About Euler's constant (1)...

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October 26th 2009, 12:18 AM
chisigma

About Euler's constant (1)...

It is not a secret for anyone that I'm a 'fan' of Leonhard Euler (Clapping) . One of his most beautiful 'discoveries' is the complex of properties of the constant that brings his name...

$\gamma = \lim_{k \rightarrow \infty} 1 + \frac{1}{2} + \dots + \frac{1}{k} - \ln k = .57721566490153 \dots$ (1)

The first problem concening Euler's constant I will propose to You is : starting from definition (1) demonstrate that is...

$\int_{0}^{\infty} \ln t\cdot e^{-t}\cdot dt = - \gamma$ (2)

Kind regards

$\chi$ $\sigma$
November 2nd 2009, 05:13 AM
PaulRS

See my 2 posts here.
November 2nd 2009, 11:26 PM
simplependulum

Here is another way to obtain this identity :
$\int_{0}^{\infty} \ln t\cdot e^{-t}\cdot dt = - \gamma$

Consider

$\int_0^{\infty} e^{-t} \ln(t) ~dt$

$= \lim_{n\to\infty} \int_0^{n} ( 1 - \frac{t}{n} )^n \ln(t) ~dt$

Sub $1 - \frac{t}{n} = y$ , $n \to \infty$

$dt = -ndy$

the integral becomes

$\lim_{n\to\infty} n \int_0^1 y^n [ \ln(n) + \ln(1-y)]~dy$

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

$= \lim_{n\to\infty} \left (\ln(n)- [ \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{n+1} ] \right )$

ps : There is a proof which states that $\int_0^{\infty} e^{-t} \ln(t) ~dt = \lim_{n\to\infty} \int_0^n ( 1- \frac{t}{n} )^n \ln(t) ~dt$ but i don't think it is important really ...
November 3rd 2009, 05:36 AM
chisigma

Congratulations for Paul and SP!...

A similar but more 'general purpose' way to arrive at the result is the use of the following 'nice lilttle formula'. From the well known definition of [negative] exponential function...

$e^{-t} = \lim_{k \rightarrow \infty} (1-\frac{t}{k})^{k}$ (1)

... derives 'spontaneously' the formula...

$\int_{0}^{\infty} f(t)\cdot e^{-t}\cdot dt= \lim_{k \rightarrow \infty} \int_{0}^{k} f(t)\cdot (1-\frac{t}{k})^{k}\cdot dt$ (2)

... that with the substitution $u=\frac{t}{k}$ becomes...

$\int_{0}^{\infty} f(t)\cdot e^{-t}\cdot dt= \lim_{k \rightarrow \infty} k \cdot \int_{0}^{1} f(ku)\cdot (1-u)^{k}\cdot du$ (3)

... and with the substitution $u=1-\frac{t}{k}$ becomes...

$\int_{0}^{\infty} f(t)\cdot e^{-t}\cdot dt= \lim_{k \rightarrow \infty} k \cdot \int_{0}^{1} f \{k(1-u)\}\cdot u^{k}\cdot du$ (4)

Now if in (4) we set $f(t)= \ln t$ we obtain...

$\int_{0}^{\infty}\ln t\cdot e^{-t}\cdot dt= \lim_{k \rightarrow \infty} k\cdot \ln k \cdot \int_{0}^{1}u^{k}\cdot du + k\cdot \int_{0}^{1} \ln (1-u)\cdot u^{k}\cdot du=$

$= \lim_{k \rightarrow \infty} \frac{k}{k+1}\cdot \ln k - k \cdot \sum_{n=1}^{\infty} \frac{1}{n}\cdot \int_{0}^{1} u^{k+n}\cdot du =$

$= \lim_{k \rightarrow \infty} \ln k - k\cdot \sum_{n=1}^{\infty} \frac{1}{n\cdot (k+n)}$ (5)

If now we remember the identity...

$\sum_{n=1}^{\infty} \frac{1}{n\cdot (k+n)} = \frac{1}{k} \sum_{n=1}^{k} \frac{1}{n}$ (6)

... we arrivwe at the final result...

$\int_{0}^{\infty} \ln t\cdot e^{-t}\cdot dt = \lim_{k \rightarrow \infty} (\ln k - \sum_{n=1}^{k}\frac{1}{n}) = -\gamma$ (7)

Kind regards

$\chi$ $\sigma$
November 4th 2009, 05:06 PM
Drexel28

Quote:

Originally Posted by chisigma

$\int_{0}^{\infty} f(t)\cdot e^{-t}\cdot dt= \lim_{k \rightarrow \infty} \int_{0}^{k} f(t)\cdot (1-\frac{t}{k})^{k}\cdot dt$ (2)

Isn't the above kind of a deep statement? Wouldn't you need to at least supply some reasoning for why it's true?
November 4th 2009, 08:08 PM
chisigma

$e^{-t}=\lim_{k \rightarrow \infty} (1-\frac{t}{k})^{k} \rightarrow$

$\rightarrow \int_{0}^{\infty} f(t)\cdot e^{-t}\cdot dt = \lim_{k \rightarrow \infty} \int_{0}^{\infty} f(t)\cdot (1-\frac{t}{k})^{k}\cdot dt \rightarrow$

$\rightarrow \int_{0}^{\infty} f(t)\cdot e^{-t}\cdot dt = \lim_{k \rightarrow \infty} \int_{0}^{k} f(t)\cdot (1-\frac{t}{k})^{k}\cdot dt$

Kind regards

$\chi$ $\sigma$
November 4th 2009, 08:41 PM
Bruno J.

Why can you pull the limit out of the integral sign? I also believe you can but I'm not quite sure I know how to justify it.
November 4th 2009, 08:51 PM
Drexel28

Quote:

Originally Posted by Bruno J.

Why can you pull the limit out of the integral sign? I also believe you can but I'm not quite sure I know how to justify it.

That was my point. When it is permissable for extracting a limit from an integrand to the exterior is nontrivial.

And saying that $\lim_{n\to\infty}\int_0^{\infty}f_n(x)=\lim_{n\to\ infty}\int_0^{n}f_n(x)$ is also highly suspect.

Take for example $f_n(x)=\frac{1}{n}$

Then $\lim_{n\to\infty}\int_0^{\infty}\frac{dx}{n}=\inft y$

but $\lim_{n\to\infty}\int_0^{n}\frac{dx}{n}=1$
November 4th 2009, 10:54 PM
chisigma

Quote:

Originally Posted by Drexel28

That was my point. When it is permissable for extracting a limit from an integrand to the exterior is nontrivial.

And saying that $\lim_{n\to\infty}\int_0^{\infty}f_n(x)=\lim_{n\to\ infty}\int_0^{n}f_n(x)$ is also highly suspect.

Take for example $f_n(x)=\frac{1}{n}$

Then $\lim_{n\to\infty}\int_0^{\infty}\frac{dx}{n}=\inft y$

but $\lim_{n\to\infty}\int_0^{n}\frac{dx}{n}=1$

The correct procedure requires to remember that by definition is...

$a_{n} = \int_{0}^{\infty} f_{n} (x)\cdot dx = \lim_{t \rightarrow \infty} \int_{0}^{t} f_{n}(x)\cdot dx$ (1)

... so that the limit...

$\lim_{n \rightarrow \infty} a_{n}$ (2)

... exists if $\forall n$ the limit (1) exists. In your example is...

$a_{n} = \lim_{t \rightarrow \infty} \int_{0}^{t} \frac{1}{n}\cdot dx$ (3)

... and none of the $a_{n}$ exists so that the expression...

$\lim_{n\to\infty}\int_0^{\infty}\frac{dx}{n}=\inft y$ (4)

... is a nonsense. Anyway your observation is probably correct so that I propose to overcame any difficulty with the following formal statement...

$\lim_{t \rightarrow \infty} \int_{0}^{t} f(x)\cdot e^{-x}\cdot dx = \lim_{k \rightarrow \infty} \int_{0}^{k} f(x)\cdot (1- \frac{x}{k})^{k}\cdot dx$ (5)

Kind regards

$\chi$ $\sigma$
November 4th 2009, 11:36 PM
simplependulum

Let $f_n(z) = \int_0^n \left ( 1 - \frac{t}{n} \right )^n t^{z-1} ~dt$

and we have

$\Gamma(z) - f_n(z) = \int_0^n \left ( e^{-t} - ( 1 - \frac{t}{n} )^n \right ) t^{z-1}~dt + \int_n^{\infty} e^{-t} t^{z-1}~dt$

now , we prove this inequality

$0 \leq e^{-t} - \left ( 1 - \frac{t}{n} \right )^n \leq \frac{t^2}{n}e^{-t}$

consider

$1 + y \leq \ e^{y} \leq \frac{1}{1 - y }$

Replace $y$ with $\frac{t}{n}$

$\left ( 1 - \frac{t}{n} \right )^n \leq e^{-t} \leq \left ( 1 + \frac{t}{n} \right )^{-n}$

Therefore ,

$0 \leq e^{-t} - \left ( 1 - \frac{t}{n} \right )^n \leq e^{-t} \left ( 1 - ( 1 - \frac{t^2}{n^2} )^n \right )$

but $( 1 - x )^n \geq 1 - nx$

therefore

$1 - \left ( 1 - \frac{t^2}{n^2} \right )^n \leq \frac{t^2}{n^2}$

$0 \leq e^{-t} - \left ( 1 - \frac{t}{n} \right )^n \leq \frac{t^2}{n}e^{-t}$

After proving the inequality , Take limit $n \to \infty$

we can obtain $\lim_{n\to\infty} f_n(z) = \Gamma(z)$
November 5th 2009, 01:04 AM
Laurent

Quote:

Originally Posted by chisigma

Anyway your observation is probably correct so that I propose to overcame any difficulty with the following formal statement...

$\lim_{t \rightarrow \infty} \int_{0}^{t} f(x)\cdot e^{-x}\cdot dx = \lim_{k \rightarrow \infty} \int_{0}^{k} f(x)\cdot (1- \frac{x}{k})^{k}\cdot dx$ (5)

This is in not really more self evident than the previous assertion. However, simplependulum provides us with a proof. Indeed,

Quote:

Originally Posted by simplependulum

now , we prove this inequality

$0 \leq e^{-t} - \left ( 1 - \frac{t}{n} \right )^n \leq \frac{t^2}{n}e^{-t}$

(this also holds for non-integer values of $n$ as soon as they're greater than 1, cf. the proof)

Hence, from this inequality, for $t>1$ ,

$\left|\int_{0}^{t} f(x)\cdot e^{-x}\cdot dx - \int_{0}^{t} f(x)\cdot (1- \frac{x}{t})^t\cdot dx\right|$ $\leq \frac{1}{t}\int_0^t f(x)x^2 e^{-x} dx\leq \frac{1}{t}\int_0^\infty f(x) x^2e^{-x}dx\to_{t\to\infty} 0$ .

(Provided the integral is finite, so under mild hypotheses on $f$ )
November 5th 2009, 02:55 AM
simplependulum

I copied this proof from a book which talks about elements of complex analysis , almost the last chapter .

Perhaps this excellent proof is completed by some GREAT Mathematicians , if i have enough time , i will complete the proof here . We can see that some parts of the original proof are missing ( e.g. proof of $( 1 - x )^n \geq 1 - nx$ ) .
November 5th 2009, 03:01 AM
simplependulum

Quote:

Hence, from this inequality, for $t>1$ ,

$\left|\int_{0}^{t} f(x)\cdot e^{-x}\cdot dx - \int_{0}^{t} f(x)\cdot (1- \frac{x}{t})^t\cdot dx\right|$ $\leq \frac{1}{t}\int_0^t f(x)x^2 e^{-x} dx\leq \frac{1}{t}\int_0^\infty f(x) x^2e^{-x}dx\to_{t\to\infty} 0$ .

(Provided the integral is finite, so under mild hypotheses on $f$ )

This inequality appears at the end of the proof in my book ! This is the main point , when n tends to infinity , their difference ( $f_n(z) - \Gamma(z)$ ) tends to zero .

Great !
November 5th 2009, 11:10 AM
Moo

Quote:

Originally Posted by Bruno J.

Why can you pull the limit out of the integral sign? I also believe you can but I'm not quite sure I know how to justify it.

Dominated convergence theorem - Wikipedia, the free encyclopedia
or Monotone convergence theorem - Wikipedia, the free encyclopedia