I would like to prove that $$B(p,p) = 2 \int_0^{1/2} (t - t^2) ^ {p - 1} dt$$ where B is the Euler Beta function. I know I have to prove that B is even at $1/2$. Some hints are welcome, thank you.
Prove that $B(p,p) = 2 \int_0^{1/2} (t - t^2) ^ {p - 1} dt$
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calculus
integration
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1What's the definition of Beta function you have learned? – 2017-01-21
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1Hint: $t-t^2=t(1-t)$. – 2017-01-21
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0I know that, it doesn't help. Yet. I was taught $B(p,q) = \int_0^1 x^{p-1}(1-x)^{q-1}dx$. – 2017-01-21
1 Answers
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Start with the definition $$B(p,p) = \int_0^1 x^{p-1}(1-x)^{p-1} dx.$$ They want you to show the integrand is even around $x=1/2.$ This means that $f(x-1/2) = f(1/2-x)$ where $f(x)$ is the integrand $f(x) = x^{p-1}(1-x)^{p-1}.$ To establish this, just plug in: $$f(x-1/2) = (x-1/2)^{p-1}(1-(x-1/2))^{p=1}=(x-1/2)^{p-1}(1/2-x)^{p-1}.$$
Then if you calculate $f(1/2-x)$ you'll see it's the same thing.
Since the integrand is even about $x=1/2,$ we can write $$ B(p,p) = 2\int_0^{1/2}x^{p-1}(1-x)^{p-1} dx=2\int_0^{1/2}(x-x^2)^{p-1} dx$$
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2since I can't upvote your answer yet I thank you here – 2017-01-21