Talk:Aufgaben:Problem 15

Let \(t>0\) fixed. Define \( f(x) := x^{t-1}e^{-x} \), non-negative function on (0,\(\infty\)).

Part a)

Problem

For \(k = 1,2,... \) let $$ f_k(x) = \left\{ \begin{array}{l l} x^{t-1}(1-\frac{x}{k})^{k} & \quad 0<x<k\\ 0 & \quad k\leq x<\infty\end{array} \right. $$ Show that \(f_k(x) \rightarrow f(x) \) and \(f_k(x) \leq f_{k+1}(x), \forall x>0.\)

Hint: use the arithmetic-geometric mean inequality. You may not use the definition of the exponential function as a limit.

Solution

To show:

$$ \lim_{k \to \infty} f_{k}(x) = f(x), \forall x>0 $$

Proof:

Take any \( x \in (0 ; \infty) \) and \( k > x \) ; consider:

$$ \ln((1-\frac{x}{k})^{k}) = k\ln(1-\frac{x}{k}) = \frac{x}{x}k(\ln(1-\frac{x}{k})-\ln(1)) = -x\dfrac{(\ln(1)-\ln(1-\frac{x}{k}))}{\frac{x}{k}} $$

The last fraction is the difference quotient of ln evaluated in 1, thus:

$$ \lim_{k \to \infty} \ln((1-\frac{x}{k})^{k}) = -x\dfrac{d}{dy}\ln(y)\mid_{y=1} = -x $$

An alternative version would be: \( \lim_{k \rightarrow \infty} \ln((1-\frac{x}{k})^{k}) = \lim_{k \rightarrow \infty} k\ln(1-\frac{x}{k}) = \lim_{k \rightarrow \infty} \frac{\ln(1-\frac{x}{k})}{\frac{1}{k}} \) and then use Bernoulli-L'Hopital to get: \( \lim_{k \rightarrow \infty} \frac{\ln(1-\frac{x}{k})}{\frac{1}{k}} = \lim_{k \rightarrow \infty} \frac{\frac{x}{k^2}}{(\frac{x}{k} - 1) \frac{1}{k^2}} = -x \)

Since exp is continuous, we can "bring the limes inside":

$$ \Rightarrow \lim_{k \to \infty} f_{k}(x) = x^{t-1} \lim_{k \to \infty} \exp ( \ln((1-\frac{x}{k})^{k}) = x^{t-1} \exp(\lim_{k \to \infty} ln((1-\frac{x}{k})^{k}) = x^{t-1} e^{-x} = f(x) $$

\( \blacksquare \)

To show:

$$ f_{k}(x) \leq f_{k+1}(x), \forall x > 0 $$

Proof:

Case 1: \( x \in \: ]0;k[ \)

The arithmetic-geometric mean equality says that \( \forall x_{1},...,x_{n} \geq 0 \) it holds: $$ (x_{1}...x_{n})^{\frac{1}{n}} \leq \dfrac{x_{1}+...+x_{n}}{n} \tag{$\dagger$}$$

Consider:

$$ ((1-\dfrac{x}{k})^{k}\cdot1)^{\frac{1}{k+1}} \overset{(\dagger)}{\leq} \dfrac{1}{k+1}(k(1-\dfrac{x}{k})+1) = \dfrac{1}{k+1}(k + 1 - x) = (1 - \dfrac{x}{k+1}) $$

$$\Rightarrow (1-\dfrac{x}{k})^{k} \leq (1 - \dfrac{x}{k+1})^{k+1} $$

Where we don't have any problem with the k+1-root because \( (1-\dfrac{x}{k})^{k} \) only takes real positive values, since \( 0 < x < k. \)

Case 2: \( x \in [k;k+1[ \)

The inequality holds since \( f_{k}(x) = 0 \) and \( f_{k+1}(x) = x^{t-1}(1 - \dfrac{x}{k+1})^{k+1} \) is a positve function \( \forall x \in [k;k+1[. \)

Case 3: \( x \in [k+1;\infty[ \)

Both functions are equal zero so the inequality still holds.

\( \blacksquare \)

Part b)

Problem

Use the monotone convergence theorem to show that $$ \Gamma(t) = \lim_{k \to \infty} \frac{k!k^{t}}{t(t+1)...(t+k)}, t>0 $$ where \( \Gamma(t) \) is the Euler Gamma function.

Hint: consider \( I_{k} = k^{t} \int_0^1 \! u^{t-1}(1-u)^{k}du \).

Solution

Idea: In a first step we show that \( I_{k} \) is equal to the fraction in the right hand side of the equality. Then we rewrite \( I_{k} \) in a more convenient way, looking for something containing the functions given in part a). Eventually we let \( k\rightarrow\infty \) and apply Lebesgue's monoton convergence theorem to conclude.

$$ I_{k} = k^{t} \int_0^1 \! u^{t-1}(1-u)^{k} \, \mathrm{d}u \overset{P.I.}{=} \\ =k^{t}\lbrace\dfrac{u^{t}}{t}(1-u)^{k}\vert_{0}^{1} - \int_0^1 \! \dfrac{u^{t}}{t}(-1)k(1-u)^{k-1} \, \mathrm{d}u \rbrace = \\ = k^{t}\lbrace 0 + \dfrac{k}{t}\int_0^1 \! u^{t}(1-u)^{k-1} \, \mathrm{d}u \rbrace =\\ = \dfrac{k^{t+1}}{t}\int_0^1 \! u^{t}(1-u)^{k-1} \, \mathrm{d}u \overset{P.I.}{=}\\ =\dfrac{k^{t+1}}{t}\lbrace\dfrac{u^{t+1}}{t+1}(1-u)^{k-1}\vert_{0}^{1} - \int_0^1 \! \dfrac{u^{t+1}}{t+1}(-1)(k-1)(1-u)^{k-2} \, \mathrm{d}u \rbrace = \\ = \dfrac{k^{t+1}}{t}\lbrace 0 + \dfrac{k-1}{t+1}\int_0^1 \! u^{t+1}(1-u)^{k-2} \, \mathrm{d}u \rbrace =\dfrac{k^{t+1}(k-1)}{t(t+1)} \int_0^1 \! u^{t+1}(1-u)^{k-2} \, \mathrm{d}u \overset{P.I.}{=}\\ = ... = \frac{(k-1)!k^{t+1}}{t(t+1)...(t+k-1)} \int_0^1 \! u^{t+k-1} \, \mathrm{d}u = \frac{(k-1)!k^{t+1}}{t(t+1)...(t+k-1)}\dfrac{u^{t+k}}{t+k}\vert^{1}_{0} = \frac{k!k^{t}}{t(t+1)...(t+k)} $$

Where we have used k times partial integration to reach the last but one step.

$$ I_{k} = k^{t} \int_0^1 \! u^{t-1}(1-u)^{k} \, \mathrm{d}u = k^{t-1} \int_0^k \! (\dfrac{x}{k})^{t-1}(1-\dfrac{x}{k})^{k} \, \mathrm{d}x = \\ = \int_0^k \! x^{t-1}(1-\dfrac{x}{k})^{k} \, \mathrm{d}x = \int_0^k \! x^{t-1}(1-\dfrac{x}{k})^{k} \, \mathrm{d}x + \int_k^\infty \! 0 \, \mathrm{d}x = \int_0^k \! f_{k}(x) \, \mathrm{d}x + \int_k^\infty \! f_{k}(x) \, \mathrm{d}x =\\ = \int_0^\infty \! f_{k}(x) \, \mathrm{d}x $$

Where the first step is a change of variable \( u = \dfrac{x}{k} \).

Letting \( k \rightarrow\infty \), using the monoton convergence theorem on \( f_{k}(x) \), we obtain:

$$ I_{\infty} = \int_0^\infty \! x^{t-1}e^{-x}\, \mathrm{d}x = \Gamma(t) $$

Where the conditions to apply the monoton convergence theorem on \( \Omega := ]0,\infty[ \) are satisfied (see part a), because: \( f_{k}(x) \) are non-negative, monotonically increasing functions \( \in L^1(\Omega) \) (because non-zero only on a bounded subset \( ]0,k[ \) and continuous).

Lebesgue monotone convergence theorem:

Let \( m \in \mathbb{N}, \Omega \subset \mathbb{R}^m, f_n \) be a sequence of functions \( \in L^1(\Omega) \) so that \( 0 \leq f_n(x) \leq f_{n+1}(x) \: \forall n \in \mathbb{N}, \: \forall x \in \Omega \) and \( \lim_{n \to \infty} f_n(x) =: f(x) \: \forall x \in \Omega \).

Then \( f \in L^1(\Omega) \) and \( \lim_{n \to \infty} \int_\Omega f_n(x)dx = \int_\Omega f(x)dx. \)

Note: some versions of the monotone convergence theorem require \( f \) to be integrable. For a proof that \( f \in L^1(\Omega) \) see the discussion.

\( \blacksquare \)