How do you integrate $\int_0^1 xe^{(\log(x))^k}dx ~?$ (for $k=7$).
For $k=3$ Wolfram alpha says the closed form is in terms of the generalized hypergeometric function and the Bi-airy function. For $k=5$ Wolfram alpha says the closed form is in terms of the gamma function and generalized hypergeometric function.
For $k=7$ Wolfram alpha says that the standard computation time exceeded.
I think the closed form, if there is one, will involve the generalized hypergeometric function and some other special function.
The reason I ask about this is because I want to know what the closed form of the integral is for $k=7.$
One thought I had while thinking about this problem is:
"For $k=3,5$ the generalized hypergeometric function is present both times for the closed form but for $k=3$ we have the Bi-Airy function whereas for $k=5$ we have the gamma function. I'm not sure why $k=5$ should have the gamma function and not the Bi-airy function again."
Best Answer
For $k=7$ Wolfram gives the exact value of the integral as $$\Gamma(8/7) _0 F_5(;2/7, 3/7, 4/7, 5/7, 6/7;-128/823543) + (112 _1 F_6(1;8/7, 9/7, 10/7, 11/7, 12/7, 13/7;-128/823543) + 840 \Gamma(5/7) _0 F_5(;6/7, 8/7, 9/7, 10/7, 11/7;-128/823543) - 336 \Gamma(6/7) _0 F_5(;8/7, 9/7, 10/7, 11/7, 12/7;-128/823543) - 315 \sqrt{7} \csc{(\pi/7)} \sec{(\pi/14)} \sec{((3\pi)/14)} \Gamma(2/7) _0 F_5(;3/7, 4/7, 5/7, 6/7, 8/7;-128/823543) - 180 \sqrt{7} \csc{(\pi/7)} \sec{(\pi/14)} \sec{((3 \pi)/14)} \Gamma(-4/7) _0 F_5(;4/7, 5/7, 6/7, 8/7, 9/7;-128/823543) - 210 \sqrt{7} \csc{(\pi/7)} \sec{(\pi/14)} \sec{((3 \pi)/14)} \Gamma(4/7) _0 F_5(;5/7, 6/7, 8/7, 9/7, 10/7;-128/823543))/8820$$ Which is approximately equal to