The following integral was posted a few days back on Integrals and Series forum:
∫2π0∫2π0∫2π0dk1dk2dk31−13(cosk1+cosk2+cosk3)=√64Γ(124)Γ(524)Γ(724)Γ(1124)
I am curious if there is a closed form solution for:
∫[0,2π]ndk1dk2dk3⋯dkn1−1n(cosk1+cosk2+cosk3+⋯+coskn)
Since |cosk1+cosk2+cosk33|<1,
∫2π0∫2π0∫2π0dk1dk2dk31−13(cosk1+cosk2+cosk3)
=8∫π0∫π0∫π0dk1dk2dk31−13(cosk1+cosk2+cosk3)
=8∞∑n=013n∫π0∫π0∫π0(cosk1+cosk2+cosk3)ndk1dk2dk3
We can ignore the odd values of n as the integral is zero for them. Also, for even values of n, the exponents of cosines in the expansion of (cosk1+cosk2+cosk3)2n must be even. Hence, from multinomial therem, we can write:
8∞∑n=0∑m1+m2+m3=n132n(2n)!(2m1)!(2m2)!(2m3)!∫π0∫π0∫π0cos2m1k1cos2m2k2cos2m3k3dk1dk2dk3
=16∞∑n=0∑m1+m2+m3=n132n(2n)!(2m1)!(2m2)!(2m3)!∫π/20∫π/20∫π/20cos2m1k1cos2m2k2cos2m3k3dk1dk2dk3
Using the result: ∫π/20cos2kxdx=(2k)!4k(k!)2π2, the integral is,
2π3∞∑n=0∑m1+m2+m3=n136n(2n)!(m1!)2(m2!)2(m3!)2
I am stuck here.
Any help is appreciated. Thanks!
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