Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can anyone give me a hint / help with the next integral? Thanks!


share|cite|improve this question
do we know anything about $a$ and $b$? – john Jun 6 '13 at 13:22
It is the Beta function of $a$ and $b$ if $t=1$. – Babak S. Jun 6 '13 at 13:23
Only that they are integers – Salieri Jun 6 '13 at 13:23
up vote 0 down vote accepted

This integral equals $t^{(a+b-1)}\frac{\Gamma(a)\Gamma(b)}{\Gamma (a+b)}$

Take $y=xt$

Then we have,

$\displaystyle\int_{0}^{t}{x^{a-1}(t-x)^{b-1}dx}=t^{(a+b-1)}\displaystyle\int_{0}^{1}{y^{a-1}(1-y)^{b-1}dx}=t^{(a+b-1)}\frac{\Gamma(a)\Gamma(b)}{\Gamma (a+b)}$

share|cite|improve this answer
This really works, I completely missed the Beta Function, thank you! – Salieri Jun 6 '13 at 13:32
You are welcome @Cardonai – Abhra Abir Kundu Jun 6 '13 at 13:35

Sub $x=t u$ and get

$$t^{a+b-1} \int_0^1 du \, u^{a-1} (1-u)^{b-1} = t^{a+b-1} \frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.