A summation problem

Question

Here is the question: My professor made this equations in a proof, however I did not understand how he gets these steps, and which technic he uses $$ \sum_{k=1}^n \int_1^k\frac {1}{x}dx=\sum_{k=1}^n \sum_{j=1}^{k-1} \int_{j}^{j+1} \frac {1}{x}dx = \sum_{j=1}^{n-1} \sum_{k=j+1}^{n} \int_{j}^{j+1}\frac {1}{x}dx $$

Answer 1

I assume that the first expression $$\color{red}{\sum_{k=1}^n \int_1^k \frac{1}{x}dx}$$ is the original problem. To get to the second expression, your professor uses the property of integrals $$\int_a^b f(x)dx+\int_b^c f(x)dx=\int_a^c f(x)dx$$ To split the integral $$\int_1^k \frac{1}{x}dx$$ into the sum of integrals $$\int_1^2 \frac{1}{x}dx+\int_2^3 \frac{1}{x}dx+...+\int_{k-1}^k \frac{1}{x}dx$$ which he then expresses using summation notation as $$\sum_{j=1}^{k-1} \int_j^{j+1} \frac{1}{x}dx$$ thus transforming the original expression into $$\color{red}{\sum_{k=1}^n \sum_{j=1}^{k-1} \int_j^{j+1} \frac{1}{x}dx}$$ The next transformation is fairly straightforward... your professor is merely utilizing the commutativity of addition to rearrange the sums into a more useful format.

Answer 2

The first equation holds because $$ \int_1^k\frac {1}{x}dx = \sum_{j=1}^{k-1} \int_{j}^{j+1} \frac {1}{x}dx $$ Since in general, $\int_a^b + \int_b^c = \int_a^c$ (this is lazy notation but hopefully you know what I mean).

The second is a reordering of the summation. Note that both sides can be written as $$ \sum_{(j,k) \in S}\int_{j}^{j+1} \frac {1}{x}dx $$ Where $S = \{(j,k): j,k \in \{1,2,\dots,n\} \text{ and }j<k\}$