Re: Find the integral of the function sqrt(1 + 1/(x^2/3))

In article <230320061030241247%bruck@xxxxxxxxxxxx>, Ronald Bruck
<bruck@xxxxxxxxxxxx> wrote:

In article <dvuncu$82m$1@xxxxxxxxxxxxxxxxxxxxxx>, Robert Israel
<israel@xxxxxxxxxxx> wrote:

In article <230320060837184841%bruck@xxxxxxxxxxxx>,
Ronald Bruck <bruck@xxxxxxxxxxxx> wrote:
In article <1143129670.614015.107150@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Brablo <gestureofrespect@xxxxxxxxx> wrote:

Find the integral of the function sqrt(1 + 1/(x^2/3)) from 0 to 8


No integration is required. The integrand is > 1/x, hence the integral
is infinite.

Unless the OP meant sqrt(1 + 1/x^(2/3)). In which case Maple
gives an answer in terms of 3F2 hypergeometric functions.

On the other hand, maybe the 8 was rotated 90 degrees...

Hmmm. Didn't think of that. (One must apply SOME limit to one's
assumptions about dunciness of posts.)

But it's curious Maple gives it in terms of a 3F2. Mathematica gives
the indefinite integral as

Sqrt[1 + x^(-2/3)]*(1 + x^(2/3))*x^(1/3)

Maple agrees with this (for the indefinite integral),
but gives the definite integral in terms of 3F2s.

Exercise: Show
179/16*hypergeom([-1/2, -1/6, 1/6], [4/3, 5/3], -1/64)
-249/40960*hypergeom([1/2, 5/6, 7/6], [7/3, 8/3], -1/64)
+117/33554432*hypergeom([3/2, 11/6, 13/6], [10/3, 11/3], -1/64) - 1
= 5^(3/2) - 1

more generally, does

179/16*hypergeom([-1/2, -1/6, 1/6], [4/3, 5/3], x)
-249/40960*hypergeom([1/2, 5/6, 7/6], [7/3, 8/3], x)
+117/33554432*hypergeom([3/2, 11/6, 13/6], [10/3, 11/3], x)

have an elementary representation?




which I admit isn't very elegant, but is a damned sight better than a
3F2. Just one more proof that one needs multiple CAS's.

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
.

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