# Gamma function is convergent and continuous

• MHB
• mathmari
In summary, the conversation discusses the convergence and continuity of the Gamma function for values of x greater than 0. The speaker presents the integral representation of the Gamma function and uses the comparison test to show its convergence. They also use the Mean value theorem to prove the function's continuity, with the conclusion that the function is indeed continuous for values of x greater than 0. The conversation ends with appreciation and confirmation of the results.
mathmari
Gold Member
MHB
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)

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)

Last edited by a moderator:
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)

I like Serena said:
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)

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

I like Serena said:
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)

Yep. (Happy)

I like Serena said:
Yep. (Happy)

Great! Thank you very much! (Clapping)

## 1. What is the Gamma function?

The Gamma function is a mathematical function that extends the factorial function to real and complex numbers.

## 2. Why is it important that the Gamma function is convergent and continuous?

The Gamma function being convergent and continuous means that it can be used in a wide range of mathematical applications, including probability theory, statistics, and physics. It also allows for easier integration and approximation of complex functions.

## 3. How is the convergence of the Gamma function proven?

The convergence of the Gamma function is proven using the Euler integral of the Gamma function, which is a special case of the more general Mellin transform.

## 4. What is the significance of the Gamma function being continuous?

The continuity of the Gamma function allows for smooth and continuous changes in its output as the input values change. This makes it a useful tool in analyzing and modeling various phenomena in science and engineering.

## 5. Can the Gamma function still be useful if it is not convergent or continuous?

While the Gamma function is most commonly used in its convergent and continuous form, there are alternative versions of the function that can be used in certain scenarios where the traditional version is not applicable. However, these versions may have limitations and may not be as widely used or accepted in the scientific community.

• Topology and Analysis
Replies
29
Views
2K
• Topology and Analysis
Replies
4
Views
508
• Topology and Analysis
Replies
3
Views
1K
• Topology and Analysis
Replies
4
Views
544
• Topology and Analysis
Replies
4
Views
1K
• Topology and Analysis
Replies
4
Views
485
• Topology and Analysis
Replies
11
Views
1K
• Calculus
Replies
4
Views
997
• Calculus and Beyond Homework Help
Replies
2
Views
374
• Topology and Analysis
Replies
2
Views
2K