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Does anyone knows how to show

$$\int_0^1 x^a(1-x)^b dx = \frac{\Gamma(a+b-2)}{\Gamma( a -1)\Gamma(b-1)} $$ for $a,b \gt -1?$

Also why it doesn't work for $a$ or $b\lt -1$ ?

SAJ
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2 Answers2

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Does anyone knows how to show...

I cannot, because it is wrong. Using the definition of Beta function you have that

$$\int_0^1 x^{a-1}(1-x)^{b-1}dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$

that in your case gives the following result

$$\frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)}$$

to realize that your result does not make sense, set $a,b=1$ and you get

$$\int_0^1 dx=\frac{\Gamma(-2)}{\Gamma(-1)\Gamma(-1)}$$

??????

While using my result you get

$$\int_0^1 dx=\frac{\Gamma(1)\Gamma(1)}{\Gamma(2)}=\frac{0!\cdot0!}{1!}=1$$

tommik
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  • The title shows the original question (before the OP edited it) which was correct. – Gary Mar 09 '21 at 14:02
  • @Gary ops, sorry, I didn't read the whole title...my fault, anyway my answer is still valid. Thanks for your comment – tommik Mar 09 '21 at 14:04
  • In some way, you are right. I don't know why the OP edited the question into a non-sense. – Gary Mar 09 '21 at 14:06
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The formula you wrote in the question is incorrect. In the title you have the correct one. To prove it, let's consider \begin{align} \Gamma(a+1)\Gamma(b+1) &= \int_0^\infty t^a e^{-t}dt \int_0^\infty s^b e^{-s}ds = \\ &= \int_0^\infty \Big( \int_0^\infty t^a s^b e^{-t-s}ds \Big) dt = \\ &= \int_0^\infty \Big( \int_t^\infty t^a (u-t)^b e^{-u}du \Big) dt = \\ &= \int_0^\infty \Big( \int_0^u t^a (u-t)^b e^{-u} dt \Big) du = \\ &= \int_0^\infty \Big( \int_0^1 (ux)^a (u-ux)^b udx \Big) e^{-u} du = \\ &= \int_0^\infty \Big( \int_0^1 x^a (1-x)^b dx \Big) u^{a+b+1} e^{-u} du = \\ &= \int_0^\infty u^{a+b+1} e^{-u} du \int_0^1 x^a (1-x)^b dx = \\ &= \Gamma(a+b+2)\int_0^1 x^a (1-x)^b dx \end{align}

It doesn't work if $a<-1$ or $b<-1$ because this integral are not converegnt for such values of $a$ and $b$.

Adam Latosiński
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