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Let$$L(x)=\int_{1}^{x}\frac{1}{t}dt.$$How can I show that $L(ab)=L(a)+L(b)?$ This is what I have tried: If we let$$\begin{align}L(ax)&=\int_{1}^{ax}\frac{1}{t}dt,\text{ and}\\L(x)&=\int_{1}^{x}\frac{1}{t}dt,\end{align}$$then, by the FTC,$$\begin{align}F(ax)-F(1)&\Longrightarrow aF^{'}(ax)=a\frac{1}{ax}=\frac{1}{x},\text{ and}\\F(x)-F(1)&\Longrightarrow F^{'}(x)=\frac{1}{x}.\end{align}$$Hence, $L(ax)=L(x)+C$, where $C$ is some constant, and it must be $L(a)$.

Is this enough to show that $L(ab)=L(a)+L(b)$? How could I go about doing the same for $L(a^r)=rL(a)$?

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You might use a substition to show $$L(a) =\int_{t=b}^{ab}\frac{1}{t}dt$$ –  Henry Apr 30 '12 at 13:26
    
Useless comment given the title. (It is a way to remove my previous comment which was useless!) –  Giovanni De Gaetano Apr 30 '12 at 13:26
    
@GiovanniDeGaetano: It looks like the whole point of the exercise is to prove that $(\ln x)' = \frac1{x}$, and establish its properties. –  Javier Badia Apr 30 '12 at 13:28
    
@Javier, I noticed it, indeed I "removed" my comment. Anyway thank you for pointing it out! –  Giovanni De Gaetano Apr 30 '12 at 13:29

1 Answer 1

up vote 3 down vote accepted

Let $a,b$ be positive reals. Then $$ L(a b) = \int_1^{a b} \frac{\mathrm{d} t}{t} = \int_{a}^{a b} \frac{\mathrm{d} t}{t} + \int_{1}^{a} \frac{\mathrm{d} t}{t} $$ In the first integral, perform the change of variables $t = a u$, then $\frac{\mathrm{d} t}{t} = \frac{a \mathrm{d} u}{a u} = \frac{\mathrm{d} u}{u}$ and we have $$ L(a b) = \int_{1}^{b} \frac{\mathrm{d} u}{u} + \int_{1}^{a} \frac{\mathrm{d} t}{t} = L(b) + L(a) $$ For the second part, use $\frac{\mathrm{d}(t^r)}{t^r} = \frac{r t^{r-1} \mathrm{d} t}{t^r} = r \frac{\mathrm{d} t}{t}$, giving $$ L(a^r) = \int_{1}^{a^r} \frac{\mathrm{d} t}{t} \stackrel{\color\maroon{t = u^r}}{=} \int_1^a r \frac{\mathrm{d} u}{u} = r L(a) $$

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I understand now. I will go and prove the rest of the logarithm properties. Thank you very much! –  Josué Apr 30 '12 at 13:50

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