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$\displaystyle T=\sqrt{\frac{1}{2g}}\int_{0}^{b}\frac{\sqrt{1+y'(x)^2}}{\sqrt{-y(x)}} $

I need to find the intervals it is convergent for $y=-2x^p$ , where g and p are constants and $p > 0$. I've tried substituting, but the 1 is making it impossible.

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Can you define $g$? And is $p$ a prime, integer, real, nonnegative, etc? – Dimitrije Kostic Nov 13 '11 at 20:40
This reminds me of the brachistochrone problem... – David Mitra Nov 13 '11 at 22:57
up vote 1 down vote accepted

We have $y'(x)=2px^{p-1}$. Now we do something that is not absoutely necessary, but that may give additional clarity. We divide our problem into $2$ cases: (i) $p \ge 1$; (ii) $0<p<1$. We will assume that $b>0$.

Case (i): We are interested in $\displaystyle\int_0^b \frac{\sqrt{1+4p^2x^{2p-2}}}{\sqrt{2}x^{p/2}}\,dx$.

The numerator is bounded above in our interval. If $p<2$, then $\displaystyle\int_0^b \frac{dx}{x^{p/2}}$ converges, so our integral converges.

What about if $p \ge 2$? Then our integrand is bigger than $\dfrac{1}{\sqrt{2}x^{p/2}}$, so the integral diverges.

Case (ii): The numerator can be rewritten as $\dfrac{1}{x^{1-p}}\sqrt{x^{2-2p}+4p^2}$. The part $\sqrt{x^{2-2p} +4p^2}$ is bounded above, since $0<p<1$. So we really only care whether the integral of $\dfrac{1}{x^{1-p}x^{p/2}}$ converges. The integrand simplifies to $\dfrac{1}{x^{1-p/2}}$, and $1-p/2<1$, so the integral converges.

In summary, our integral converges when $0<p<2$, and diverges if $p \ge 2$.

Comment: Here is a slicker way of doing things uniformly for all $p<2$. Note that $$1+4p^2x^{2p-2}<(1+2px^{p-1})^2.$$ So our integrand is less than $$\frac{1}{x^{p/2}} +\frac{2p}{x^{1-p/2}}.$$ If $p<2$, then each of $\displaystyle\int\frac{dx}{x^{p/2}}$ and $\displaystyle\int\frac{2p\,dx}{x^{1-p/2}}$ converges.

The downside of this approach is that it is probably less natural than the first approach we gave.

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+1. But I do not understand the "not absolutely necessary but may give clarity" bit. What other method(s) could be used for this problem? – Srivatsan Nov 13 '11 at 21:53
@Srivatsan: The method you were using works nicely, if you make a little change, and observe that $1+4p^2x^{2p-2}<(1+2px^{p-1})^2$. I will write a comment about that. – André Nicolas Nov 13 '11 at 22:02
Aw, that indeed is a nice idea. :) – Srivatsan Nov 13 '11 at 22:07

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