0

Is there any way to prove: $$f(x)=\int_1^x {1\over t} dt $$ Increases without bound as $x \to\infty$ and is monotonically increasing on $(0,\infty)$. Without knowing it is the logarithm

I think you can prove it is monotonically increasing on $(0,\infty)$ by the fact that
$1/x > 0$ for $x$ in $(0,\infty) $

To prove it is unbounded :
> I know you can somehow use the divergence harmonic series but i don't know how.

So how to prove it is unbounded ?(Using harmonic series)

Paramanand Singh
  • 87,309
Vivaan Daga
  • 5,531
  • 1
    Don't use the same variable as upper bound of integration and as integration variable. This is highly confusing. – mrtaurho Jun 15 '20 at 14:12
  • 1
    Prove that $f(xy) =f(x) +f(y) $ and then $f(2^n)=nf(2)$ for positive integers $n$. Note that $f(2)>0$ and done! I had mentioned about $f(xy) =f(x) +f(y) $ in a comment to your previous question. You should have tried that. – Paramanand Singh Jun 15 '20 at 20:15
  • @ParamanandSingh basically can we say f(x) exp(x) satisfies f’(x)=f(x) with f(0)=1 and is the only function to have these properties because a function which satisfies these properties has the inverse function (shown easily in blog )as the integral(which is log) and is also the inverse function of log and so log can only have one inverse function hence only one function satisfies these properties is that correct – Vivaan Daga Jun 17 '20 at 14:21
  • 1
    All your comments about $\exp (x) $ are already handled in this answer of mine: https://math.stackexchange.com/a/1292586/72031 However this thread is about $\log x$. – Paramanand Singh Jun 17 '20 at 14:34
  • Well you can do that, but then the mystery remains as to why would one think of integral of $1/x$. My answers gives the genesis of $1/x$ by directly starting from the differential equation. Essentially what you say is same as in my answer. And I don't see a fundamental difference here. – Paramanand Singh Jun 17 '20 at 14:42
  • 1
    Well, I wanted to prove uniqueness just using differential equation rather than using any property of any known functions. And what you say is also right. – Paramanand Singh Jun 17 '20 at 14:52
  • The FTC says that if the integrand is continuous at some point then the integral is differentiable at that point. Here the integrand is continuous at all positive points including $1$ and $x$. – Paramanand Singh Jun 25 '20 at 08:48
  • @ParamanandSingh are you refering to : Theorem : let f be a riemann integrable function on [a,b] for a$\le x \le$b, put F(x)=$\int_a^b f(t) dt $ . Then F is continous on [a,b] , furthermore , if f is continous at a point $x_{0}$ of [a,b] , then F is differentaible at $x_{0}$, and F'($x_{0}$)=f($x_{0}$) .. This implies that our integral is differentiable [1,x] and hence one am i correct or incorrect? – Vivaan Daga Jun 25 '20 at 11:22
  • Yes that's what I am referring to. – Paramanand Singh Jun 25 '20 at 17:41
  • @ParamanandSingh your proof utilises the Archimedean property is that correct? – Vivaan Daga Jun 26 '20 at 05:58
  • I think you should ask a separate question for more details on Fundamental Theorem of Calculus or you can create a chat room and discuss. – Paramanand Singh Jun 26 '20 at 07:51

1 Answers1

1

The monotonicity, as you say, is obvious. For any $x,y \ge 1$ such that $x>y$, you can just note that

$$ f(x)-f(y) = \int_y^x \frac 1t\, dt > 0.$$

As for the second part, using Taylor's formula you can easily obtain $$ f(2x) \ge f(x) + \frac 12. $$

If the limit as $x \to \infty$ existed (and since $f$ is monotonous, it either exists or is infinite), it would satisfy $L \ge L + \frac 12$, which is not possible.

$$ f(2x) = f(x) + f'(x) (2x-x) + \frac{f''(\xi)}{2}(2x-x)^2, \quad \xi \in (x, 2x) $$

Since $f'(x)=\frac 1x$ and $\frac{f''(\xi)}{2} x^2 = -\frac{x^2}{2 \xi^2}$, by substituting $\xi$ by $x$ we get $$ f(2x) \ge f(x) + 1 - \frac 12 = f(x)+\frac 12 $$

PierreCarre
  • 20,974
  • 1
  • 18
  • 34
  • @VivaanDaga You compute $f(2x)$ using Taylor's formula centered at $x$: $$f(2x) = f(x) + f'(x) (2x - x) + \frac{f''(\xi)}{2}(2x-x)^2, \quad \xi \in (x, 2x). $$ – PierreCarre Jun 15 '20 at 15:07
  • I need no convergence... I'm just going up to order one with remainder. I just require $f$ to be $C^2$ – PierreCarre Jun 15 '20 at 15:10
  • $$\frac{f''(\xi)}{2} x^2 = -\frac{x^2}{2 \xi^2} $$

    Replacing $\xi$ by $2x$ gives o lower bound for this term. So,

    $$ f(2x) = f(x) + \frac 1x (2x-x) - \frac{x^2}{2\xi^2} \ge f(x) + 1-\frac 18$$

     – PierreCarre Jun 15 '20 at 15:22
  • @VivaanDaga Sure... but if $>$ holds, "$\ge$" also holds. – PierreCarre Jun 15 '20 at 15:45
  • @VivaanDaga Yes, but it is not wrong to use "$\ge$", and is in fact irrelevant for the main argument. – PierreCarre Jun 15 '20 at 15:51
  • The fundamental theorem of calculus guarantees the diferentiability $f$ and provides its derivative, which is also differentiable. So $f \in C^2(0,+\infty)$ and Taylor's theorem can be used. – PierreCarre Jun 16 '20 at 08:25
  • If x is negative then log does not exist so f should not exist so although there exists an antiderivative you can’t evaluate f(x) because log of a negative no is undefined(for real case) how do we know the same won’t happen for positive x? – Vivaan Daga Jun 16 '20 at 10:03
  • @VivaanDaga $f$ exists for negative $x$, because the antiderivative of $\frac 1x$ is $\ln |x|$, but this is irrelevant to the problem at hands since we are studying the behaviour for large positive $x$. – PierreCarre Jun 16 '20 at 10:06
  • Is there a theorem that says that the antiderivative of a continuous function must exist I think fundamental theorem of calc says that I am i right ? – Vivaan Daga Jun 16 '20 at 10:53
  • Let us continue this discussion in chat. – PierreCarre Jun 16 '20 at 11:46
  • How do you know the integral is differentiable at 1 FTC does not guarantee that – Vivaan Daga Jun 25 '20 at 07:41
  • @VivaanDaga The FTC does guarantee that $f$ is differentiable at $x=1$. As long as the integrand is continuous, the integral is differentiable. This is how you see that continuous functions have anti-derivatives. – PierreCarre Jun 25 '20 at 13:37
  • is that because of :: Theorem : let f be a riemann integrable function on [a,b] for a≤x≤b, put F(x)=$\int_a^x f(t) dt$ . Then F is continous on [a,b] , furthermore , if f is continous at a point $x_{0}$ of [a,b] , then F is differentaible at $x_{0}$, and F'($x_{0}$)=f($x_{0}$) .. This implies that our integral is differentiable [1,x] and hence one am i correct or incorrect? – – Vivaan Daga Jun 25 '20 at 13:40
  • is the above theorem the reason for that? – Vivaan Daga Jun 25 '20 at 13:42
  • @VivaanDaga Yes, that is the result. – PierreCarre Jun 25 '20 at 19:35