#1749 integrate(sqrt(t)*log(t)^(1/2),t,0,1) wrong sign


The following two integrals have the wrong sign:

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

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)

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


  • Raymond Toy

    Raymond Toy - 2010-02-04

    The definite integral is computed by doing the indefinite integral via rischint. The limits are then taken. For some reason limit cannot evaluate the limit, which explains the noun form in the result.

    In addition, the limit at 0 is done by breaking the result into real and imaginary parts and taking the limit of each and putting them back together. The limit of the real part is 0, but the limit of the imaginary part has the incorrect sign. Perhaps the imaginary part is computed incorrectly?

  • Dan Gildea

    Dan Gildea - 2010-03-16

    Fixed in sin.lisp 1.58.

  • Dan Gildea

    Dan Gildea - 2010-03-16
    • status: open --> closed

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