- #1

mathmari

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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)