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Using $$\ln(x) = \int_1^x \frac{1}{t} dt$$

Show that for $x > 0$, $\ln\left(\frac{1}{x}\right) = -\ln(x)$

I am following a provided answer and didn't quite understand the following transformation and why/how it is done:

$$\ln\left(\frac{1}{x}\right) = \int_1^{\frac{1}{x}} \frac{1}{t}dt$$

$$~u = \frac{1}{t}, du = \frac{-1}{t^2}dt$$

$$= -\int_1^{\frac{1}{x}} t\frac{dt}{t^2}$$

$$=-\int_1^x \frac{1}{u}du$$

$$= -\ln(x)$$

I don't understand how the limits of integration were changed in the $2nd$ and $3rd$ term (from $\frac{1}{x}$ to $x$). Is there a property that makes this correct or is there some other reasoning behind the change?

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6  
$\qquad t=1\iff u=1;\quad t=1/x\iff u=x$ –  anon Jun 13 '12 at 6:50

2 Answers 2

up vote 2 down vote accepted

\begin{align} \ln\left(\frac{1}{x}\right) & = \int_1^{\frac{1}{x}} \frac{1}{t}dt & & {\text{Replace $x$ by $\dfrac1x$ in the definition}}\\ u & = \frac{1}{t} & &{\text{This is the substitution you are making}}\\ du & = \frac{-1}{t^2}dt & &{\text{This is because }\dfrac{du}{dt} = - \dfrac1{t^2}}\\ \ln\left(\frac{1}{x}\right) & = \int_1^{\frac{1}{x}} \frac{1}{t}dt & = \int_1^{\frac{1}{x}} -t \times \left(-\frac{1}{t^2}dt \right) & \text{Multiply and divide the integrand by $-t$} \end{align} In the above integrand, we can replace $-\dfrac{1}{t^2}dt$ by $du$ and $-t$ by $- \dfrac1u$.

Note that we are making the change of variable, the integrand is in terms of $u$ now.

Hence, we need to look at the limits for the variable $u$ in the integral.

We have the transformation that $u = \dfrac1t$. Hence, if $t$ goes from $1$ to $1/x$, then $u$ goes from $1$ to $x$. This is because when $t=1$, $u = \dfrac11 = 1$. Similarly, if $t = 1/x$, then $u = \dfrac1{1/x} = x$.

(For example, if $t$ goes from $1$ to $2$, then $1/t$ goes from $1$ to $1/2$.)

Hence, we get that \begin{align} \ln \left( \dfrac1x\right) & = \int_1^{\frac{1}{x}} -t \times \left(-\frac{1}{t^2}dt \right) & = \int_1^x \left(-\dfrac1u \right) \times du = - \int_1^x \dfrac{du}{u} = - \ln(x) \end{align} where the last equality is obtained since we have defined $\ln(x)$ as $\displaystyle \int_1^x \dfrac{dt}{t}$.

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I like to add that this change of integral limts is due specifically to the terminals as follows: If $g$ is a function continuously differentiable on $[a,b]$ and $f$ continuous on $J=g ([a, b​​])$, let $F$ is a primitive of $f$ on $J$ then: $$\forall t \in [a,b] \quad (F \circ g)'(t)=F'(g(t)) g'(t) = f(g(t)) g'(t)$$ Then : $$\begin{align*}\int_a^b f(g(t)) g'(t) dt &= \int_a^b (F \circ g)'(t) dt \\ &=\left[(F \circ g) (t)\right]_a^b \\ &= F(g(b))-F(g(a)) \\ &= [F(t)]_{g(a)}^{g(b)} \\ &= \int_{g(a)}^{g(b)} f(t) dt \end{align*}$$ Now try to compare this with the special case above.

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