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Does there exist an easy method for proving that

$$ \int_1^2\int_1^2\int_1^2\arcsin\left(\frac{\sqrt{x^2-1}+\sqrt{y^2-1}+\sqrt{z^2-1}-\sqrt{x^2-1}\sqrt{y^2-1}\sqrt{z^2-1}}{xyz}\right)\,dx\,dy\,dz =3\log(2+\sqrt{3})-\pi. $$

Can someone give me an easy method?

Daniel W. Farlow
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  • Given the form, have you tried playing with the various trigonometric coordinate changes? – Steven Stadnicki Dec 02 '13 at 16:48
  • Just to be clear , have you solved it using any method or you are directly asking for a quicker method ? – abkds Dec 02 '13 at 16:49
  • Triple integration across the arcsin of a giant quotient, with multiple square root terms. And you want an easy method for solving it? – Cruncher Dec 02 '13 at 20:18

1 Answers1

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The radical $\sqrt{x^2-1}$ is one leg of a right triangle whose other leg is $1$ and whose hypotenuse is $x$. Hence

$$\sqrt{x^2-1}=\tan\alpha\text{ and }x=\sec\alpha$$

where you'll see which angle $\alpha$ is if you draw the picture. Similarly $$ \begin{align} \sqrt{y^2-1} = \tan\beta, & &y = \sec\beta, \\[6pt] \sqrt{z^2-1} = \tan\gamma, & &z = \sec\gamma. \end{align} $$ Now $$ \tan(\alpha+\beta+\gamma)=\frac{\tan\alpha+\tan\beta+\tan\gamma-\tan\alpha\tan\beta\tan\gamma}{1-\tan\alpha\tan\beta-\tan\alpha\tan\gamma-\tan\beta\tan\gamma}, $$ and $$ \sec(\alpha+\beta+\gamma)=\frac{\sec\alpha\sec\beta\sec\gamma}{1-\tan\alpha\tan\beta-\tan\alpha\tan\gamma-\tan\beta\tan\gamma}. $$ So the function in your integral is $$ \arcsin\left(\frac{\tan(\alpha+\beta+\gamma)}{\sec(\alpha+\beta+\gamma)}\right) = \arcsin(\sin(\alpha+\beta+\gamma))=\alpha+\beta+\gamma. $$ Then you have $$ dx = \sec\alpha\tan\alpha\,d\alpha $$ and similarly for $dy$ and $dz$. And $$ \int_1^2 \cdots\cdots dx $$ becomes $$ \int_0^{\pi/3} \cdots\cdots (\sec\alpha\tan\alpha\,d\alpha) $$ since $\sec0=1$ and $\sec\dfrac\pi3=2$.

Then make it a sum of three integrals, all with the same numerical value, one of which is $$ \iiint \alpha \sec\alpha \tan\alpha \sec\beta \tan\beta \sec\gamma \tan\gamma \,d\alpha \,d\beta \,d\gamma $$ $$ = \int_0^{\pi/3} \alpha \sec\alpha \tan\alpha\,d\alpha\cdot \int_0^{\pi/3} \sec\beta \tan\beta \,d\beta \cdot \int_0^{\pi/3} \sec\gamma\tan\gamma\,d\gamma. $$

One of these three will require integration by parts. The other two are routine.

Postscript inspired by comments below:

Clearly $$ \int_0^{\pi/3} \sec\beta\tan\beta\,d\beta = \sec\frac\pi3-\sec0 = 2-1=1. $$ Then: $$ \begin{align} & \phantom{{}=}\int_0^{\pi/3}\alpha\Big(\sec\alpha\tan\alpha\,d\alpha\Big) \\[10pt] & =\int\alpha\,d\delta=\alpha\delta-\int\delta\,d\alpha \\[10pt] & =\alpha\sec\alpha-\int\sec\alpha\,d\alpha \\[10pt] & =\left[\phantom{\int}\!\!\!\!\! \alpha\sec\alpha-\log \left| \sec\alpha + \tan\alpha \right| \, \right]_0^{\pi/3} \\[10pt] & =\frac{2\pi}{3}-\log(2+\sqrt{3}) \end{align} $$

Post-post-script inspired by further comments:

Remember that $\log(2+\sqrt{3})$ is the same thing as $-\log(2-\sqrt{3})$.

. . . and yet another addendum inspired by comments below:

Alright, so we've reduced the integral via trigonometric substitutions to $$ \iiint \arcsin(\sin(\alpha+\beta+\gamma)) (\sec\alpha\tan\alpha\,d\alpha) (\sec\beta\tan\beta\,d\beta) (\sec\gamma\tan\gamma\,d\gamma) $$ over the set $(\alpha,\beta,\gamma)\in[0,\pi/3]^3$.

If we can change $\arcsin(\sin(\bullet))$ to just $\bullet$, then the rest of what appears above solves the problem. But it is objected that within $(\alpha,\beta,\gamma)\in[0,\pi/3]^3$ there are points where $\alpha+\beta+\gamma>\pi/2$, and so we'd have $\arcsin(\sin(\alpha+\beta+\gamma))= \pi-(\alpha+\beta+\gamma)$.

One can look at that, but before I do, I have to say I suspect it's quite unnatural to consider such a thing, so I wonder why it's being done.

To be continued . . .

Michael Hardy
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  • Thanks. But by Mathematica $$ 3\int_0^{\pi/3} \alpha \sec\alpha \tan\alpha,d\alpha\cdot \int_0^{\pi/3} \sec\beta \tan\beta ,d\beta \cdot \int_0^{\pi/3} \sec\gamma\tan\gamma,d\gamma\neq3\log(2+\sqrt{3})-\pi $$ I think you have a typo. I don't think that always $\arcsin(\sin(u))=u$. –  Dec 02 '13 at 18:21
  • @Ilan : We are dealing with values of $\alpha$ between $0$ and $\pi/3$, so your concern need not concern us in this case. – Michael Hardy Dec 02 '13 at 20:10
  • @Ilan : What output did Mathematica give you, and how did you conclude it's not equal to $3\log(2+\sqrt{3})-\pi$? – Michael Hardy Dec 02 '13 at 20:11
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    @Ilan: Mathematica's result is $2\pi + 3 \log(2-\sqrt{3})$, which is the same as three times the answer above. – rogerl Dec 02 '13 at 20:33
  • @MichaelHardy: $\alpha$ is between $0$ and $\pi/3$, but $\alpha+\beta+\gamma$ not. –  Dec 03 '13 at 14:06
  • OK, if that's a problem, we just need to look at things piecewise. The idea of the method above still applies. I'll be back....... – Michael Hardy Dec 03 '13 at 19:41
  • @MichaelHardy: Can we write $$\int_0^{\pi/3}\int_0^{\pi/3}\int_0^{\pi/3} ...=\int_0^{\pi/6}\int_0^{\pi/6}\int_0^{\pi/6} ...+\int_{\pi/6}^{\pi/3}\int_{\pi/6}^{\pi/3}\int_{\pi/6}^{\pi/3} ...$$ $$=\int_0^{\pi/6}\int_0^{\pi/6}\int_0^{\pi/6} (\alpha+\beta+\gamma)...+\int_{\pi/6}^{\pi/3}\int_{\pi/6}^{\pi/3}\int_{\pi/6}^{\pi/3} (\pi-\alpha-\beta-\gamma)...?$$ –  Dec 11 '13 at 07:06
  • No. That is certainly incorrect. If you write $\int_0^1\left(\int_0^1\cdots\cdots,dy\right),dx + \int_1^2\left(\int_1^2\cdots\cdots,dy\right),dx$, then you're integrating over the region wheren $x$ and $y$ are both between $0$ and $1$, which is a square, and then over the region where both are between $1$ and $2$, which is another square. Each is a quarter of the larger square, which is the region where $x$ and $y$ are both between $0$ and $1$. The boundary between the region where $\alpha+\beta+\gamma\le\pi/2$ and the region where the same sum is $\ge\pi/2$ is different from that. – Michael Hardy Dec 11 '13 at 19:40