I'm looking at the following problem:

Let $L:(0,\infty)\rightarrow\mathbb{R}$ be defined by $$L(x)=\int_1^x\frac1t dt, \qquad x>0$$

a) Prove: $L$ strictly increasing and differentiable on $(0,\infty)$.

b) Show that $L(2)>0$

c) Prove that $L(xy)=L(x)+L(y),\quad x,y>0$

d) Show that for $n\in \mathbb{Z}$, $L(2^n)=nL(2)$

e) Prove that $L(x)$ is both injective and surjective.

Now, a and b are simple enough (I'll still include them) but starting from c I'm having some serious trouble. It'd be easy to note that $L(x)$ is the logarithmic function, so all these things hold, but since we haven't discussed logarithms in class, I suppose this exercise is built to derive some of its properties.

My work:

a) Note how $\frac 1t$ is continuous on $(0,\infty)$; it follow immediately that $L(x)$ is differentiable and $L'(x)=1/x$. Note also how $1/x>0$ for all $x>0$. Because of this, $L(x)$ is strictly increasing.

b) Consider $L(1)=\int_1^1\frac1t dt=0$. (Fundamental theorem of calculus) Since $L(x)$ is strictly increasing, $L(2)>L(1)=0$.

c) No significant results, as I don't want to use the logarithm for this.

d) Let $n\geq 0$, then this follows immediately from c). For $n<0$ I've tried to play around with the fundamental theorem of calculus but didn't get to any significant result.

e) Note how $L(x)$ is strictly increasing and thus injective. To prove that $L(x)$ is on to, we have to take an arbitrary $y\in\mathbb{R}$ and show that $L(x)=y$ for some $x\in (0,\infty)$. I can intuitively see how this would be the case and that it could probably be proven without use of logarithms, but find it difficult to come up with any sort of formal argument.

Now I'm very curious to know whether these can be proven without making use of logarithms, and if so, what their proofs look like.


c) \begin{align*}L(xy)&=\int_1^{xy}\frac1t\,dt\\&=\int_1^x\frac1t\,dt+\int_x^{xy}\frac1t\,dt\end{align*}and so all you have to prove is that $\int_x^{xy}\frac1t\,dt=\int_1^y\frac1t\,dt$. You can do it doing the substitution $t=xu$ and $dt=x\,du$

d) If $n<0$, then $L(2^n2^{-n})=L(1)=0$ and $L(2^n2^{-n})=L(2^n)+L(2^{-n})=L(2^n)+(-n)L(2)$. And $L(2^n)+(-n)L(2)=0\Longleftrightarrow L(2^n)=nL(2)$.

e) Take $y\in[0,+\infty)$. Pick $n\in\mathbb N$ such that $nL(2)>y\Longleftrightarrow L(2^n)>y$. So, by the intermediate value theorem, there is some $x\in[1,2^n]$ such that $L(x)=y$.

On the other hand, if $y\in(-\infty,0]$, $-y=L(x)$ for some $x$ and therefore $y=-L(x)=L\left(\frac1x\right)$.


let $g(x):=f(xy)$ and differentiate with respect to $x$. You arrive at $$g^\prime(x) = f^\prime(xy) y = \frac{1}{xy}y=\frac{1}{x}$$ So $f$ and $g$ satisfy the same differential equation as a function of $x$. This implies the differ by a constant $c(y)$ which depends only on $y$. So

$$c(y) = \int_1^{xy}\frac{1}{t}dt-\int_1^{x}\frac{1}{t}dt =\int_x^{xy}\frac{1}{t}dt$$ No apply the change of variable theorem to show that $c(y)= L(y)$


How about changing the integration variable in $L(y)$ by $t' = xt$? It seems $$ L(y) = \int_{x}^{xy} \frac{x}{t'} \frac{dt'}{x} = \int_{x}^{xy} \frac{1}{t'} dt'$$ and that this gives $L(xy) = L(x) + L(y)$ without using logarithms.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.