{(a + (b-a)k/n)(a + (b-a)(k+1)/n)}$ – RRL Oct 05 '20 at 16:23
  • (+1) ... as always, well done! – Mark Viola Nov 12 '20 at 16:22