Laplace transform to prove relationship between the Gamma and Beta functions

Question

I am stuck in the following theoretical part: I have the function: $$f(t,\mu,\nu) = \int_{0}^{t}x^{\mu-1}(t-x)^{\nu-1}dx$$ Now I have to perform a Laplace transform to deduce that: $$B(\mu,\nu) = \frac{\Gamma(\mu)\Gamma(\nu)}{\Gamma(\mu+\nu)}$$

A "normal" Laplace transform I know how to use, but in this case: I don't know where to start.

Answer 1

In the ensuing analysis, we will make use of the Laplace Transform Pair

$$\begin{align} f(t) &\leftrightarrow F(s)\\\\ t^{a-1} &\leftrightarrow \frac{\Gamma(a)}{s^a}\tag1 \end{align}$$

where $F(s)=\int_0^\infty f(t)e^{-st}\,dt$ and $f(t)=\frac1{2\pi i}\int_{\sigma -i\infty}^{\sigma +i\infty}F(s)e^{st}\,ds$.

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Let $f(t,\mu,\nu)$ be represented by the convolution integral

$$f(t,\mu,\nu)=\int_0^t \tau^{\mu-1}(t-\tau)^{\nu-1}\,d\tau\tag2$$

Note that $f(1,\mu,\nu)=B(\mu,\nu)$, where $B(x,y)$ is the Beta function, which can be represented by the integral

$$B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\,dt$$

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Applying the Convolution Theorem of the Laplace Transform to $(2)$ and using $(1)$ reveals that

$$\begin{align} \int_0^\infty f(t,\mu,\nu)e^{-st}\,dt&=\left(\int_0^\infty t^{\mu-1}e^{-st}\,dt\right)\left(\int_0^\infty t^{\mu-1}e^{-st}\,dt\right)\\\\ &=\left(\frac{\Gamma(\mu)}{s^\mu}\right)\left(\frac{\Gamma(\nu)}{s^\nu}\right)\\\\ &=\frac{\Gamma(\mu)\Gamma(\nu)}{s^{\mu+\nu}}\\\\ &=\frac{\Gamma(\mu)\Gamma(\nu)}{\Gamma(\mu+\nu))}\int_0^\infty t^{\mu+\nu-1}e^{-st}\,dt\tag3 \end{align}$$

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Taking the inverse Laplace Transform of both sides of $(3)$, we find that

$$f(t,\mu,\nu)=\frac{\Gamma(\mu)\Gamma(\nu)}{\Gamma(\mu+\nu)}\,t^{\mu+\nu-1}\tag4$$

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Finally, setting $t=1$ in $(4)$ yields the coveted relationship

$$\bbox[5px,border:2px solid #C0A000]{B(\mu,\nu)=\frac{\Gamma(\mu)\Gamma(\nu)}{\Gamma(\mu+\nu)}}$$

as was to be shown!

Answer 2

Although there is already a good answer, I needed to fill the missing steps for my understanding. Perhaps will help others as well.

The convolution of $f(x) * g(x) = \int_0^x f(y)g(x-y)\,dy$

For functions $f(x)=x^{\alpha-1}$ and $g(x)=x^{\beta-1}$ we have

$$ f*g = \int_0^x y^{\alpha-1} (x-y)^{\beta-1}\,dy = x^{\alpha+\beta-1}\int_0^1 u^{\alpha-1} (1-u)^{\beta-1}\,du=x^{\alpha+\beta-1}B(\alpha,\beta) $$

Now, the Laplace transform of $f(x)$

$$ \mathcal{L}\{x^{\alpha-1}\} = \int_0^\infty x^{\alpha-1} e^{-sx}\,dx = s^\alpha \int_0^\infty x^{\alpha-1} e^{-x} \, dx = s^\alpha \Gamma(\alpha) $$

Using the fact that $\mathcal{L}\{f*g\} = \mathcal{L}\{f\} \mathcal{L}\{g\} $

$$ \mathcal{L}\{x^{\alpha+\beta-1}B(\alpha,\beta)\} = \mathcal{L}\{x^{\alpha-1}\} \mathcal{L}\{x^{\beta-1}\} $$ substituing left and right hand side with Laplace transforms $$ s^{\alpha+\beta} \Gamma(\alpha+\beta) B(\alpha,\beta) = s^\alpha \Gamma(\alpha) s^\beta \Gamma(\beta) $$

eliminating $s$ gives you the sought relationship

$$ \Gamma(\alpha+\beta) B(\alpha,\beta) = \Gamma(\alpha) \Gamma(\beta) $$