The following two integrals have the wrong sign:

integrate(sqrt(t)*log(t)^(1/2),t,0,1) and

integrate(sqrt(t)*log(t)^(-1/2),t,0,1)

It is interesting that Maxima is able to solve the more general type:

(%i164) declare(s,noninteger);

(%o164) done

(%i165) expr:integrate(sqrt(t)*log(t)^s,t,0,1);

(%o165) 3^(-s-1)*(-1)^s*2^(s+1)*gamma_incomplete(s+1,0)

For s=1/2 and s=-1/2 we get the answers:

(%i167) expr,s=1/2;

(%o167) sqrt(2)*sqrt(%pi)*%i/(2*sqrt(3))

(%i168) expr,s=-1/2;

(%o168) -sqrt(2)*sqrt(%pi)*%i/sqrt(3)

Both solutions can be checked to be correct.

Now we do it directly:

(%i4) integrate(sqrt(t)*log(t)^(1/2),t,0,1);

(%o4) -%i*('limit(sqrt(2)*sqrt(%pi)*erf(sqrt(3)*sqrt(-log(t))/sqrt(2))/3^(3/2)

-2*t^(3/2)*sqrt(-log(t))/3,t,0,plus))

We need an extra evaluation, but this is another problem:

(%i5) %,nouns;

(%o5) -sqrt(2)*sqrt(%pi)*%i/3^(3/2)

Now the integral for s=-1/2:

(%i6) integrate(sqrt(t)*log(t)^(-1/2),t,0,1);

(%o6) sqrt(2)*sqrt(%pi)*%i/sqrt(3)

These solutions differ by the sign with the answers from above.

I have checked it for a lot of other values for the parameter s. In all other cases the result of the integral and the more general solution are equal.

Remark: The integral is divergent for s a negative integer. For these cases the gamma_incomplete function is not defined.

Dieter Kaiser