# Riemann sum of $1/x^3$

Evaluate the following integral using Riemann's sums: $$\int_1^4 {1\over x^3} \mathop{dx}$$

$$\Delta x = \frac{3}{n},\ \ \ x_i=1+\frac{3i}{n}.$$

we have $$\sum_{i=1}^n f(x_i)\Delta x .$$

I am at this point but I don't know what to do with $$\sum_{i=1}^n-\frac{3n^2}{(n + 3 i)^3}$$.

• Now it's just plugging in what $n$ is, plugging in the corresponding values of $i$, and adding them up. You're not likely to get a nice symbolic answer.
– Ian
Commented Aug 29, 2016 at 23:16
• It wasn't until this question that I realized.. I've never done an explicit Riemann sum calculation with rational functions. Now I can see why.. Commented Aug 29, 2016 at 23:56
• Possibly of interest: $\int_{0}^{a} x^{1/n}\, dx$ without antiderivative for $n>0$, if you're willing to use a "geometric progression" partition instead of an equal-length partition. Commented Aug 30, 2016 at 0:04

There is actually no need to express the result in terms of digamma function. A neat way to solve the problem is to choose points in the segments so that they form a geometric progression.

So we are going to find the following integral:

$$I = \int_1^4 {1\over x^3}\mathop{dx}$$

Consider the segment $$[1, 4]$$. Let's split it into $$n$$ parts with points $$x_0, x_1,\dots x_n$$ so that the points form a geometric progression. Let $$q$$ be the denominator of the geometric progression, then the segment in split by the points: $$\{1, q, q^2, \dots, q^n\}$$ and therefore: $$q^n = 4 \iff q =\sqrt[n]{4}$$

By choosing those point in such a way we can see that the length of each interval is as follows: \begin{align} \Delta x_1 &= q - 1\\ \Delta x_2 &= q^2 - q\\ &\dots \\ \Delta x_n &= q^n - q^{n-1} \end{align}

Now we have to choose points $$\zeta_k$$ in every subsegment. Let stick to the right side of each segment for simplicity. This gives: $$\zeta_k = \{q, q^2, q^3, \dots, q^n\}$$

Let's calculate the value of the function in those points: $$f(\zeta_k) = \left\{{1\over q^3}, {1\over q^6}, \dots, {1\over q^{3k}} \right\}$$ We are now ready to build the Riemann's sum: \begin{align} S_n &= \sum_{k=1}^n f(\zeta_k)\Delta x_k\\ &= \sum_{k=1}^n {1\over q^{3k}}(q^k - q^{k-1})\\ &= {1\over q^3}(q-1) + {1\over q^6}(q^2 - q) + \cdots + {1\over q^{3n}}(q^n - q^{n-1})\\ &= {1\over q^3}(q-1) + {1\over q^5}(q - 1) + \cdots + {1\over q^{2n-1}}(q - 1)\\ &= (q - 1)\left( {1\over q^3} + {1\over q^5} + \cdots + {1\over q^{2n+1}} \right)\\ &= (q-1)\sum_{k=1}^n {1\over q^{2k+1}} \end{align}

But the sum in last expression is nothing but a geometric series and we know how to find its sum. So the expression above becomes: $$S_n = (q-1) \cdot \frac{{1\over q} - {1\over q^{2n+1}}}{q^2 - 1} = \frac{{1\over q} - {1\over q^{2n+1}}}{q+1}$$

Remember our $$q = \sqrt[n]{4}$$: $$I = \lim_{n\to\infty} \frac{{1\over \sqrt[n]{4}} - {1\over (\sqrt[n]{4})^{2n+1}}}{\sqrt[n]{4}+1} = \boxed{{15\over 32}}$$

• Yes, this is basically Fermat's method of quadrature. But it's not the "standard" Riemann sum which is based on equal intervals. Commented Nov 21, 2019 at 22:24

Your Riemann sum is $$\frac{n^2}{18} \left( \Psi''(n/3+1) - \Psi''(4n/3+1)\right)$$ where $\Psi$ is the digamma function and $\Psi''$ is its second derivative, the polygamma function of order $2$. It is not an elementary function.

If this is from a Calculus homework question, I suggest taking a closer look at the question. Assuming your instructor is not sadistic, you are not expected to find a closed-form formula for this Riemann sum.

• There is actually a way to solve this using a somewhat elementary method. I've added an answer below Commented Nov 21, 2019 at 18:59

If you are only asked to compute the integral using Riemann sums (and not calculating this specific sum) then there is a trick: In Riemann sums you have the right to choose the point in each interval, see e.g. Riemann sums

First show that there is $a_i\in [x_{i-1},x_i]$ so that $$f(a_i) = \frac{x_{i-1}+x_i}{2x_{i-1}^2x_i^2} =\frac{1}{x_{i}-x_{i-1}}\left(\frac{1}{2x_{i-1}^2} - \frac{1}{2x_i^2}\right)$$ Then $\sum_{i=1}^{n} f(a_i)(x_{i}-x_{i-1})$ is a Riemann sum which is telescopic and (very) easy to calculate.