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I have a problem asking me to show that $$\frac{3}{8}<\int_0^\frac{1}{2}\sqrt{\frac{1-x}{1+x}}dx <\frac{\sqrt3}{4}.$$ The left side of the equation is clear since $1-x<\sqrt{\frac{1-x}{1+x}}$ for $x \in (0,1/2)$. I cannot see a clean way to obtain the bound on the right. Any asistance would be appreciated.

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    Evaluating the integral is a good way to show that. Evaluating the integral is easy by putting $x=\sin(\theta)$. – Hussain-Alqatari Jul 26 '19 at 20:25
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    $$\int_0^\frac{1}{2}\sqrt{\frac{1-x}{1+x}}dx <\int_0^\frac12\frac1{1+x}dx=\ln\left(\frac32\right)<\frac{\sqrt3}{4}.$$ – J. W. Tanner Jul 26 '19 at 20:44
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    Maybe $$\int_0^\frac{1}{2}\sqrt{\frac{1-x}{1+x}}dx <\int_0^\frac12\left[\left(\frac2{\sqrt3}-2\right)x+1\right] dx=\dfrac1{4\sqrt3}+\dfrac14<\frac{\sqrt3}{4}.$$ – J. W. Tanner Jul 26 '19 at 21:15
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    I.e., $$\int_0^\frac{1}{2}\sqrt{\frac{1-x}{1+x}}dx <\int_0^\frac12\left[\left(\frac2{\sqrt3}-2\right)x+1\right] dx=\dfrac1{4\sqrt3}+\dfrac14=\dfrac{\sqrt3}{12}+\dfrac3{12}<\dfrac{\sqrt3}{12}+\dfrac{2\sqrt3}{12}=\frac{\sqrt3}{4}.$$ – J. W. Tanner Jul 26 '19 at 21:21

3 Answers3

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For the right side compare it with $$\int _0^{1/2} \sqrt {1-x} dx =\frac {2(4-\sqrt 2)}{12} \approx .3779<\frac {\sqrt 3}{4}.$$

Mohammad Riazi-Kermani
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  • I had seen this, though I kept looking for an integral that evaluated directly to $\frac{\sqrt 3}{4}$ to compare with. Is there no such integral? –  Jul 26 '19 at 20:54
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May be using the series expansion $$\sqrt{\frac{1-x}{1+x}}=1-x+\frac{x^2}{2}-\frac{x^3}{2}+\frac{3 x^4}{8}+O\left(x^5\right)$$ which is alternating (this is a key point). So, for $0 \leq x \leq 1$, $$\sqrt{\frac{1-x}{1+x}}<1-x+\frac{x^2}{2}$$ $$\int_0^\frac{1}{2}\sqrt{\frac{1-x}{1+x}}dx <\int_0^\frac12\left(1-x+\frac{x^2}{2}\right)\,dx=\frac12-\frac1{8}+\frac{1}{48}=\frac{19}{48} \approx 0.395833 < \frac{\sqrt3}{4}$$

Claude Leibovici
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$$0\leq x\leq \frac{1}{2}$$

$$1-x^2\geq \frac{3}{4}$$

$$\sqrt{1-x^2}\geq \frac{\sqrt{3}}{2}$$

$$1 \leq \frac{1}{\sqrt{1-x^2}}\leq \frac{2}{\sqrt{3}}$$

Multiply by $1-x$

$$0 \leq 1-x\leq \sqrt{\frac{1-x}{1+x}}\leq \frac{2}{\sqrt{3}}(1-x)$$

$$\int_0^{1/2} (1-x) \, dx\leq \int_0^{1/2} \sqrt{\frac{1-x}{1+x}} \, dx \leq \int _0^{1/2}\frac{2}{\sqrt{3}}(1-x)dx$$

$$\frac{3}{8}\leq \int_0^{1/2} \sqrt{\frac{1-x}{1+x}} \, dx\leq \frac{\sqrt{3}}{4}$$

Lozenges
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