# [Maxima] limits of psi

Dieter Kaiser drdieterkaiser at web.de
Thu Aug 27 17:41:34 CDT 2009

```Am Freitag, den 28.08.2009, 00:08 +0200 schrieb Dieter Kaiser:
> Am Donnerstag, den 27.08.2009, 14:35 -0700 schrieb Richard Fateman:
> > Have you, or someone, looked at the thesis by Dominik Gruntz?
>
> I have a copy of the thesis. But I have not studied the thesis in detail
> up to now.

By the way, here is an example of a definite integral. But the answer is

(%i6) integrate(x*log(sin(x)),x,0,%pi);
(%o6) %i*(('limit(12*'imagpart(li[3](%e^(%i*x)))
-12*x*'realpart(li[2](%e^(%i*x)))
+12*'imagpart(li[3](-%e^(%i*x)))
-12*x*'realpart(li[2](-%e^(%i*x)))
+6*x^2*atan(sin(x)/(cos(x)+1))
+6*x^2*atan(sin(x)/(cos(x)-1))-2*x^3,x,0,plus))
/12
-('limit(12*'imagpart(li[3](%e^(%i*x)))
-12*x*'realpart(li[2](%e^(%i*x)))
+12*'imagpart(li[3](-%e^(%i*x)))
-12*x*'realpart(li[2](-%e^(%i*x)))
+6*x^2*atan(sin(x)/(cos(x)+1))
+6*x^2*atan(sin(x)/(cos(x)-1))-2*x^3,x,%pi,minus))
/12)
-%pi*'imagpart(%pi^2/6)-%pi*'imagpart(-%pi^2/12)-%pi^2*log(4)/4

When we evalute it two times with the evflag nouns, we get:

(%i7) %,nouns;
(%o7) %i*((12*'imagpart(zeta(3))+12*'imagpart(-3*zeta(3)/4))/12
-(12*'imagpart(zeta(3))+12*'imagpart(-3*zeta(3)/4)
-12*%pi*'realpart(%pi^2/6)
-12*%pi*'realpart(-%pi^2/12)+%pi^3)
/12)
-%pi^2*log(4)/4

(%i8) %,nouns;
(%o8) -%pi^2*log(4)/4

I think the last result is correct. Wolfram alpha does not give a
symbolic result, but the numerical approximation of Maximas last answer.

I think the problem is that Maxima does not evaluate and simplify the
limits of the infinite integral accordingly:

(%i10) integrate(x*log(sin(x)),x);
(%o10) x^2*log(sin(x))/2-(3*x^2*log(sin(x)^2+cos(x)^2+2*cos(x)+1)
+3*x^2*log(sin(x)^2+cos(x)^2-2*cos(x)+1)
+6*%i*x^2*atan2(sin(x),cos(x)+1)
-6*%i*x^2*atan2(sin(x),1-cos(x))+12*li[3](%e^(%
i*x))
-12*%i*x*li[2](%e^(%i*x))+12*li[3](-%e^(%i*x))
-12*%i*x*li[2](-%e^(%i*x))-2*%i*x^3)
/12

Dieter Kaiser

```