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 $\sum_{i=1}^n-\frac{3n^2}{(n + 3 i)^3}$.
 A: 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 $q = \sqrt[n]{4}$:
$$
I = \lim_{n\to\infty} \frac{{1\over \sqrt[n]{4}} - {1\over (\sqrt[n]{4})^{2n+1}}}{\sqrt[n]{4}+1} = \boxed{{15\over 32}}
$$
A: 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.
A: 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. 
