A Special integration
Dave L. Renfro
[sci.math May 2 2003 9:47:19:000AM]
http://mathforum.org/discuss/sci.math/t/504855
wrote
> INTEGRATE[(x^3-1)/(lnx)] from 0 to 1..ans in ln4....but how?
> Hope some expert could tell me the solution....
This was also posted in alt.math.undergrad at
http://mathforum.org/epigone/alt.math.undergrad/strehchandbe
and I've already replied there, but in case anyone here
is interested -->
This (improper) integral can be evaluated using a method
sometimes called "differentiating under the integral sign".
Feynman talks about this method on pp. 86-87 of his 1985
book "Surely You're Joking Mr. Feynman", which you may recall
if you've read the book.
I discussed this method and used basically the same integral
you're asking about in this April 8, 2000 sci.math post -->
http://mathforum.org/discuss/sci.math/m/264661/264663
The same integral is also evaluated on p. 284 of Edwin Bidwell
Wilson's 1912 book "Advanced Calculus", which you can find a
digital copy of at the Cornell Digital Mathbook Collection -->
http://makeashorterlink.com/?W1E661C64
Dave L. Renfro
Axel Vogt
I am looking for Almquist & Zeilberger, the method of differentiating
under the integral sign, Journal of Symbolic Computing 10(1990).
Do you know where to get it on the www (i am off a library)?
Dave L. Renfro
[sci.math May 4 2003 2:19:34:000PM]
http://mathforum.org/discuss/sci.math/m/504855/505640
wrote (in part):
>> INTEGRATE[(x^3-1)/(lnx)] from 0 to 1..ans in ln4....but how?
>> Hope some expert could tell me the solution....
>
> This (improper) integral can be evaluated using a method
> sometimes called "differentiating under the integral sign".
Ooops! This isn't an improper integral.
Both x=0 and x=1 are removable discontinuities for (x^3 - 1) / (ln x).
The integrand clearly approaches zero as x --> 0 (numerator stays
bounded and the denominator's magnitude gets infinitely large).
Regarding x=1, note that (x^3 - 1) / (ln x)
= (x^2 + x + 1) * (x-1) / (ln x) approaches 3 as x --> 1 since
(x-1) / (ln x) approaches 1 as x --> 1. This last limit is
immediate from L'Hopital's rule, or note that
u / [ln(u+1)] = u / [u - (1/2)*u^2 + ...] approaches 1
as u = x-1 --> 0.
In fact, the discontinuity is still removable when 3 is
replaced with any positive real number p.
If we let u = x^p and note that u --> 0 if and only if
x --> 0, then
the limit as x --> 0 of (x^p - 1) / (ln x)
is equal to
the limit as u --> 0 of (u-1) / [ln(u^(1/p))]
or,
the limit as u --> 0 of p*(u-1) / (ln u),
which is p.
Therefore, the function f defined by
f(x) = (x^p - 1) / (ln x) if 0 < x < 1
f(0) = 0
f(1) = p
is continuous on the closed interval [0,1].
Dave L. Renfro
Dave L. Renfro
wrote
> I am looking for Almquist & Zeilberger, the method of
> differentiating under the integral sign, Journal of
> Symbolic Computing 10(1990). Do you know where to get
> it on the www (i am off a library)?
Our library doesn't have "Journal of Symbolic Computing", but
it does have, online in fact, "Journal Of Symbolic Computation".
However, the available issues only go back to volume 15 (1993).
Dave L. Renfro