# 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?

• Did you forget limits: $$\Gamma (t) = \frac 1t x^t e^{-x}\Big|_0^\infty + \frac 1t \int_0^\infty x^t e^{-x} dx$$ – Nosrati Sep 3 '17 at 2:19
• Oh my bad! Thanks a lot! If you post it as an answer, I'll give you a star or something – David Bang Sep 3 '17 at 2:20
• Did you try differentiating the other term and integrating the exponential ? – reuns Sep 3 '17 at 2:21

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*}
• Why $\lim-t^ze^{-t}=0$? can you elaborate. – Nosrati Sep 3 '17 at 8:23
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.