For what values $p,q$ does the improper integral $\int_0^1 x^p (1-x^2)^q dx$ converge?

Question

Question:

For what values $p,q$ does the improper integral $\int_0^1 x^p (1-x^2)^q dx$ converge?

I am struggling as I'm not sure where to start. What is the best way to approach this question?

Let $t=x^2$, and then see beta function. – Lucian Nov 19 '14 at 03:58 — Lucian, Nov 19 '14 at 03:58

score 3 · Accepted Answer · answered Nov 19 '14 at 03:53

3

Set $x=\sin(t)$. We then get the integral as $$\int_0^{\pi/2}\sin^p(t)\cos^{2q}(t)\cos(t)dt = \int_0^{\pi/2}\sin^p(t)\cos^{2q+1}(t)dt=\dfrac12\beta((p+1)/2,q+1)$$ which exists whenever $(p+1)/2>0$ and $q+1>0$, i.e., $p,q>-1$.

answered Nov 19 '14 at 03:53

Adhvaitha

5,441

Doesn't $(1-\sin(t)^2)^q = \cos^{2q}(t)$ rather than $(1-\sin(t)^2)^q = \cos^{2q}(t)\cos(t)$? What am I missing? – mathjacks Nov 20 '14 at 03:03
@flapjackery You are missing the fact that $dx = \cos(t)dt$. – Adhvaitha Nov 20 '14 at 03:14

For what values $p,q$ does the improper integral $\int_0^1 x^p (1-x^2)^q dx$ converge?

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