Laplace transform to prove relationship between the Gamma and Beta functions

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!