Integral hangs G2
03-29-2020, 06:38 PM
Post: #1
 lrdheat Senior Member Posts: 809 Joined: Feb 2014
Integral hangs G2
The integral from 0 to 1 of 3Root(1-x^7) - 7Root(1-x^3) hangs the G2, requiring a reset. The virtual unit gives a warning message, pressing enter returns original integral (which one can then ~ans). Virtual and G2 work in Home...almost produce the correct answer of zero.
03-29-2020, 09:56 PM
Post: #2
 Albert Chan Senior Member Posts: 2,230 Joined: Jul 2018
RE: Integral hangs G2
(03-29-2020 06:38 PM)lrdheat Wrote:  The integral from 0 to 1 of 3Root(1-x^7) - 7Root(1-x^3) ... correct answer of zero.

Googled "wallis product", I found this: Mathematical Analysis and the Mathematics of Computation, p.404, eqn. 7.87

$$\Large {1 \over \int_0^1 (1-x^{1/p})^q\;dx} = \binom{p+q}{p} = \binom{p+q}{q} = {1 \over \int_0^1 (1-x^{1/q})^p\;dx}$$

P.S. I do not know how above is derived. Any insight is appreciated.
03-30-2020, 12:16 AM
Post: #3
 toml_12953 Senior Member Posts: 2,031 Joined: Dec 2013
RE: Integral hangs G2
(03-29-2020 09:56 PM)Albert Chan Wrote:
(03-29-2020 06:38 PM)lrdheat Wrote:  The integral from 0 to 1 of 3Root(1-x^7) - 7Root(1-x^3) ... correct answer of zero.

Googled "wallis product", I found this: Mathematical Analysis and the Mathematics of Computation, p.404, eqn. 7.87

$$\Large {1 \over \int_0^1 (1-x^{1/p})^q\;dx} = \binom{p+q}{p} = \binom{p+q}{q} = {1 \over \int_0^1 (1-x^{1/q})^p\;dx}$$

P.S. I do not know how above is derived. Any insight is appreciated.

The G2 shouldn't crash over it, though.

Tom L
Cui bono?
03-30-2020, 12:27 AM
Post: #4
 roadrunner Senior Member Posts: 437 Joined: Jun 2015
RE: Integral hangs G2
My G2 didn't crash; it behaved exactly like the simulator. Perhaps you have x defined somewhere?
03-30-2020, 05:16 PM (This post was last modified: 06-30-2022 07:31 PM by Albert Chan.)
Post: #5
 Albert Chan Senior Member Posts: 2,230 Joined: Jul 2018
RE: Integral hangs G2
(03-29-2020 09:56 PM)Albert Chan Wrote:  Googled "wallis product", I found this: Mathematical Analysis and the Mathematics of Computation, p.404, eqn. 7.87

$$\Large {1 \over \int_0^1 (1-x^{1/p})^q\;dx} = \binom{p+q}{p} = \binom{p+q}{q} = {1 \over \int_0^1 (1-x^{1/q})^p\;dx}$$

Just realized the integral is simply the beta function (see equation 18) Let $$(1-u)^p = x\quad →\quad -p(1-u)^{p-1}\;du=dx$$

$$\large \int_0^1 (1-x^{1/p})^q\;dx = p \int_0^1 u^q (1-u)^{p-1}\;du = p\;B(q+1,p) = {\Gamma(p+1)\Gamma(q+1) \over \Gamma(p+q+1)} = 1 / \binom{p+q}{p}$$

Update:

We can prove u integral is beta function by induction (via integration by parts)
Foundations of Combinatorics with Applications:       Appendix A, Example A.4
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