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The Maxima on-line user's manual

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Intanalysis Calculator

Intanalysis

-- Option variable: intanalysis Default value: true

When true, definite integration tries to find poles in the integrand in the interval of integration. If there are, then the integral is evaluated appropriately as a principal value integral. If intanalysis is false, this check is not performed and integration is done assuming there are no poles.

assume (t>0);
assume (T>0);
erfflag:false;
erfflag;
intanalysis;
integrate (j*(2*j+1)*exp(-(t/T)*j*(j+1)), j, 0, inf);

See also ldefint.

Examples:

Maxima can solve the following integrals, when intanalysis is set to false:

          (%i1) integrate(1/(sqrt(x)+1),x,0,1);
                                          1
                                         /
                                         [       1
          (%o1)                          I  ----------- dx
                                         ]  sqrt(x) + 1
                                         /
                                          0

          (%i2) integrate(1/(sqrt(x)+1),x,0,1),intanalysis:false;
          (%o2)                            2 - 2 log(2)

          (%i3) integrate(cos(a)/sqrt((tan(a))^2 +1),a,-%pi/2,%pi/2);
          The number 1 isnt in the domain of atanh
           -- an error. To debug this try: debugmode(true);

          (%i4) intanalysis:false$
          (%i5) integrate(cos(a)/sqrt((tan(a))^2+1),a,-%pi/2,%pi/2);
                                                %pi
          (%o5)                                 ---
                                                 2

(%o1)                                true
(%i2) 

Intanalysis Example

Related Examples