Improper Integral Question

Express $$\int_0^1x^m(1-x^n)^pdx$$ in terms of gama function and hence evaluate the integral.

I used the substitution $x^n=y$ and solving got this integral as equal to the beta-function $${1\over n}\beta({{{m+1}\over n},{p+1}})$$ solving futher by converting to gamma function ($\gamma)$ i get

$${1 \over n}[{{\Gamma({m+1 \over n})\Gamma(p+1)}} \over \Gamma({{m+1} \over n}+p+1)$$

and the answer i finally get is

$$p\over (m+p-1)!(m+np+1)$$

Is it correct ? One doubt i had in the solution is can we write

$$\Gamma(a+p)={{(a+p-1)!} \over (a-1)!} \Gamma(a)$$ ??

• At first I thought it was a question about an improper integral. But you integral does not seem improper, so perhaps it is an improper question about an integral? (English is such a weird language sometimes.) – Harald Hanche-Olsen May 2 '13 at 16:35
• Your answer looks suspicious, because unless you write it in terms of gamma functions, it can't possibly be true for the case when m and/or n and/or p aren't positive whole numbers. – Stochastically May 2 '13 at 16:36
• For your last question, no you can't do that. $\Gamma(a+p)=(a+p-1)(a+p-2)\cdots a\cdot\Gamma(a)$, and you can't simplify that the way you suggested. – Harald Hanche-Olsen May 2 '13 at 16:36
• Lets say yes, coz nothing is mentioned in the question. – Aman Mittal May 2 '13 at 16:44
• I think i figured it out. Should it not be $$\Gamma(a+p)={{(a+p-1)!} \over (a-1)!} \Gamma(a)$$ – Aman Mittal May 2 '13 at 16:50

\begin{align} \int_0^1x^m(1-x^n)^p\,\mathrm{d}x &=\int_0^1u^{m/n}(1-u)^p\,\mathrm{d}u^{1/n}\\ &=\frac1n\int_0^1u^{(m+1)/n-1}(1-u)^p\,\mathrm{d}u\\ &=\frac1n\frac{\Gamma\left(\frac{m+1}{n}\right)\Gamma(p+1)}{\Gamma\left(\frac{m+1}{n}+p+1\right)}\\ &=\frac{p}{m+1+pn}\frac{\Gamma\left(\frac{m+1}{n}\right)\Gamma(p)}{\Gamma\left(\frac{m+1}{n}+p\right)} \end{align} So your initial answer looks good. However, I don't see how you get from there to $$\frac{p}{(m+p-1)!(m+np+1)}$$
In the last question, for integer $n$, $\Gamma(n)=(n-1)!$. The rest is simply cancellation.