Proving Gamma function relation

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

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 $$\begin{align*}\Gamma(z+1) & =z\int\limits_{0}^{\infty}e^{-t}t^{z-1}\,\mathrm dt\\ & =z\Gamma(z)\end{align*}$$