Given:
  f =: 3 'x^_0.001'

The integral of f as a J expression ignoring an arbitrary constant is
  if =: _1000*^&_0.001

So the integral of f between a lower limit (ll) and an upper limit (ul)
  -~/if ll,ul

  -~/if 1 _
1000

  -~/if 1 1e300
498.813

So the calculation using adapt is correct as far as it goes. But as
the integral is close to becoming unbounded, going to 1e300 is not
nearly far enough to be correct as the upper bound goes to unbounded.

Gotta know the math behind the problem and not just blindly accept an
answer from a computer.


On Thu, Jan 7, 2010 at 12:26 AM, Oleg Kobchenko <[email protected]> wrote:
>> From: DIETER ENSSLEN <[email protected]>
>
>>
>> yes, intriguing, mysterious, baffling
>>
>> but the answer is 1000, not 498.813
>
> Tools don't contain answers, they merely enable
> people to derive them. That's exactly what I did.
> I was too lazy to recall a paper and pencil method or
> find the answer on the internet. I used J, and only J
> to find the answer. And the answer, obtained with J,
> is correct.
>
>
>> From: Oleg Kobchenko <[email protected]>
>>
>> > From: DIETER ENSSLEN
>> >
>> > Also, I just found out to my disappointment  that J won't handle relatively
>> > simple definite integrals like 1/x^1.001 from 1 to infinity
>>
>> Using induction and forgiving numerical limitations,
>> it should probably be 1000.
>                        ^^^^
>>
>>    load'~system/packages/math/integrat.ijs'
>>
>>    ^&_1.5 adapt 1 1e200
>> 2
>>    ^&_1.2 adapt 1 1e200
>> 5
>>    ^&_1.1 adapt 1 1e200
>> 10
>>
>>    ^&_1.05 adapt 1 1e300
>> 20
>>    ^&_1.02 adapt 1 1e300
>> 50
>>    ^&_1.01 adapt 1 1e300
>> 99.9
>>
>>    ^&_1.005 adapt 1 1e300
>> 193.675
>>    ^&_1.002 adapt 1 1e300
>> 374.406
>>    ^&_1.001 adapt 1 1e300
>> 498.813
>
>
>
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