Show an integration equality without using $\int\frac{\mathrm dx}{x}=\ln|x|+c$

Question

Show that $$\int_{1}^{t_1t_2}\frac{\mathrm{d} x}{x}=\int_1^{t_1}\frac{\mathrm{d} x}{x}+\int_1^{t_2}\frac{\mathrm{d} x}{x}$$

without using

$$\int\frac{\mathrm{d} x}{x}=\ln|x|+c.$$

Answer 1

We have that

$$\int_1^{t_1 t_2} \frac 1 x \, dx = \int_1^{t_1} \frac{1}{x} \, dx +\int_{t_1}^{t_1t_2} \frac{1}{x} \, dx. $$

Now, what happens to $$\int_a^b \frac 1 x \, dx$$ when you do a $u$ substitution of the form $u=cx$? Do such a $u$-substitution to the third integral in the line above.

Answer 2

Disclaimer: I'd definitely say Aaron's proof is simpler but...here goes anyway.

Define $$F(t)=\int_{1}^{t}\frac{1}{x}\,dx$$ Let $\nabla$ be the gradient operator defined by, $\nabla u=(\frac{\partial u}{\partial t_1},\frac{\partial u}{\partial t_2})$ where we're viewing $t_1$ and $t_2$ as making up $(t_1,t_2)$ Cartesian coordinates (think $x$ and $y$ if you'd like). Then we have $$\nabla F(t_1 t_2)=\bigg(\frac{\partial}{\partial t_1}F(t_1 t_2),\frac{\partial}{\partial t_2}F(t_1 t_2)\bigg)=\bigg(\frac{1}{t_1},\frac{1}{t_2}\bigg)$$ where the second equality is easily verifiable via chain rule (and the fundamental theorem of calculus and the definition of $F$). On the other hand, similar computations show $$\nabla F(t_1)=\bigg(\frac{1}{t_1},0\bigg)$$ and $$\nabla F(t_2)=\bigg(0,\frac{1}{t_2}\bigg)$$ Then, we have \begin{align}\nabla F(t_1 t_2)=\nabla F(t_1)+\nabla F(t_2)&\Rightarrow \nabla (F(t_1 t_2)-F(t_1)-F(t_2))=0\\ &\Rightarrow F(t_1 t_2)-F(t_1)-F(t_2)=C\end{align} where $C$ is some constant. Plugging in $t_1=t_2=1$, we see that $C=0$ and your claim is proved.

Above, we're essentially viewing $F(t_1 t_2)$, $F(t_1)$ and $F(t_2)$ as "different" functions defined on the $(t_1,t_2)$ plane (i.e. $\mathbb{R}^2$) that just happen to satisfy the above relationship due to the relation between their gradients.

Answer 3

Let us assume $t_1,t_2>0$.
Since $\frac{dx}{x}$ is invariant with respect to substitutions of the form $x\mapsto \lambda x$, we have:

$$ \color{blue}{\int_{1}^{t_1}\frac{dx}{x}}+\color{green}{\int_{1}^{t_2}\frac{dx}{x}} = \color{blue}{\int_{1}^{t_1}\frac{dx}{x}}+\color{green}{\int_{t_1}^{t_2 t_2}\frac{dx}{x}} = \color{red}{\int_{1}^{t_1 t_2}\frac{dx}{x}} $$ sic et simpliciter.