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#2426 No result for definite integral (II)

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nobody
5
2012-11-18
2012-06-21
No

Enter in Maxima:

integrate(sin(x)/x*log(x),x,0,inf)

and Maxima returns:

integrate((log(x)*sin(x))/x,x,0,inf)

The correct result is: -(%pi*gamma)/2, gamma=Euler's constant.

build_info("5.27.0","2012-04-24 08:52:03","i686-pc-mingw32","GNU Common Lisp (GCL)","GCL 2.6.8")

Regards

Chris

Discussion

  • Aleksas

    Aleksas - 2012-06-22

    Problem: integrate(sin(x)*log(x)/x,x,0,inf)

    We use laplace integral transformations metrhod.

    (%i1) declare(integrate,linear)$
    (%i2) assume(s>0)$

    (%i3) S:'integrate(sin(x)*log(x*a)/x,x,0,inf)$
    (%i4) f:first(S);
    (%o4) (sin(x)*log(a*x))/x
    (%i5) laplace(%,a,s);
    (%o5) ((log(x)-log(s)-%gamma)*sin(x))/(s*x)
    (%i6) integrate(%,x,0,inf),expand;
    (%o6) integrate((log(x)*sin(x))/x,x,0,inf)/s-(log(s)*integrate(sin(x)/x,x,0,inf))/s-(%gamma*integrate(sin(x)/x,x,0,inf))/s
    (%i7) ev(%, nouns);
    (%o7) integrate((log(x)*sin(x))/x,x,0,inf)/s-(%pi*log(s))/(2*s)-(%gamma*%pi)/(2*s)
    (%i8) eq1:-S=ilt(-%,s,a);
    (%o8) -integrate((sin(x)*log(a*x))/x,x,0,inf)=-integrate((log(x)*sin(x))/x,x,0,inf)+ilt((%pi*log(s))/(2*s),s,a)+(%gamma*%pi)/2

    (%i9) S1:'integrate(sin(a*x)*log(x)/x,x,0,inf)$
    (%i10) f:first(S1);
    (%o10) (log(x)*sin(a*x))/x
    (%i11) laplace(%,a,s);
    (%o11) log(x)/(x^2+s^2)
    (%i12) integrate(%,x,0,inf);
    (%o12) (%pi*log(s))/(2*s)
    (%i13) eq2:S1=ilt(%,s,a);
    (%o13) integrate((log(x)*sin(a*x))/x,x,0,inf)=ilt((%pi*log(s))/(2*s),s,a)

    (%i14) subst(a=1,[eq1,eq2]);
    (%o14) [-integrate((log(x)*sin(x))/x,x,0,inf)=-integrate((log(x)*sin(x))/x,x,0,inf)+ilt((%pi*log(s))/(2*s),s,1)+(%gamma*%pi)/2,integrate((log(x)*sin(x))/x,x,0,inf)=ilt((%pi*log(s))/(2*s),s,1)]
    (%i15) solve(%,[integrate((log(x)*sin(x))/x,x,0,inf),ilt((%pi*log(s))/(2*s),s,1)]);
    (%o15) [[integrate((log(x)*sin(x))/x,x,0,inf)=-(%gamma*%pi)/2,ilt((%pi*log(s))/(2*s),s,1)=-(%gamma*%pi)/2]]

    (%i16) solution:%[1][1];
    (%o16) integrate((log(x)*sin(x))/x,x,0,inf)=-(%gamma*%pi)/2

    Best
    Aleksas D

     
  • christoph reineke

    Aleksas,

    thank you very much. Brlliant. The master of integrals... ;-))

    All the best,

    Chris

     

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