MHB Gamma function is convergent and continuous

[mathmari]

Hey!

I want to show that the Gamma function converges and is continuous for $x>0$. I have done the following:

The Gamma function is the integral \begin{equation*}\Gamma (x)=\int_0^{\infty}t^{x-1}e^{-t}\, dt\end{equation*}

Let $x>0$.

It holds that \begin{equation*}\int_0^{\infty}t^{x-1}e^{-t}\, dt=\int_0^{1}t^{x-1}e^{-t}\, dt+\int_1^{\infty}t^{x-1}e^{-t}\, dt\end{equation*}

For $t\in (0,1]$ it holds that $|t^{x-1}e^{-t}|\leq t^{x-1}$ and the integral $\displaystyle{\int_0^1t^{x-1}\,dt}$ converges.

Therefore, the integral $\displaystyle{\int_0^1t^{x-1}e^{-t}\, dt}$ converges according to the comparison test.

The function $t\mapsto t^{x+1}e^{-t}$ is continuous on $[1,\infty )$ and $\displaystyle{\lim_{t\rightarrow \infty}t^{x+1}e^{-t}=0}$. That means that this function is bounded on $[1,\infty)$, or not? (Wondering)
That would mean that there is a constant $c\in (0,\infty)$ with \begin{equation*}|t^{x-1}e^{-t}|=|t^{x+1-2}e^{-t}|=\left |\frac{1}{t^2}t^{x+1}e^{-t}\right |\leq \frac{1}{t^2}\left |t^{x+1}e^{-t}\right |\leq \frac{1}{t^2}c=\frac{c}{t^2}\end{equation*}
The integral $\displaystyle{\int_1^{\infty}\frac{c}{t^2}\, dt}$ converges. From the comparison test, the integral $\displaystyle{\int_1^{\infty}t^{x-1}e^{-t}\, dt}$ converges.

That implies that the integral $\displaystyle{\int_0^{\infty}t^{x-1}e^{-t}\, dt}$ converges for $x\in (0,\infty)$.
Is everythijng correct? Could I improve something?

But how could we show that the function is continuous?

(Wondering)

 

[mathmari]

We have that \begin{equation*}\Gamma(x)-\Gamma(y)=\int_0^{\infty}t^{x-1}e^{-t}dt-\int_0^{\infty}t^{y-1}e^{-t}dt=\int_0^{\infty}(t^{x-1}-t^{y-1})e^{-t}dt\end{equation*}

Let $g(x) = t^{x-1}$. Then it holds that $|g(x)-g(y)|=|t^{x-1}-t^{y-1}|$

We apply the Mean value theorem with the interval $[x,y]$.

Let $\xi \in (x,y)$ then it holds that:
\begin{equation*}g'(\xi)=\frac{g(x)-g(y)}{x-y}\Rightarrow t^{\xi -1}\cdot \ln t=\frac{t^{x-1}-t^{y-1}}{x-y}\end{equation*}
Therefore we get:
\begin{equation*}t^{x-1}-t^{y-1}=t^{\xi -1}\cdot \ln t\cdot (x-y) \Rightarrow |t^{x-1}-t^{y-1}|=|t^{\xi -1}\cdot \ln t\cdot (x-y)|\leq |t^{\xi -1}\cdot \ln t|\cdot |x-y|\end{equation*}

So, we get:
\begin{align*}|\Gamma(x)-\Gamma(y)|&=\left |\int_0^{\infty}(t^{x-1}-t^{y-1})e^{-t}dt\right | \\ & \leq \int_0^{\infty}|(t^{x-1}-t^{y-1})e^{-t}|dt \\ & =\int_0^{\infty}|t^{x-1}-t^{y-1}|e^{-t}dt \\ & \leq \int_0^{\infty}|t^{\xi -1}\cdot \ln t|\cdot |x-y|\cdot e^{-t}dt \\ & =|x-y|\cdot \int_0^{\infty}|t^{\xi -1}\cdot \ln t|\cdot e^{-t}dt\end{align*}

We want to show that the integral is finite, or not?

We have that \begin{equation*}\int_0^{\infty}|t^{\xi -1}\cdot \ln t|\cdot e^{-t}dt=\int_0^{1}|t^{\xi -1}\cdot \ln t|\cdot e^{-t}dt+\int_1^{\infty}|t^{\xi -1}\cdot \ln t|\cdot e^{-t}dt\end{equation*}

When $0\leq t\leq 1$ then $t^{\xi-1}\geq 0$ and $\ln t\leq 0$. So we get \begin{equation*}\int_0^{1}|t^{\xi -1}\cdot \ln t|\cdot e^{-t}dt=-\int_0^{1}t^{\xi -1}\cdot \ln t\cdot e^{-t}dt\end{equation*}

When $ t\geq 1$ then $t^{\xi-1}\geq 0$ and $\ln t\geq 0$. So we get \begin{equation*}\int_1^{\infty}|t^{\xi -1}\cdot \ln t|\cdot e^{-t}dt=\int_1^{\infty}t^{\xi -1}\cdot \ln t\cdot e^{-t}dt\end{equation*} How could we continue? (Wondering)

 

[I like Serena]

If t>1, then t>ln t, isn't it?
And if 0<t<1, then -ln t<ln 1/t<1/t, isn't it? (Wondering)

 

[mathmari]

If t>1, then t>ln t, isn't it?
And if 0<t<1, then -ln t<ln 1/t<1/t, isn't it? (Wondering)

Ah ok!

If $0\leq t\leq 1$ then $t^{\xi-1}\geq 0$ and $\ln t\leq 0$. In this case it also holds that $ -\ln t=\ln \frac{1}{t}<\frac{1}{t}$. So we get \begin{align*}\int_0^{1}|t^{\xi -1}\cdot \ln t|\cdot e^{-t}dt&=\int_0^{1}t^{\xi -1}\cdot \left (-\ln t\right )\cdot e^{-t}dt \\ & <\int_0^{1}t^{\xi -1}\cdot \frac{1}{t}\cdot e^{-t}dt \\ & =\int_0^{1}t^{\xi -2}\cdot e^{-t}dt \end{align*}

If $ t\geq 1$ then $t^{\xi-1}\geq 0$ and $\ln t\geq 0$. In this case it also holds that $\ln t<t$. So we get \begin{equation*}\int_1^{\infty}|t^{\xi -1}\cdot \ln t|\cdot e^{-t}dt=\int_1^{\infty}t^{\xi -1}\cdot \ln t\cdot e^{-t}dt<\int_1^{\infty}t^{\xi -1}\cdot t\cdot e^{-t}dt=\int_1^{\infty}t^{\xi }\cdot e^{-t}dt\end{equation*}

Can we calculate these two integrals? (Wondering)

 

[I like Serena]

Can't we do the same thing as you did in oost #1? (Wondering)

 

[mathmari]

Can't we do the same thing as you did in oost #1? (Wondering)

Oh yes!

So, we have that \begin{equation*}|\Gamma(x)-\Gamma(y)| \leq|x-y|\cdot \int_0^{\infty}|t^{\xi -1}\cdot \ln t|\cdot e^{-t}dt\end{equation*} For $|x-y|\rightarrow 0$ the right side goes to $0$, as the integral is finite. That means that $|\Gamma(x)-\Gamma(y)|$ goes to $0$ and this implies that the function $\Gamma (x)$ is continuous, right? (Wondering)

 

[I like Serena]

Yep. (Happy)

 

[mathmari]

Yep. (Happy)

Great! Thank you very much! (Clapping)