We have a definite integral $$\int_1^2x \ln x~dx$$
Using the Newton-Leibniz formula, we find that the solution to this integral is $2\ln2-\frac{3}{4}$.
But I come to an obstacle whenever I try to solve this using the definition of an integral, Riemman sums.
What I tried doing was making a partition $P=\{1=x_0<x_1<…<x_n=2\}$ such that $x_i=2^{\frac{i}{n}}$, and let's take $\theta=\{x_i\ |\ x_i\in[x_{i-1},\,x_i]\}$ as set of selected points. That led me to following limit:
$$\begin{array}{r c l}
\lim\limits_{n\to\infty}\sum\limits_{i=0}^{n}f(x_i)\Delta x_i & =& \lim\limits_{n\to\infty}\sum\limits_{i=0}^{n}\left(2^{\frac{i}{n}}\cdot\ln(2^\frac{i}{n})\right)\cdot\left(2^\frac{i+1}{n}-2^\frac{i}{n}\right) \\
& = & \ln 2\cdot \lim\limits_{n\to\infty}\sum\limits_{i=0}^{n}\frac{i}{n}\cdot 2^\frac{i}{n}\cdot\left(2^\frac{i+1}{n}-2^\frac{i}{n}\right)
\end{array}$$
This is the part where I get stuck. Is this the correct approach or not? If it is then how to find the limit of this sum?
Best Answer
The computation can be made a little easier by considering instead a generalization of your choice of partition, namely let $$x_k = z^k, \quad k \in \{0, 1, 2, \ldots, n\}.\tag{1}$$ (I am using $k$ instead of $i$ to avoid any confusion with complex numbers.) Then the Riemann sum for the partition $\bigcup_{k=0}^{n-1} [x_k, x_{k+1})$ is $$\begin{align} \sum_{k=0}^{n-1} f(x_k) \Delta x_k &= \sum_{k=0}^{n-1} z^k \log z^k (z^{k+1} - z^k) \\ &= (z-1) \log z \sum_{k=0}^{n-1} k z^{2k} \\ &= \frac{z(z-1)}{2} \log z \sum_{k=0}^{n-1} 2k z^{2k-1} \\ &= \frac{z(z-1)}{2} \log z \sum_{k=0}^{n-1} \frac{d}{dz}\left[z^{2k}\right] \\ &= \frac{z(z-1)}{2} \log z \frac{d}{dz}\left[ \sum_{k=0}^{n-1} z^{2k} \right] \\ &= \frac{z(z-1)}{2} \log z \frac{d}{dz}\left[ \frac{z^{2n} - 1}{z^2 - 1} \right] \\ &= \frac{z(z-1)}{2} \frac{2((n-1)z^{2n+2} - n z^{2n} + z^2)}{z(z^2-1)^2} \log z \\ &= \frac{(n-1)z^{2n+2} - nz^{2n} + z^2}{(z+1)(z^2-1)} \log z. \tag{2} \end{align}$$ Now, the choice of $z$ that satisfies $x_0 = 1$ and $x_n = 2$ is clearly $z = 2^{1/n}$, so the Riemann sum becomes $$\begin{align} \int_{x=1}^2 x \log x \, dx &= \lim_{n \to \infty} \frac{4n(4^{1/n} - 1) - 3(4^{1/n})}{n(2^{1/n} - 1)(2^{1/n} + 1)^2} \log 2. \end{align}$$ The evaluation of this limit is...not fun, so I leave it as an exercise; but it does exist and yields the desired result.