Triple Integration

Calculus Level 5

0 2 π 0 2 π 0 2 π d x d y d z 1 1 3 ( cos x + cos y + cos z ) \large \int_{0}^{2\pi} \int_{0}^{2\pi} \int_{0}^{2\pi} \dfrac{ dx \; dy \; dz}{ 1- \frac 13 ( \cos x + \cos y + \cos z) }

If the closed form of the integral above can be expressed as:

A 4 Γ ( B 24 ) Γ ( C 24 ) Γ ( D 24 ) Γ ( E 24 ) \dfrac{\sqrt{A}}{4} \Gamma \left( \dfrac{B}{24}\right)\Gamma\left(\dfrac{C}{24}\right) \Gamma \left(\dfrac{D}{24}\right) \Gamma \left(\dfrac{E}{24}\right)

Find A + B + C + D + E A+B+C+D+E .

Also try


The answer is 30.

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1 solution

Mark Hennings
Jun 4, 2017

This is basically one of the three famous Watson triple integrals 0 2 π 0 2 π 0 2 π 1 1 1 3 ( cos x + cos y + cos z ) d x d y d z = 24 π 3 I 3 \int_0^{2\pi}\int_0^{2\pi}\int_0^{2\pi} \frac{1}{1 - \frac13(\cos x + \cos y + \cos z)}\,dx\,dy\,dz \; = \; 24\pi^3I_3 where I 3 = 1 π 3 0 π 0 π 0 π 1 3 cos x cos y cos z d x d y d z = 1 16 6 π 3 Γ ( 1 24 ) Γ ( 5 24 ) Γ ( 7 24 ) Γ ( 11 24 ) I_3 \; = \; \frac{1}{\pi^3}\int_0^\pi \int_0^\pi \int_0^\pi \frac{1}{3 - \cos x - \cos y - \cos z}\,dx\,dy\,dz \; =\; \frac{1}{16\sqrt{6}\pi^3}\Gamma\big(\tfrac{1}{24}\big)\Gamma\big(\tfrac{5}{24}\big)\Gamma\big(\tfrac{7}{24}\big)\Gamma\big(\tfrac{11}{24}\big) The proof is in the original 1939 paper, but some details can be found here .

This makes this integral equal to 6 4 Γ ( 1 24 ) Γ ( 5 24 ) Γ ( 7 24 ) Γ ( 11 24 ) \frac{\sqrt{6}}{4}\Gamma\big(\tfrac{1}{24}\big)\Gamma\big(\tfrac{5}{24}\big)\Gamma\big(\tfrac{7}{24}\big)\Gamma\big(\tfrac{11}{24}\big) and so the desired answer is 6 + 1 + 5 + 7 + 11 = 30 6+1+5+7+11 = \boxed{30} .

Yay! I did it! My brother couldnt!

Md Junaid - 3 years, 11 months ago

According to the Wolfram MathWorld source provided by @Mark Hennings , "to obtain an entirely closed form, it is necessary to do perform some analytic wizardry." This is way over my head ...

But nice solution anyway!!

Christopher Criscitiello - 3 years, 9 months ago

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