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Topic: The Charlwood Fifty
Replies: 52   Last Post: Jun 24, 2013 10:24 PM

 Messages: [ Previous | Next ]
 Waldek Hebisch Posts: 267 Registered: 12/8/04
Re: The Charlwood Fifty
Posted: Jun 17, 2013 7:56 PM

clicliclic@freenet.de wrote:
>
> "Nasser M. Abbasi" schrieb:

> >
> > free version of reduce transformed
> >
> > arcsin(x)*log(x)
> >
> > to
> >
> > arcsin( sin(g0) ) * cos(g0) * log( sin(g0) )
> >
> > by replacing x with sin(g0).
> >
> > i.e Where does cos(g0) term come from in the above transformation?
> >
> > Here is a link to the reduce trace for integral #1, which it
> > could not do btw. And the above was the first step in the process.
> >
> > http://12000.org/my_notes/ten_hard_integrals/reduce_logs/1/HTML/trace_1.html
> >

>
> In the original INT(ASIN(x)*LN(x), x), Reduce makes the variable
> substitution x = SIN(y), dx = COS(y)*dy (I've written y for g0). So the
> factor COS(y) just represents the derivative dx/dy that must be included
> in the transformed integrand.
>
> It looks like the integration process is simply restarted after the
> variable substitution. While the first step is easy to understand, I
> have no clear idea what Reduce is trying to do later - and without
> success. It may be the Risch-Norman heuristic again.
>
> By the way, Derive 6.10 attacks the original integral using integration
> by parts. It can also evaluate INT(x*COS(x)*LN(SIN(x)), x) which is
> obtained when ASIN(SIN(x)) in the variable-substituted integral is
> replaced by x (this replacement is not valid for all complex x, nor for
> all real x, however).

I am not sure why Reduce fares so poorly on this problem. It
can do:

3: int(x*cos(x)*log(sin(x)), x);

x
2*tan(---)
2 x 2
cos(x)*log(---------------) - 2*cos(x) - log(tan(---) + 1)
x 2 2
tan(---) + 1
2

x x
2*tan(---) 2*tan(---)
2 2
+ log(---------------)*sin(x)*x - log(---------------) - sin(x)*x + 2
x 2 x 2
tan(---) + 1 tan(---) + 1
2 2

so maybe it fails at backsubstitution x -> asin(x). Or maybe
it got doubts like you against replacing asin(sin(x)) by x.
But such replacement is fine:

(21) -> integrate(x*cos(x)*log(sin(x)), x)

(21)
cos(x) + 1 - cos(x) + 1
(2x sin(x) + 2cos(x))log(sin(x)) + log(----------) - log(------------)
2 2
+
- 2x sin(x) - 4cos(x)
/
2
Type: Union(Expression(Integer),...)
(22) -> eval(%, x = asin(x))

(22)
+--------+ +--------+
| 2 | 2 +--------+
\|- x + 1 + 1 - \|- x + 1 + 1 | 2
log(---------------) - log(-----------------) + (2log(x) - 4)\|- x + 1
2 2
+
2x asin(x)log(x) - 2x asin(x)
/
2
Type: Expression(Integer)
(23) -> ii := D(%, x)

(23) asin(x)log(x)
Type: Expression(Integer)

In general, since original integrand is analytic except for {-1, 0, 1},
it is enough for substitution to be valid in neighbourhood of a
single point. You just need to be careful to choose desired
branch after backsubstition.

--
Waldek Hebisch
hebisch@math.uni.wroc.pl