Timeline for Integrate takes long time
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 30 at 4:30 | comment | added | Bill Watts |
I think you are right about the integral, because I can't reproduce my results from yesterday. I may have had the value a assigned without realizing it.
|
|
| Apr 30 at 2:30 | comment | added | Steve237 | oh mine's 12.1.1.0 -- didn't realize that makes a difference! (They have improved it that much? wow) However I tried with different x^a powers, like 2,3, 4,5,6 ...and each yields a different looking result! Try it! | |
| Apr 30 at 2:27 | comment | added | Bill Watts | With v14, I got a solution for the indefinite integral. | |
| Apr 30 at 2:23 | comment | added | Steve237 | @Bill Watts: I tried that (with a) but it just spits out the same input for me with the Integral sign! lol | |
| Apr 29 at 1:54 | comment | added | Bob Hanlon |
int = RootApproximant[NIntegrate[1/((1 + x^2015) (1 + x^2)), {x, 0, Infinity}] / Pi] * Pi evaluates to Pi/4
|
|
| Apr 29 at 0:38 | comment | added | Nasser | fyi, I tried this on few other CAS systems (Maple, Fricas, Rubi, Mupad) and none of them can do it. | |
| Apr 29 at 0:23 | comment | added | Steve237 | Gocha, Thx! So mathematica barfs after 10^7 bytes of mem usage normally? .... | |
| Apr 28 at 22:18 | history | edited | David G. Stork | CC BY-SA 4.0 |
deleted 20 characters in body
|
| Apr 28 at 22:10 | comment | added | Goofy |
Not complicated? There are 2,017 poles....1,009 real factors...It may not be complicated, but it's quite a large computation. For instance, this quits after 4 sec: MemoryConstrained[Integrate[1/((1 + x^2015) (1 + x^2)), {x, 0, 1, Infinity}], 10^7]
|
|
| Apr 28 at 21:30 | comment | added | Steve237 | The graphs or all other powers look different, so I am sure there is a mathematical way of proving the area is preserved, independent of the exponent. | |
| Apr 28 at 21:23 | comment | added | Steve237 | Naturally - is there a way to solve the original one | |
| Apr 28 at 21:22 | comment | added | JimB |
But if your original Integrate returned $\pi/4$, would you consider that a proof?
|
|
| Apr 28 at 21:09 | comment | added | Steve237 | Yes it's always Pi/4, however it's not a proof for the original question | |
| Apr 28 at 21:08 | history | edited | Steve237 | CC BY-SA 4.0 |
added 1 character in body
|
| Apr 28 at 21:07 | comment | added | azerbajdzan |
Try to replace 2015 with some small number and then some other small number...
|
|
| Apr 28 at 21:05 | comment | added | Steve237 | Thanks, but how? ... | |
| Apr 28 at 21:02 | comment | added | azerbajdzan | $\frac{\pi }{4}$ | |
| Apr 28 at 20:58 | history | asked | Steve237 | CC BY-SA 4.0 |