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Let $K>L>0$. I would like to find a good upper bound for the integral $$\int_0^L \sqrt{x \left(1 + \frac{1}{K-x}\right)} \,dx.$$ An explicit expression for the antiderivative would have to involve elliptic functions; thus, if we want to stick to simple expressions, a bound is the best we can do.

One obvious approach goes as follows: $$\begin{aligned} \int_0^L \sqrt{x \left(1 + \frac{1}{K-x}\right)} \, dx &\leq \int_0^L \sqrt{x} \left(1 + \frac{1/2}{K-x}\right)\, dx\\ &\leq \frac{2}{3} L^{3/2} + \frac{1}{2} \sqrt{\int_0^L x \,dx \cdot \int_0^L \frac{dx}{(K-x)^2}}\\ &= \frac{2}{3} L^{3/2} + \frac{1}{\sqrt{8}} \frac{L^{3/2}}{\sqrt{K (K-L)}} \end{aligned}$$ The second inequality (Cauchy-Schwarz) does not seem too bad, though one can easily improve on the constant $1/\sqrt{8}$ by proceeding more directly.

I dislike the first step, however, as it makes the integral diverge as $L\to K^-$, whereas the original integral did not.

What other simple approximations are there? Is it obvious that a bound of type $(2/3) L^{3/2} + O(L^{3/2}/K)$, say, could not be valid?

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2 Answers 2

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In general one has the following bound on your integral $$ \frac{2}{3}L^{3/2}+ \frac{L^{1/2}}{\sqrt{2}}\ln \left(\frac{1+2K}{2+2(K-L)}\right)\leq \mathrm{integral}\leq \frac{2}{3}L^{3/2}+ L^{1/2} \ln\left(\frac{3+4K}{1+4(K-L)}\right) $$

Let us make change of variables $x=Ly$. Then we want to estimate from above $$ L^{3/2}\int_{0}^{1}\sqrt{y \left(1+ \frac{1}{K-Ly}\right)} - \sqrt{y}\; dy $$ We have $$ \int_{0}^{1}\sqrt{y}\left(\frac{\sqrt{K-Ly+1}-\sqrt{K-Ly}}{\sqrt{K-Ly}} \right)\; dy = \frac{1}{K}\times \int_{0}^{1}\sqrt{y}\left(\frac{1}{\sqrt{1-(L/K)y}(\sqrt{1-(L/K)y+(1/K)}+\sqrt{1-(L/K)y})} \right)\; dy \leq \frac{1}{K} \times \int_{0}^{1}\left(\frac{1}{\sqrt{1-(L/K)y}\sqrt{1-(L/K)y+(1/K)}} \right)\; dy = \frac{1}{L}\ln\left( \frac{1+2K+2\sqrt{K}\sqrt{K+1}}{1+2(K-L)+2\sqrt{K-L}\sqrt{K-L+1}}\right)\leq \frac{1}{L}\ln\left(\frac{3+4K}{1+4(K-L)}\right) $$

Update (lower bound): $$ \int_{0}^{1}\sqrt{y}\left(\frac{1}{\sqrt{1-(L/K)y}(\sqrt{1-(L/K)y+(1/K)}+\sqrt{1-(L/K)y})} \right)\; dy \geq \frac{1}{\sqrt{2}}\int_{1/2}^{1}\left(\frac{1}{\sqrt{1-(L/K)y}\sqrt{1-(L/K)y+(1/K)}} \right)\; dy = \frac{1}{\sqrt{2}}\frac{K}{L}\int_{1-\frac{L}{K}}^{1-\frac{L}{2K}} \frac{1}{\sqrt{z}\sqrt{z+\frac{1}{K}}}dz $$ Now let us use the fact that $$ \int \frac{1}{\sqrt{z^{2}+\frac{z}{K}}} = \ln \left(1+ 2Kz +2\sqrt{Kz}\sqrt{Kz+1}\right) - \ln K, \quad z>0 $$ which can be checked by direct differentiation (but I will skip it), then we can continuo our chain of inequalities $$ =\frac{1}{\sqrt{2}}\frac{K}{L}\,\ln \frac{1+2(K-\frac{L}{2}) +2\sqrt{K-\frac{L}{2}}\sqrt{K-\frac{L}{2}+1}}{1+2(K-L) +2\sqrt{K-L}\sqrt{K-L+1}}\geq \frac{1}{\sqrt{2}}\frac{K}{L}\,\ln \frac{1+4(K-\frac{L}{2})}{2+2(K-L)}\geq \frac{1}{\sqrt{2}}\frac{K}{L}\,\ln \frac{1+2K}{2+2(K-L)} $$

which is pretty much the same.

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  • $\begingroup$ Is it absolutely clear that one isn't losing much in the first inequality? $\endgroup$ Apr 30, 2018 at 5:51
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    $\begingroup$ I have updated the answer. Now it has the lower bound $\endgroup$ Apr 30, 2018 at 12:06
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After $x=Kt^{2/3}$, using an obvious inequality $\sqrt{A+B}\le\sqrt{A}+\sqrt{B}$, we have $$\int_0^L \sqrt{x\left(1+\frac{1}{K-x}\right)}\,dx\le \frac{2}{3}L^{3/2}+K\int_0^{L/K}\sqrt{\frac{t}{1-t}}\,dt.$$ The integral may be computed in elementary functions, but there is not much point in this. For $L$ close to $K$, it is better to just replace it by $\int_0^{1}\sqrt{\frac{t}{1-t}}\,dt=\frac{\pi}{2}$. For $L\ll K$, it is $\sim \frac{2}{3}(L/K)^{3/2}$.

Very simple and we do not lose too much as there is an estimate from below, $\sqrt{A+B}\ge\sqrt{A/2}+\sqrt{B/2}$.

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  • $\begingroup$ If you consider the regime $L=K$ and $L$ goes to infinity, then the asymptotic of the integral is $\frac{2}{3} L^{3/2} + L^{1/2}\ln(L)$ but you get upper bound $\frac{2}{3} L^{3/2} + \frac{\pi}{2}L$, so there is logarithmic and power loss. $\endgroup$ Apr 30, 2018 at 12:13
  • $\begingroup$ I agree. (Also, I did not answer the last question.) $\endgroup$ Apr 30, 2018 at 13:04

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