# Evaluate $\int_1^e\frac{1}{x}dx$ using upper and lower sums

Using the upper and lower sums, I've tried to solve $$\int_1^e\frac{1}{x}dx$$ in the following way.

Let $$\Delta x = \frac{\mathit e -1}{n}$$ and let a partition $$P$$ be given by $$P = \{1,1+\Delta x, 1+2\Delta x,...,1+(n-1)\Delta x,\mathit e\}$$

Since $$1/x$$ is a monotonically decreasing function, noting the definition of the upper and lower partitions is $$M_i=\mathrm {sup}\{f(x) \lvert \text{P is a partition} \}$$ and $$m_i=\mathrm {inf}\{f(x)\lvert \text{P is a partition} \}$$, this means $$M_i = f(x_{i-1}) = \frac{a}{1+\frac{(i-1)(e-1)}{n}} \quad \mathrm {and} \quad m_i = f(x_i) = \frac{a}{1+\frac{i(e-1)}{n}}$$

So the upper sum is $$U(f,P) = \sum_{i=1}^n M_i\Delta x = \sum_{i=1}^n \frac{e-1}{n+(i-1)(e-1)}$$ and the lower sum is $$L(f,P)=\sum_{i=1}^n m_i\Delta x=\sum_{i=1}^n\frac{e-1}{n+i(e-1)}$$

But I can't evaluate these sums. I'm not sure if my process is wrong, so I am getting difficult sums or what.

• The partial sums are a digamma function. You also can factor $\sum_n\frac a{b n+c}=\frac ab\sum_n \frac1{n+\frac cb}$ Aug 5 at 17:00
• Use the partition with points in geometric progression. Thus define $x_i=e^{i/n}$ for $i=0,1,2,\dots, n$ and then evaluate upper and lower sums. Aug 5 at 17:40
• @Paramanand Singh Thank you! Solved it! Aug 5 at 18:51