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From: J. Clarke on 14 May 2010 17:30
On 5/14/2010 3:46 PM, Robert Israel wrote:
> A N Niel<anniel(a)nym.alias.net.invalid> writes:
>
>> In article<AOdHn.100119$ae7.98318(a)unlimited.newshosting.com>, Greg
>> Neill<gneillRE(a)MOVEsympatico.ca> wrote:
>>
>>> J. Clarke wrote:
>>>> On 5/14/2010 9:53 AM, andrews wrote:
>>>>> I am not a student, thus it is not for a homework.
>>>>> With the answers I could find the solution.
>>>>> It is based on int(df/f) = lnf where f is any function
>>>>> Thanks for this
>>>>
>>>> Just for future reference, Maxima (GPL computer algebra system) solved
>>>> this instantly. If you put it in the form "1/(x*ln(x))" then it's in
>>>> the Schaums mathematical handbook ($12.95 from Amazon) and the Ti-89
>>>> calculator solved it as well. I think you'll find having one or all of
>>>> those on hand will be a tremendous convenience if you run into this
>>>> kind
>>>> of question often.
>>>
>>> Wolfram's online integrator is also handy.
>>>
>>> Interestingly, it can resolve 1/(x*ln(x)) but
>>> not 1/ln(x^x).
>>>
>>>
>>
>>> http://integrals.wolfram.com/index.jsp?expr=1%2F%28x*ln%28x%29%29&random=fal
>>> se
>>>
>>>
>>
>> So wolfram is smart, like Maple ... here is Maple:
>>
>> int(1/x/log(x),x);
>> ln(ln(x))
>> int(1/log(x^x),x);
>> / 1 \
>> int|------, x|
>> | / x\ |
>> \ln\x / /
>> int(1/log(x^x),x) assuming x>0;
>> ln(ln(x))
>
> Indeed: if you're using principal branches, log(x^x) = x log(x) only when
> -pi< Im(x log(x))<= pi. For example, it's false for real x< -1, where
> Im(x log(x)) = pi x. The general formula is log(x^x) = x log(x) + 2 pi i n
> where n is chosen so -pi< Im(x log(x)) + 2 pi n<= pi.
> And for n<> 0, 1/(x log(x) + 2 pi i n) does not have an elementary
> antiderivative.

Learn something every day.

Wrapped "Assuming[x>0 . . .]" around the integral in Mathematica and it
came out with the answer. Doesn't seem to be a way to do that with the
online integrator though.

Also tried this on an HP49 emulator (emulates the hardware, uses the HP
ROMs) and it took 1/ln(x^x).

Now I need to see if I can tell the Ti how to restrict the range.



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