Proving Gamma function relation

Question

Prompt. Using integration by parts, show that the gamma function $$\Gamma (t) = \int_0^\infty x^{t-1} e^{-x} \, dx $$ satisfies the relation $t\Gamma (t) = \Gamma (t+1)$ for $t > 0$.

My solution. Let $ u = e^{-x}$, $du = - e^{-x}\,dx$, $v = \frac 1t x^t$ , $dv = x^{t-1} \,dx$. Then after applying integration by parts, we get $$\Gamma (t) = \frac 1t x^t e^{-x} + \frac 1t \int_0^\infty x^t e^{-x} \, dx$$ and subsequently $t\Gamma(t) = x^t e^ {-x} + \int_0^\infty x^t e^{-x} \, dx $ . Now, $\Gamma(t + 1) = \int_0^\infty x^t e^{-x} \, dx$.

We can rewrite $t\Gamma (t) = \Gamma'(t) + \Gamma (t + 1)$ .

Am I doing something wrong? Am I on the right track?

Answer 1

I'm sorry to say, but those are the wrong substitutiongs. By integration by parts, set$$u=x^t\qquad\qquad\mathrm du=tx^{t-1}\,\mathrm dx$$And therefore, we have$$\begin{align*}\Gamma(z+1) & =\int\limits_{0}^{\infty}e^{-t}t^z\,\mathrm dt\\ & =-t^ze^{-t}\,\biggr\rvert_{0}^{\infty}+\int\limits_{0}^{\infty}ze^{-t}t^{z-1}\,\mathrm dt\end{align*}$$The first term evaluates to zero. You can see this by taking the limit, and then using L'Hopital's. Therefore, we're left with$$\begin{align*}\Gamma(z+1) & =z\int\limits_{0}^{\infty}e^{-t}t^{z-1}\,\mathrm dt\\ & =z\Gamma(z)\end{align*}$$

Answer 2

You need to do it the other way around: \begin{align} u & = x^t \\[6pt] du & = t x^{t-1} \, dx \\[6pt] dv & = e^{-x}\,dx \\[6pt] v & = -e^{-x} \end{align}

That way you get $t$ as a factor in front of the integral instead of $1/t,$ and that's what you need.

You will also need to show that $\displaystyle \left. \phantom{\frac 1 1} uv\, \right|_{x\,=\,0}^{x\,=\,\infty} = 0.$ One way of doing that uses L'Hopital's rule.