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I have run a FORTRAN code and I have obtained strong evidence that $$\int_0^{\pi/3} \!\! \big((\sqrt{3}\cos\vartheta-\sin\vartheta)\sin\vartheta\big)^{\!1/2}\!\cos\vartheta \,d\vartheta =\frac{\pi\sqrt{3}}{8\sqrt{2}}. $$ In fact, it looks like my numerical integration method (the trapezoid rule) converges to the value $\dfrac{\pi\sqrt{3}}{8\sqrt{2}}$, with at least $12$ significant digits.

Any ideas how to prove this result analytically?

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    $\begingroup$ Have you tried the Weierstrass substitution ? $\endgroup$ – Lucian Jan 16 '14 at 15:50
  • $\begingroup$ @Lucian: I am afraid I am not familiar with it. Could you be more specific? $\endgroup$ – Yiorgos S. Smyrlis Jan 16 '14 at 15:53
  • $\begingroup$ @Lucian: Yes I did try it ($t=\tan(x/2)$) and it doesn't look any better. $\endgroup$ – Yiorgos S. Smyrlis Jan 16 '14 at 15:54
  • $\begingroup$ @Jonathan - You are right - I missed some factor. See update. $\endgroup$ – Yiorgos S. Smyrlis Jan 16 '14 at 16:25
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    $\begingroup$ FORTRAN? why would you do that to yourself? $\endgroup$ – nbubis Jan 16 '14 at 23:03
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First, an identity: $$ \begin{align} (\sqrt{3}\cos(\theta)-\sin(\theta))\sin(\theta) &=2\sin(\tfrac\pi3-\theta)\sin(\theta)\\ &=\cos(\tfrac\pi3-2\theta)-\cos(\tfrac\pi3)\\ &=\tfrac12-2\sin^2(\tfrac\pi6-\theta)\\ &=\tfrac12-2\sin^2(\theta-\tfrac\pi6)\tag{1} \end{align} $$ Now, the integral $$ \begin{align} &\int_0^{\pi/3} \left((\sqrt{3}\cos(\theta)-\sin(\theta))\sin(\theta)\right)^{1/2}\cos(\theta)\,\mathrm{d}\theta\tag{2}\\ &=\int_0^{\pi/3} \left(\tfrac12-2\sin^2(\theta-\tfrac\pi6)\right)^{1/2}\cos(\theta)\,\mathrm{d}\theta\tag{3}\\ &=\frac1{\sqrt2}\int_{-\pi/6}^{\pi/6} \left(1-4\sin^2(\theta)\right)^{1/2}\cos(\theta+\tfrac\pi6)\,\mathrm{d}\theta\tag{4}\\ &=\frac1{\sqrt2}\int_{-\pi/6}^{\pi/6} \left(1-4\sin^2(\theta)\right)^{1/2}\left(\tfrac{\sqrt3}{2}\cos(\theta)-\tfrac12\sin(\theta)\right)\,\mathrm{d}\theta\tag{5}\\ &=\frac{\sqrt3}{2\sqrt2}\int_{-\pi/6}^{\pi/6} \left(1-4\sin^2(\theta)\right)^{1/2}\cos(\theta)\,\mathrm{d}\theta\tag{6}\\ &=\frac{\sqrt3}{2\sqrt2}\int_{-\pi/6}^{\pi/6} \left(1-4\sin^2(\theta)\right)^{1/2}\,\mathrm{d}\sin(\theta)\tag{7}\\ &=\frac{\sqrt3}{4\sqrt2}\int_{-\pi/2}^{\pi/2} \left(1-\sin^2(\theta)\right)^{1/2}\,\mathrm{d}\sin(\theta)\tag{8}\\ &=\frac{\sqrt3}{4\sqrt2}\int_{-\pi/2}^{\pi/2} \cos^2(\theta)\,\mathrm{d}\theta\tag{9}\\ &=\frac{\pi\sqrt3}{8\sqrt2}\tag{10} \end{align} $$ Explanation:
$\ \:(3)$: Apply $(1)$
$\ \:(4)$: subsitute $\theta\mapsto\theta+\frac\pi6$
$\ \:(5)$: cosine of a sum formula
$\ \:(6)$: toss the odd part over a symmetric domain
$\ \:(7)$: $\mathrm{d}\sin(\theta)=\cos(\theta)\,\mathrm{d}\theta$
$\ \:(8)$: subsitute $\sin(\theta)\mapsto\frac12\sin(\theta)$
$\ \:(9)$: $\sqrt{1-\sin^2(\theta)}=\cos(\theta)$ and $\mathrm{d}\sin(\theta)=\cos(\theta)\,\mathrm{d}\theta$
$(10)$: $\int_{-\pi/2}^{\pi/2}\cos^2(\theta)\,\mathrm{d}\theta=\frac\pi2$

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OK, you can start by noting that

$$\sqrt{3} \cos{x} - \sin{x} = 2 \sin{\left ( \frac{\pi}{3}-x\right)}$$

which means that the integral is

$$ \sqrt{2} \int_0^{\pi/3} dx \, \cos{x} \sqrt{\sin{x} \sin{\left ( \frac{\pi}{3}-x\right)}} = \sqrt{2} I$$

You can make a substitution $x \mapsto \frac{\pi}{3}-x$ and see that

$$I = \sqrt{3} \int_0^{\pi/3} dx \, \sin{x} \sqrt{\sin{x} \sin{\left ( \frac{\pi}{3}-x\right)}}$$

EDIT

Integrate by parts:

$$\begin{align}I &= \left [\sin{x} \sqrt{\sin{x} \sin{\left ( \frac{\pi}{3}-x\right)}} \right ]_0^{\pi/3} - \int_0^{\pi/3} dx \, \sin{x} \frac{d}{dx} \sqrt{\sin{x} \sin{\left ( \frac{\pi}{3}-x\right)}}\\ &= -\frac12 \int_0^{\pi/3} dx \, \sin{x} \frac{\cos{x} \sin{\left ( \frac{\pi}{3}-x\right)} - \sin{x} \cos{\left ( \frac{\pi}{3}-x\right)}}{\sqrt{\sin{x} \sin{\left ( \frac{\pi}{3}-x\right)}}}\\ &= -\frac12 I + \frac12 \int_0^{\pi/3} dx \, \sin{x} \sqrt{\frac{\sin{x}}{\sin{\left ( \frac{\pi}{3}-x\right)}} } \cos{\left ( \frac{\pi}{3}-x\right)} \end{align}$$

This means that

$$\begin{align}3 I &= \int_0^{\pi/3} dx \, \frac{\cos{x}}{\sin{x}} \sqrt{\sin{x} \sin{\left ( \frac{\pi}{3}-x\right)}} \cos{\left ( \frac{\pi}{3}-x\right)}\\ &= \frac{\sqrt{3}}{2} \int_0^{\pi/3} dx \, \frac{\cos^2{x}}{\sin{x}} \sqrt{\sin{x} \sin{\left ( \frac{\pi}{3}-x\right)}} - \frac12 I \end{align}$$

Combining again...

$$7 I = \sqrt{3} \int_0^{\pi/3} dx \sqrt{\frac{\sin{\left ( \frac{\pi}{3}-x\right)}}{\sin{x}}} - \underbrace{\sqrt{3} \int_0^{\pi/3} dx \, \sin{x} \sqrt{\sin{x} \sin{\left ( \frac{\pi}{3}-x\right)}}}_{\text{We know from above this equals } I}$$

Thus

END EDIT

$$I = \frac{\sqrt{3}}{8} \int_0^{\pi/3} dx \sqrt{\frac{\sin{x}}{\sin{\left ( \frac{\pi}{3}-x\right)}}}$$

Now, in a similar manipulation as in the evaluation of this integral, sub $u = \sin{x}/\sin{(\pi/3-x)}$ and find that the integral becomes

$$I = \frac{3}{16} \int_0^{\infty} du \frac{\sqrt{u}}{1+u+u^2}$$

This integral is very straightforward to evaluate via residues using, e.g., a keyhole contour about the positive real axis. By the residue theorem, the original integral is then

$$\sqrt{2} I = \frac{3 \sqrt{2}}{16} \frac12 i 2 \pi \left (\frac{e^{i \pi/3}}{i \sqrt{3}} - \frac{e^{i 2 \pi/3}}{i \sqrt{3}} \right ) = \frac{\pi}{8} \sqrt{\frac{3}{2}}$$

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  • $\begingroup$ I couldn't understand how you found that $$I = \frac{\sqrt{3}}{8} \int_0^{\pi/3} dx \sqrt{\frac{\sin{x}}{\sin{\left ( \frac{\pi}{3}-x\right)}}}$$. $\endgroup$ – Yiorgos S. Smyrlis Jan 16 '14 at 22:48
  • $\begingroup$ @YiorgosS.Smyrlis: I filled in the gap so it should be clear how I got that intermediate result. $\endgroup$ – Ron Gordon Jan 17 '14 at 1:44

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