# Thread: Reimann sum of the logarithm function?

1. ## [SOLVED] Reimann sum of the logarithm function?

Dear friends,

I tried to evaluate the integral $\displaystyle \int_{1}^{t}\ln(s)\mathrm{d}s$, where $\displaystyle t\geq1$ by making the usage of the Reimann sum, but i failed.
More precisely, how should I pick $\displaystyle \xi_{k}$ and $\displaystyle \Delta_{k}$ below?
$\displaystyle \int_{1}^{t}\ln(s)\mathrm{d}s=\ln\bigg(\lim_{n\to\ infty}\prod_{j=1}^{n}\xi_{k}^{\Delta_{k}}\bigg)$

2. Originally Posted by bkarpuz
Dear friends,

I tried to evaluate the integral $\displaystyle \int_{1}^{t}\ln(s)\mathrm{d}s$, where $\displaystyle t\geq1$ by making the usage of the Reimann sum, but i failed.
More precisely, how should I pick $\displaystyle \xi_{k}$ and $\displaystyle \Delta_{k}$ below?
$\displaystyle \int_{1}^{t}\ln(s)\mathrm{d}s=\ln\bigg(\lim_{n\to\ infty}\prod_{j=1}^{n}\xi_{k}^{\Delta_{k}}\bigg)$
You can evaluate $\displaystyle \int{\ln{s}\,ds}$ without using Riemann sums.

Have you ever used Integration by Parts before?

$\displaystyle \int{u\,dv} = uv - \int{v\,du}$.

In this case, $\displaystyle u = \ln{s}$ and $\displaystyle dv = 1$.

So $\displaystyle du = \frac{1}{s}$ and $\displaystyle v = s$.

Therefore

$\displaystyle \int{\ln{s}\,ds} = \int{\ln{s}\cdot 1\,ds}$

$\displaystyle = s\ln{s} - \int{s\cdot \frac{1}{s}\,ds}$

$\displaystyle = s\ln{s} - \int{1\,ds}$

$\displaystyle = s\ln{s} - s + C$.

Can you work out

$\displaystyle \int_1^t{\ln{s}\,ds}$ now?

3. Originally Posted by Prove It
You can evaluate $\displaystyle \int{\ln{s}\,ds}$ without using Riemann sums.

Have you ever used Integration by Parts before?

$\displaystyle \int{u\,dv} = uv - \int{v\,du}$.

In this case, $\displaystyle u = \ln{s}$ and $\displaystyle dv = 1$.

So $\displaystyle du = \frac{1}{s}$ and $\displaystyle v = s$.

Therefore

$\displaystyle \int{\ln{s}\,ds} = \int{\ln{s}\cdot 1\,ds}$

$\displaystyle = s\ln{s} - \int{s\cdot \frac{1}{s}\,ds}$

$\displaystyle = s\ln{s} - \int{1\,ds}$

$\displaystyle = s\ln{s} - s + C$.

Can you work out

$\displaystyle \int_1^t{\ln{s}\,ds}$ now?
Many thanks my friend Prove It, you really made me feel like a kid .
I already knew that $\displaystyle \big[t\ln(t)-t\big]^{\prime}=\ln(t)$ but I just wanted to show it via Remiann sum and the integration by parts tool is not allowed for use.

4. Originally Posted by bkarpuz
Many thanks my friend Prove It, you really made me feel like a kid .
I already knew that $\displaystyle \big[t\ln(t)-t\big]^{\prime}=\ln(t)$ but I just wanted to show it via Remiann sum and the integration by parts tool is not allowed for use.
Why would Integration by Parts not be allowed to be used?

What level maths are you taking?

Anyway, incase you were wondering, you can prove Integration by Parts using the Product Rule.

If you can prove this relationship works, I'm sure you'd be allowed to use it.

If $\displaystyle u = u(x)$ and $\displaystyle v = v(x)$, by the product rule...

$\displaystyle \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$.

So $\displaystyle \int{\frac{d}{dx}(uv)\,dx} = \int{\left(u\frac{dv}{dx} + v\frac{du}{dx}\right)\,dx}$

$\displaystyle \int{u\frac{dv}{dx}\,dx} + \int{v\frac{du}{dx}\,dx} = uv$

$\displaystyle \int{u\,dv} + \int{v\,du} = uv$

$\displaystyle \int{u\,dv} = uv - \int{v\,du}$.

5. Originally Posted by Prove It
Why would Integration by Parts not be allowed to be used?

What level maths are you taking?

Anyway, incase you were wondering, you can prove Integration by Parts using the Product Rule.

If you can prove this relationship works, I'm sure you'd be allowed to use it.

If $\displaystyle u = u(x)$ and $\displaystyle v = v(x)$, by the product rule...

$\displaystyle \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$.

So $\displaystyle \int{\frac{d}{dx}(uv)\,dx} = \int{\left(u\frac{dv}{dx} + v\frac{du}{dx}\right)\,dx}$

$\displaystyle \int{u\frac{dv}{dx}\,dx} + \int{v\frac{du}{dx}\,dx} = uv$

$\displaystyle \int{u\,dv} + \int{v\,du} = uv$

$\displaystyle \int{u\,dv} = uv - \int{v\,du}$.
I myself tried to solve it by Reimann sum but I could not pick $\displaystyle \xi_{k},\Delta_{k}$ in a suitable way to solve it easily.
Btw, I am a PhD student.

In the line below I tried to make it look similar to the Taylor's formula of the exponential function but not worked.
$\displaystyle \int_{1}^{t}\ln(s)\mathrm{d}s=\lim_{n\to\infty}\su m_{j=1}^{n}\ln\big(\xi_{j}^{\Delta_{j}}\big)=\lim_ {n\to\infty}\sum_{j=0}^{n-1}\frac{\Big[\ln\Big(t\big(\ln(t)-1\big)+1\Big)\Big]^{j}}{j!}.$

6. If I let $\displaystyle x_0=1, x_1=1+\Delta x,\cdots,x_{n-1}=1+(n-1)\Delta x$ and $\displaystyle \Delta x=\frac{t-1}{n}$. Then:

$\displaystyle \int_{1}^{t} \ln(s)ds=\lim_{n\to\infty}\sum_{i=1}^{\infty} f(x_{n-1}) \Delta x=\lim_{n\to\infty}\sum_{i=1}^{\infty} \ln\big(1+(i-1)\Delta x\big)\Delta x$

$\displaystyle =\lim_{n\to\infty}\sum_{i=1}^{\infty} \ln\left(1+\frac{(i-1)(t-1)}{n}\right)\frac{t-1}{n}$

and I can't go further with it.

7. I think it's enough to provide a proof, using Riemann sums, that

$\displaystyle \int_{a}^{b}{f(x)\,dx} = F(b) - F(a)$, where $\displaystyle \frac{d}{dx}[F(x)] = f(x)$

and then to directly evaluate $\displaystyle \int_1^t{\ln{x}\,dx}$.

Technically speaking, if you're evaluating a definite integral, you ARE using a Riemann sum.

I think you're making the problem harder than it needs to be...

8. Originally Posted by Prove It
I think it's enough to provide a proof, using Riemann sums, that

$\displaystyle \int_{a}^{b}{f(x)\,dx} = F(b) - F(a)$, where $\displaystyle \frac{d}{dx}[F(x)] = f(x)$

and then to directly evaluate $\displaystyle \int_1^t{\ln{x}\,dx}$.

Technically speaking, if you're evaluating a definite integral, you ARE using a Riemann sum.

I think you're making the problem harder than it needs to be...
Because what I want is to show that it can be proved by Reimann sum by picking the terms appropriately, my way is not only calculating that the solution of the integral is $\displaystyle t\ln(t)-t+1$.

May be the relation
$\displaystyle t\ln(t)-t+1=(t-1)^{2}\sum_{j=0}^{\infty}\frac{(-1)^{j+1}(t-1)^{j}}{(j+2)!}$
should be used. :S

9. OK, I've done it, but I don't think you'll like the solution...

For this to work, we need to make use of the Mean Value Theorem.

i.e. for some function $\displaystyle F(x)$, which is continuous on $\displaystyle x \in [a, b]$, we have

$\displaystyle f(c) = \frac{F(b) - F(a)}{b - a}$, where $\displaystyle a < c < b$ and $\displaystyle \frac{d}{dx}[F(x)] = f(x)$.

We can rearrange the formula so that it is of the form

$\displaystyle (b - a)f(c) = F(b) - F(a)$.

For this function, $\displaystyle f(x) = \ln{x}$, suppose we take some rectangular strip, between $\displaystyle x_{k - 1}$ and $\displaystyle x_k$.

Then this strip's area is

$\displaystyle (x_k - x_{k - 1})f(c)$, for some $\displaystyle c \in (x_{k - 1}, x_k)$.

Using the Mean Value Theorem, we can rewrite this as

$\displaystyle F(x_k) - F(x_{k - 1})$, where $\displaystyle \frac{d}{dx}[F(x)] = f(x)$.

Clearly, since $\displaystyle f(x) = \ln{x}$, this means $\displaystyle F(x) = x\ln{x} - x$.

So the area of one strip is...

$\displaystyle x_k\ln{x_k} - x_k - (x_{k - 1}\ln{x_{k - 1}} - x_{k - 1})$

$\displaystyle = x_k\ln{x_k} - x_k - x_{k - 1}\ln{x_{k - 1}} + x_{k - 1}$.

If we have $\displaystyle n$ strips, then the Area between $\displaystyle x_0$ and $\displaystyle x_n$ is

$\displaystyle A = \sum_{k = 1}^n{(x_k\ln{x_k} - x_k - x_{k - 1}\ln{x_{k - 1}} + x_{k - 1})}$

$\displaystyle = x_n\ln{x_n} - x_n - x_0\ln{x_0} + x_0$, since this is a telescopic series.

Since $\displaystyle x_0 = 1$ and $\displaystyle x_n = t$, you have

$\displaystyle A = t\ln{t} - t - 1\ln{1} + 1$

$\displaystyle = t\ln{t} - t + 1$.

To fully complete this, you would need to note that you make $\displaystyle n \to \infty$ and then take the limit.

Can you evaluate

$\displaystyle \lim_{n \to \infty}(t\ln{t} - t + 1)$?

10. Originally Posted by Prove It
OK, I've done it, but I don't think you'll like the solution...

For this to work, we need to make use of the Mean Value Theorem.

i.e. for some function $\displaystyle F(x)$, which is continuous on $\displaystyle x \in [a, b]$, we have

$\displaystyle f(c) = \frac{F(b) - F(a)}{b - a}$, where $\displaystyle a < c < b$ and $\displaystyle \frac{d}{dx}[F(x)] = f(x)$.

We can rearrange the formula so that it is of the form

$\displaystyle (b - a)f(c) = F(b) - F(a)$.

For this function, $\displaystyle f(x) = \ln{x}$, suppose we take some rectangular strip, between $\displaystyle x_{k - 1}$ and $\displaystyle x_k$.

Then this strip's area is

$\displaystyle (x_k - x_{k - 1})f(c)$, for some $\displaystyle c \in (x_{k - 1}, x_k)$.

Using the Mean Value Theorem, we can rewrite this as

$\displaystyle F(x_k) - F(x_{k - 1})$, where $\displaystyle \frac{d}{dx}[F(x)] = f(x)$.

Clearly, since $\displaystyle f(x) = \ln{x}$, this means $\displaystyle F(x) = x\ln{x} - x$.

So the area of one strip is...

$\displaystyle x_k\ln{x_k} - x_k - (x_{k - 1}\ln{x_{k - 1}} - x_{k - 1})$

$\displaystyle = x_k\ln{x_k} - x_k - x_{k - 1}\ln{x_{k - 1}} + x_{k - 1}$.

If we have $\displaystyle n$ strips, then the Area between $\displaystyle x_0$ and $\displaystyle x_n$ is

$\displaystyle A = \sum_{k = 1}^n{(x_k\ln{x_k} - x_k - x_{k - 1}\ln{x_{k - 1}} + x_{k - 1})}$

$\displaystyle = x_n\ln{x_n} - x_n - x_0\ln{x_0} + x_0$, since this is a telescopic series.

Since $\displaystyle x_0 = 1$ and $\displaystyle x_n = t$, you have

$\displaystyle A = t\ln{t} - t - 1\ln{1} + 1$

$\displaystyle = t\ln{t} - t + 1$.

To fully complete this, you would need to note that you make $\displaystyle n \to \infty$ and then take the limit.

Can you evaluate

$\displaystyle \lim_{n \to \infty}(t\ln{t} - t + 1)$?
Many thanks Prove It, but as you mentioned I did not like it, cuz I dont see any Reimann sum

By the way, is it your style to talk like that all the time?
Originally Posted by Prove It
Can you evaluate
$\displaystyle \lim_{{\color{red}{n \to \infty}}}({\color{blue}{t\ln{t} - t + 1}})$?
Sorry, forgive me since I am not very used to it...

11. Something similar was posed a good while back. Here is the link if you're interested.

http://www.mathhelpforum.com/math-he...tml#post248858

12. Originally Posted by bkarpuz
Many thanks Prove It, but as you mentioned I did not like it, cuz I dont see any Reimann sum

By the way, is it your style to talk like that all the time?

Sorry, forgive me since I am not very used to it...
Yes it is my style.

I think you might need to refresh your memory as to what a Riemann sum actually is, as what I provided is, in fact, a Riemann sum.

13. Originally Posted by Prove It
Yes it is my style.

I think you might need to refresh your memory as to what a Riemann sum actually is, as what I provided is, in fact, a Riemann sum.
You are actually right that I need to refresh my mind about Reimann sum.
I think you did not read what I wrote in the topic, I dont want to calculate the integral with formulas, I want to show that the Reimann sum converges to $\displaystyle t\ln(t)-t+1$.
If you dont have any answer, please dont mess the topic, but if you really want to help then you are welcome.
thanks.

14. Bkarpuz . . . that's a nice Riemann sum above. He wrote:

Originally Posted by Prove It

Then this strip's area is

$\displaystyle (x_k - x_{k - 1})f(c)$, for some $\displaystyle c \in (x_{k - 1}, x_k)$.

Using the Mean Value Theorem, we can rewrite this as

$\displaystyle F(x_k) - F(x_{k - 1})$, where $\displaystyle \frac{d}{dx}[F(x)] = f(x)$.

Clearly, since $\displaystyle f(x) = \ln{x}$, this means $\displaystyle F(x) = x\ln{x} - x$.

So the area of one strip is...

$\displaystyle x_k\ln{x_k} - x_k - (x_{k - 1}\ln{x_{k - 1}} - x_{k - 1})$

$\displaystyle = x_k\ln{x_k} - x_k - x_{k - 1}\ln{x_{k - 1}} + x_{k - 1}$.

If we have $\displaystyle n$ strips, then the Area between $\displaystyle x_0$ and $\displaystyle x_n$ is

$\displaystyle A = \sum_{k = 1}^n{(x_k\ln{x_k} - x_k - x_{k - 1}\ln{x_{k - 1}} + x_{k - 1})}$
Also, I personally saw nothing wrong with how it was written. That's just math-speak. We all do it and it's nice, brief, concise, and to the point. Maybe you can reconsider.

15. Originally Posted by shawsend
Bkarpuz . . . that's a nice Riemann sum above. He wrote:

Also, I personally saw nothing wrong with how it was written. That's just math-speak. We all do it and it's nice, brief, concise, and to the point. Maybe you can reconsider.
@shawsend: It is not a Reimann sum it is just a (telescopic) sum.
If someone knows $\displaystyle F$ which is the antiderivative of $\displaystyle f$, then one can easily write
$\displaystyle F(x_{n})-F(x_{0})=\sum_{i=0}^{n-1}\big[F(x_{i+1})-F(x_{i})\big]$ and then let $\displaystyle x_{n}=t$ and $\displaystyle x_{0}=s$.

He did not write anything wrong but it is not related to the solution.

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