Why is the change of variable used in proving $L(1/a) = -L(a)$ in $L(x)=\int_1^x \frac1t dt$ is by letting $u = \frac1t$?

Question

The point is I really don't understand why we are letting $t = \frac1u$ or $u = \frac1t$, because there is a $\frac1t dt$ in there? The material I got says $\int_c^d f(u) du = \int_a^b f(u(x)) \frac {du}{dx} dx$ if $u(a) = c$ and $u(b) = d$, which I thought should be used in letting $\frac1t dt = du$. Please emlight me on this, I really need to understand this. Please help

Answer 1

If you let $u =1/t$, then $$\frac{du}{dt}=-1/t^2 = -u^2.$$ So formally you get $\frac{du}{-u^2}=dt$. Plugging this into your integral, you obtain $$L(x) = \int_1^x \frac{1}{t}dt = \int u \frac{du}{-u^2} = \int_{1}^{1/x} -\frac{1}{u}du$$ And the right hand side is precisely $$ -\int_{1}^{1/x} \frac{1}{t}dt = -L(1/x)$$ Note how we changed the range of integration correspondingly to the new variable $u$.

To add to my answer: So WHY did we do such a change of variables? Actually, if you know that the integral of $1/x$ is the logarithm function, then the statement is obvious. So suppose we don't know this fact.

To prove that $L(1/x) = -L(x)$, take a look at what the left hand side is. Just writing it out, you have $$L(1/x) =\int_{1}^{1/x} \frac{1}{t}dt .$$ Note that the range of integral has changed from $x$ to $1/x$; according to your material, this suggests that some kind of change of variables was performed that resulted in the upper and lower ranges of the integral to become inversed. The best guess is that we should make the substitution $1/t = u.$