I am studying the gamma and beta functions and I have seen an exercise which asks you to re-write the beta function in terms of the gamma function as follows:

$B(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha + \beta)}$

The only hint that the exercise gives is that you should start with a function of the form:


The first thing I notice is that the right-hand side can be considered as a convolution of two functions, so:


However, this just ends up with the Laplace transform of f on the LHS and then a mess of transforms on the right. The other thing is that if you set $t=1$, $f$ becomes the beta function:


Hence I am hoping that if I convert $f$ to this form we have the beta function and then take the product of the Laplace transforms of the two functions in the convolution, it somehow comes out as a set of integrals which are equivalent to the given identity once I use the integral formula for the gamma function. However, I have tried this and got a mess, could someone assist (if this is the right way of doing it)?

Edit: I have looked on Wikipedia and seen that the identity can be proved just by taking the product of two gamma functions and then changing variables to show that this is $B(\alpha,\beta)\Gamma(\alpha+\beta)$: however, the exercise is in a section on Laplace transforms so I think it wants you to take the Laplace transform of the given function $f$ and work it out that way.

  • $\begingroup$ Convolution was the correct approach, but you didn't identify the correct functions that were being convolved. $\endgroup$
    – adfriedman
    Mar 21, 2018 at 1:07
  • $\begingroup$ Yes, I just saw that after looking back at it: I did write the form of the convolution correctly in my notes but typed it out wrong for some reason. $\endgroup$
    – Tom
    Mar 21, 2018 at 1:17
  • $\begingroup$ I think it was just getting that the inverse Laplace transform of the gamma function would give the power functions on the RHS, hence the product follows when you take the Laplace transform on the RHS. $\endgroup$
    – Tom
    Mar 21, 2018 at 1:19

2 Answers 2


$$f(\alpha, \beta, t) = \int_0^t x^{\alpha -1} (t-x)^{\beta -1}\;dx = \big(t^{\alpha - 1} * t^{\beta -1}\big)(t)$$

so using the fact that convolution is multiplicative in the Laplace domain, \begin{align} \mathcal{L}\{f(\alpha, \beta, t)\}(s) &= \mathcal{L}\{t^{\alpha-1}\}(s) \cdot \mathcal{L}\{t^{\beta - 1}\}(s)\\ &= \frac{\Gamma(\alpha)}{s^{\alpha}}\cdot \frac{\Gamma(\beta)}{s^{\beta}}\\ &= \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)} \cdot\underbrace{\frac{\Gamma(\alpha+\beta)}{s^{\alpha+\beta}}}_{\mathcal{L\{t^{\alpha+\beta - 1}\}(s)}} \end{align}

Taking inverse Laplace we then find $$f(\alpha,\beta,t) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)} t^{\alpha+\beta -1}$$ hence at $t=1$: $$B(\alpha,\beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$$

Added To address the extra question in the comments, $$ \Gamma(\tfrac12)\cdot\Gamma(\tfrac12) = \Gamma(\tfrac12+\tfrac12)\cdot B(\tfrac12,\tfrac12) = (1)\int_0^1 x^{-1/ 2} (1-x)^{-1/ 2}\;dx = \int_0^1 \frac{dx}{\sqrt{x(1-x)}}$$ We can evaluate this integral by noting the symmetry about $x=\tfrac12$, so substitute $x =\frac{1+u}{2}$ to shift the symmetry about $u=0$

$$\Gamma(\tfrac12)^2 = \int_{-1}^1 \underbrace{\frac{du}{\sqrt{1-u^2}}}_{\text{even function}} = 2\int_0^1 \frac{du}{\sqrt{1-u^2}} = 2\cdot \left.\sin^{-1}(u)\right|_0^1 = 2 \cdot \left(\frac{\pi}{2} - 0\right) = \pi$$ Hence $\Gamma(\tfrac12) = \sqrt{\pi}$.

  • $\begingroup$ One question, it seems like you take the Laplace transform but then 'un-do' the effect by taking the inverse Laplace transform, so is there not a sense in which you don't need to take the Laplace transform at all in your solution? $\endgroup$
    – Tom
    Mar 21, 2018 at 1:26
  • $\begingroup$ One of the major uses of Laplace transforms is to simplify these calculations (e.g., going from integral convolution to multiplication). You really do have to take the Laplace inverse though, as not every back and forth with the Laplace domain is perfect (it isn't necessarily bijective). For instance, the Laplace transform doesn't encode any function data on $t\leq 0$. To be more rigorous the formula for $f(\alpha,\beta,t)$, arrived at by taking the inverse Laplace transform, is only valid for $t>0$. $\endgroup$
    – adfriedman
    Mar 21, 2018 at 2:05
  • $\begingroup$ The derivation of the Laplace convolution formula comes from a two-dimensional change of variables though, so you could just derive the Beta-Gamma equation directly, but the purpose of the using the Laplace transform is that it lets you avoid dealing with integrals in this case. $\endgroup$
    – adfriedman
    Mar 21, 2018 at 2:08
  • $\begingroup$ I see, thank you. Could you also clarify why when you take the inverse Laplace transform of the product the left-hand part $\Gamma(\alpha)\Gamma(\beta)/\Gamma(\alpha + \beta)$ remains the same? $\endgroup$
    – Tom
    Mar 21, 2018 at 2:11
  • $\begingroup$ Sure, it is just a multiplicative constant as it has no dependency on $s$. $\endgroup$
    – adfriedman
    Mar 21, 2018 at 2:12

Here it is another approach, assuming we are just interested in real $\alpha,\beta>0$.
For a fixed value of $\beta$ the function $\int_{0}^{1}x^{\beta-1}(1-x)^{\alpha-1}\,dx $ is clearly positive, continuous and log-convex with respect to $\alpha$: the Cauchy-Schwarz inequality proves the midpoint-log-convexity and the continuity improves this to log-convexity. In particular $$g(\alpha)=\frac{\Gamma(\alpha+\beta)}{\Gamma(\beta)}\int_{0}^{1}x^{\beta-1}(1-x)^{\alpha-1}\,dx $$ is positive, continuous and log-convex. By integration by parts we have $g(\alpha+1)=\alpha\cdot g(\alpha)$ and $g(1)=1$ is trivial. By invoking the Bohr-Mollerup theorem we have $g(\alpha)=\Gamma(\alpha)$, hence $$ \Gamma(\alpha+\beta)\,B(\alpha,\beta) = \Gamma(\alpha)\,\Gamma(\beta) $$ as wanted.


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