4
$\begingroup$

From $$\Gamma(x)\Gamma(y)=\int_0^\infty e^{-t}t^{x-1} \left( \int_0^\infty e^{-t} s^{y-1} ds \right) dt,$$ use a change of variable $s=ut$ to show $$\Gamma(x)\Gamma(y)=\Gamma(x+y)\beta(x,y).$$

Let $s=ut$ so $ds = udt + tdu$ and then

\begin{align*} \Gamma(x)\Gamma(y) &=\int_0^\infty e^{-t}t^{x-1} \left( \int_0^\infty e^{-s} s^{y-1} ds \right) dt \\ &= \int_0^\infty e^{-t} t^{x-1} \left( \int_0^{\infty} e^{-ut} (ut)^{y-1}(udt+tdu)\right)dt \\ &= \int_0^\infty e^{-t} t^{x-1} \left( \int_0^{\infty} (e^{-ut} u^y t^{y-1})dt+ \int_0^{\infty} (e^{-ut}u^{y-1}t^y)du\right)dt \\ \end{align*} My next thought was to use integration by parts, but that didn't pan out. Any suggestions?

Note that Gamma Function for $x>0$ we define: $$\Gamma(x):=\int_0^\infty e^{-t}t^{x-1}dt.$$

And for the Beta Function for $x>0$, $y>0$, we define $$\beta(x,y):=\int_0^1 t^{x-1}(1-t)^{y-1}dt.$$

$\endgroup$
0
$\begingroup$

Hint: Try the substitution $s+t=\alpha$ and $t=\alpha \gamma $ so $s=\alpha(1-\gamma)$ and $ds \, dt = \alpha d \alpha \, d \gamma$.

$\endgroup$
  • $\begingroup$ Maybe $\beta$ isn't the best symbol to use here. $\endgroup$ – Shaun Jun 3 '18 at 21:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.