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

$$\displaystyle\int_{0}^{t}{x^{a-1}(t-x)^{b-1}dx}$$

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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 $t=1$. –  B. S. Jun 6 '13 at 13:23
    
Only that they are integers –  Salieri Jun 6 '13 at 13:23
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2 Answers 2

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,

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

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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
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Sub $x=t u$ and get

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

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