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I am struggling to see how after substituting this: $$\frac{1}{(1+u)^{a+b}}=\frac{1}{\Gamma(a+b)}\int_{0}^{\infty}e^{-(1+u)t}t^{a+b-1}dt$$ into this: $$B(a,b)=\int_{0}^{\infty}\dfrac{u^{a-1}}{(1+u)^{a+b}}du$$ I will get this: $$B(a,b)=\frac{1}{\Gamma(a+b)}\int_{0}^{\infty}e^{-t}t^{a+b-1}dt\int_{0}^{\infty}e^{-ut}u^{a-1}du$$ because all I have is this: $$B(a,b)=\frac{1}{\Gamma(a+b)}\int_{0}^{\infty}u^{a-1}\left[\int_{0}^{\infty}e^{-(1+u)t}t^{a+b-1}dt\right]du$$

Is there a lot of omitted steps here or is there a trick I am failing to catch onto?

The is a part of the derivation on p.25 of this note relating the Beta function to the Gamma function.

Edit:

I give up... I still don't understand but I am starting to seriously question the correctness/necessity of the step mentioned above in the derivation. I think a much more straightforward approach is to multiple left and right side of: $$B(a,b)=\int_{0}^{\infty}\dfrac{u^{a-1}}{(1-u)^{a+b}}du$$ by the left and right side of: $${\Gamma(a+b)}=(1+u)^{a+b}\int_{0}^{\infty}e^{-(1+u)t}t^{a+b-1}dt$$ respectively. Then with the change of variable $k=(1+u)t$, arrive at essentially step 2 of the proof linked here.

crazyproof
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  • Both of your expressions for $B(a,b)$ are iterated integrals. Notice the $t$ in both integrands ($e^{-t} t^{a+b-1}$ and $e^{-ut}$) – user170231 Apr 15 '22 at 19:31
  • You have a mistake (typo?) in the integral representation of the beta function. The denominator is $(1+u)^{a+b}$. Also, check out Wikipedia for a different proof of this property. – bjorn93 Apr 15 '22 at 21:06
  • @bjorn93 Thanks for pointing out the typo. Fixed it! – crazyproof Apr 16 '22 at 04:20
  • It's a typo. The third equation doesn't even make sense as written because the dummy variable $t$ appears outside of the integral where it's defined. Instead the $dt$ is supposed to be after the second integral, not before it. So it's not a product of integrals. (Physicists also like to put differentials before integrands, but that's not what's going on here.) – anon Apr 16 '22 at 04:28
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    @runway44 So you mean the more appropriate way of writing it would be: $$B(a,b)=\frac{1}{\Gamma(a+b)}\int_{0}^{\infty}e^{-t}t^{a+b-1}\int_{0}^{\infty}e^{-ut}u^{a-1}dudt$$ I copied the equation as shown in the original question straight from the notes... guess I should take course notes with a grain of salt here on out – crazyproof Apr 16 '22 at 04:39
  • Yes, the typo is in the original notes. – anon Apr 16 '22 at 04:41

1 Answers1

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The hidden step is a change in the order of integration: $$\int_{0}^{\infty}u^{a-1}\left(\int_{0}^{\infty}e^{-(1+u)t}t^{a+b-1}\,dt\right)du=\int_0^\infty\int_0^\infty u^{a-1}e^{-(1+u)t}t^{a+b-1}\,du\,dt\\ =\int_0^\infty t^{a+b-1}e^{-t}\left(\int_0^\infty u^{a-1}e^{-ut}\,du\right)dt$$ Then, the substitution $u\mapsto u/t$ shows that $$\int_0^\infty u^{a-1}e^{-ut}\,du=\frac{\Gamma(a)}{t^a} $$ and we're left with $$\Gamma(a)\int_0^\infty t^{b-1}e^{-t}\,dt=\Gamma(a)\Gamma(b) $$

Edit: I noticed that in the title you mention a product of two integrals. The way the desired result is written could possibly be parsed as a product of integrals but if that were the case, then the equality would be wrong. The correct interpretation should be iterated integration.

bjorn93
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  • In other words $$\int_{0}^{\infty}e^{-t}t^{a+b-1}dt\int_{0}^{\infty}e^{-ut}u^{a-1}du$$ should be understood as iterated integration and not a product of integrals. Got it! Thanks – crazyproof Apr 16 '22 at 04:28