List Archives >  Maple User Group List Archive >  Archive by date >  This Month By Date >  This Month By Topic

[MUG] missing definite integral

Search email archive for  

[MUG] missing definite integral
Author: Rob Scott    Posted: Thu, 24 Jan 2002 08:48:55 -0500
>> From: Rob Scott /> The following definite integral has a solution in terms of erf: assume(a>0, b>0, c::real); int( sin(a * x) * cos(b * x) * exp(- c^2 * x^2) / x, x = 0 .. infinity); maple can't handle this though the correct solution is: Pi/4 * ( erf( (a-b)/2/c ) + erf( (a+b)/2/c ) ) P.S. Sorry if this list is the correct forum for this comment.

[MUG] Re: missing definite integral
Author: Maple User Group    Posted: Fri, 1 Feb 2002 11:46:13 -0500 (
>> From: Maple User Group /> >> From: Rob Scott /> | The following definite integral has a solution in terms of erf: | assume(a>0, b>0, c::real); | int( sin(a * x) * cos(b * x) * exp(- c^2 * x^2) / x, x = 0 .. infinity); | maple can't handle this though the correct solution is: | Pi/4 * ( erf( (a-b)/2/c ) + erf( (a+b)/2/c ) ) -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Fri, 25 Jan 2002 18:28:15 -0500 To: /> From: "Gerald A. Edgar" /> Subject: missing definite integral I do not think Maple (nor any other program) claims to do all known definite integrals. At least Maple does this: > int(sin(a*x)*cos(a*x)*exp(-c^2*x^2)/x,x=0..infinity); 2 1/2 Pi (c ) erf(a/c) 1/4 ------------------- c And with some help from me, it can do the one in question: 2 2 sin(a x) cos(b x) exp(-c x ) ----------------------------- x > combine(%); 2 2 2 2 exp(-c x ) sin(a x + b x) + exp(-c x ) sin(a x - b x) 1/2 ------------------------------------------------------- x > int((exp(-c^2*x^2)*sin((a+b)*x)+exp(-c^2*x^2)*sin((a-b)*x)) > /x/2,x=0..infinity); a + b a - b Pi c erf(1/2 -----) 1/4 Pi c erf(1/2 -----) c c 1/4 ------------------- + ----------------------- 2 1/2 2 1/2 (c ) (c ) -- Gerald A. Edgar /> -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- From: "Willard, Daniel Dr DUSA-OR" /> To: /> Subject: missing definite integral Date: Mon, 28 Jan 2002 10:16:03 -0500 But note that sin(a*x)*cos(b*x)=1/2((sin(a+b)*x)+sin((a-b)*x)) and Maple recognizes the integrals of these terms separately. -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Mon, 28 Jan 2002 18:58:41 -0600 (CST) From: Brian Blank /> To: /> Subject: missing definite integral Since you are not assuming that c is positive the expression that you say is correct should not depend on sgn(c). Maple knows how to do this integral correctly but you must prepare the integrand first. > restart; > assume(a , positive, b , positive , c , real ): > Int( sin(a*x) * cos(b*x) * exp(-c^2*x^2) / x, x = 0 .. infinity): > map(combine, %): # In Release 4 use " instead of % > e1 := student[integrand](%): > expand(expandoff()): > expandoff(sin): > expand(e1): > map(z -> Int(z, x = 0 .. infinity), % ): > map(factor, %): # Regrettable! > value(%); | c | a 1/2 | c | b 1/4 Pi erf(1/2 ------- + -----------) 2 2 c c | c | a | c | b + 1/4 Pi erf(1/2 ------- - 1/2 -------) 2 2 c c with assumptions on c, a and b Brian Blank Department of Mathematics Washington University in St. Louis, St. Louis, MO 63130

[MUG] Re: missing definite integral
Author: Carl Devore    Posted: Thu, 31 Jan 2002 21:40:23 -0500
>> From: Carl Devore /> On Thu, 24 Jan 2002, Rob Scott wrote: > The following definite integral has a solution in terms of erf: > assume(a>0, b>0, c::real); > int( sin(a * x) * cos(b * x) * exp(- c^2 * x^2) / x, x = 0 .. infinity); > > maple can't handle this though the correct solution is: > > Pi/4 * ( erf( (a-b)/2/c ) + erf( (a+b)/2/c ) ) If we apply a trivial identity to the trig factors, then Maple can do the integral. Why Maple can't apply the identity automatically is a an interesting question discussed (but not answered) afterwards. > restart; Making assumptions about a and b is unncessary and does not affect Maple's ability to do the problem. Without loss of generality, we may assume c>0. This also does not affect Maple's ability to do the problem. It just makes the answer simplify neater. > assume(c>0); The trig identity is: > combine(sin(a*x)*cos(b*x)); 1/2 sin(a x + b x) + 1/2 sin(a x - b x) Note the distributed form of the arguments. Surprisingly, this turns out to be why Maple can't do the problem. I will enter the integral with those arguments factored. > (1/2)*Int((sin((a+b)*x)+sin((a-b)*x))*exp(-c^2*x^2)/x, x= 0..infinity); infinity / 2 2 | (sin((a + b) x) + sin((a - b) x)) exp(-c~ x ) 1/2 | ---------------------------------------------- dx | x / 0 > value(%); a + b a - b 1/4 Pi erf(1/2 -----) + 1/4 Pi erf(1/2 -----) c~ c~ The form that this answer is returned in seems random, even after doing restarts. This is perhaps related to the "random returns from radsimp". I obtained all of the following forms *even after doing restarts* > value(%); a + b Pi (1/2 a + 1/2 b) erf(1/2 -----) c~ 1/2 --------------------------------- a + b a - b 1/2 Pi (1/2 a - 1/2 b) erf(1/2 -----) c~ + ------------------------------------- a - b > value(%); a - b 1/2 Pi (1/2 a - 1/2 b) erf(1/2 -----) a + b c~ 1/4 Pi erf(1/2 -----) + ------------------------------------- c~ a - b > value(%); a + b Pi (1/2 a + 1/2 b) erf(1/2 -----) c~ a - b 1/2 --------------------------------- + 1/4 Pi erf(1/2 -----) a + b c~ They can all trivially be seen to be equivalent to the simplified form. It is just a question of whether the (a+b) and/or (a-b) gets factored out of the numerators. Now I do the integral with the sine arguments distributed: > Int((sin(a*x+b*x)+sin(a*x-b*x))*exp(-c^2*x^2)/x, x= 0..infinity); infinity / 2 2 | (sin(a x + b x) + sin(a x - b x)) exp(-c~ x ) | ---------------------------------------------- dx | x / 0 > value(%); infinity / 2 2 | (sin(a x + b x) + sin(a x - b x)) exp(-c~ x ) | ---------------------------------------------- dx | x / 0 Maple decides quickly that it can't do it. Let's investigate the distinction. The distinction remains if I simplify the problem one sine term and set c=1, b=1. > restart; > infolevel[all]:= 5: > Int(sin(a*x+x)*exp(-x^2)/x, x= 0..infinity); infinity / 2 | sin(a x + x) exp(-x ) | --------------------- dx | x / 0 > value(%); Maple will try many different things below. So the following output is lengthy. (I have deleted some of it). int/elliptic: trying elliptic integration int/ellalg/elltype: Checking for an elliptic integral sin(a*x+x)*exp(-x^2)/x freeof(x) x int/indef1: first-stage indefinite integration int/indef2: second-stage indefinite integration int/trigexp: case of integrand containing exp and trigs combine: combining with respect to trig combine: combining with respect to exp int/rischnorm: enter Risch-Norman integrator sqrfree/Yun: square-free factorization in T[3] sqrfree/Yun: square-free factorization in T[1] int/rischnorm: exit Risch-Norman integrator int/risch: enter Risch integration int/risch/algebraic1: RootOfs should be algebraic numbers and functions radnormal/addrads: radicals are {`radnormal/Radical`(-1,2,0)} radnormal/addone: adding a single radical radnormal/addone: radical is `radnormal/Radical`(-1,2,0) radfield: field is {RootOf(_Z^2+1,index = 1)} radfield: extension degree is 2 radfield: substitutions are [`radnormal/Radical`(-1,2,0) = RootOf(_Z^2+1,index = 1)] radfield: backward substitutions are [RootOf(_Z^2+1,index = 1) = I] radfield: field is {RootOf(_Z^2+1,index = 1)} radfield: extension degree is 2 radfield: substitutions are [`radnormal/Radical`(-1,2,0) = RootOf(_Z^2+1,index = 1)] radfield: backward substitutions are [RootOf(_Z^2+1,index = 1) = I] solve/linear: # equations 3 evala/Reduce/nf/urad: Reducing radicals [RootOf(_Z^2+1,index = 1)] at time 157.729 solve/linear/sparse/polyalg: # equations is: 2 radfield: computing a basis for {RootOf(_Z^2+1,index = 1)} radnormal/addtop: algebraics are {`radnormal/Radical`(-1,2,0,`radnormal/notation/RootOf`)} radnormal/addrads: adding 1 radicals radnormal/addrads: radicals are {`radnormal/Radical`(-1,2,0,`radnormal/notation/RootOf`)} radnormal/addone: adding a single radical radnormal/addone: radical is `radnormal/Radical`(-1,2,0,`radnormal/notation/RootOf`) radfield: field is {RootOf(_Z^2+1,index = 1)} radfield: extension degree is 2 radfield: substitutions are [`radnormal/Radical`(-1,2,0,`radnormal/notation/RootOf`) = RootOf(_Z^2+1,index = 1)] radfield: backward substitutions are [] int/risch: the field extensions are 2 2 [x~, exp(-x~ ), exp(RootOf(_Z + 1, index = 1) (a x~ + x~))] int/risch: Introduce the namings: 2 {_th[1] = exp(-x~ ), 2 _th[2] = exp(RootOf(_Z + 1, index = 1) (a x~ + x~))} int/risch/int: integrand is / 1 \ |_th[2] - ------| _th[1] \ _th[2]/ ------------------------ x~ int/risch/int: integrand expressed as _th[1] _th[2] _th[1] ------------- - --------- x~ x~ _th[2] int/risch/exppoly: integrating _th[1] _th[2] _th[1] ------------- - --------- x~ x~ _th[2] int/risch/diffeq: solving Risch d.e. y' + f y = g where f,g are: 2 _th[1] RootOf(_Z + 1, index = 1) (a + 1), ------ x~ evala/Indep: testing independence of RootOfs evala/Indep: number of RootOfs 1 evala/Indep: number of RootOfs 1 evala/Reduce/nf/urad: Reducing radicals [RootOf(_Z^2+1,index = 1)] at time 157.800 int/risch/DEexp: solving Risch d.e. y' + f y = g where f,g are: 2 _th[1] RootOf(_Z + 1, index = 1) (a + 1), ------ x~ int/risch/int: integrand is 2 RootOf(_Z + 1, index = 1) (a + 1) int/ratpoly/horowitz: integrating 2 RootOf(_Z + 1, index = 1) (a + 1) int/risch/ratpoly: result is 2 RootOf(_Z + 1, index = 1) (a + 1) x~ radfield: computing a basis for {RootOf(_Z^2+1,index = 1)} radfield: field is {RootOf(_Z^2+1,index = 1)} radfield: extension degree is 2 radfield: substitutions are [`radnormal/Radical`(-1,2,0,`radnormal/notation/RootOf`) = RootOf(_Z^2+1,index = 1)] radfield: computing a basis for {RootOf(_Z^2+1,index = 1)} radfield: field is {RootOf(_Z^2+1,index = 1)} radfield: extension degree is 2 radfield: substitutions are [`radnormal/Radical`(-1,2,0,`radnormal/notation/RootOf`) = RootOf(_Z^2+1,index = 1)] solve/linear: # equations 3 solve/linear/sparse/polyalg: # equations is: 2 int/risch/diffeq: solving Risch d.e. y' + f y = g where f,g are: 2 1 -2 x~ + RootOf(_Z + 1, index = 1) (a + 1), ---- x~ evala/Reduce/nf/urad: Reducing radicals [RootOf(_Z^2+1,index = 1)] at time 157.879 evala/Reduce/nf/urad: Reducing radicals [RootOf(_Z^2+1,index = 1)] at time 157.879 evala/Reduce/nf/urad: Reducing radicals [RootOf(_Z^2+1,index = 1)] at time 157.879 evala/inv: heuristic succeeded evala/Reduce/nf/urad: Reducing radicals [RootOf(_Z^2+1,index = 1)] at time 157.889 evala/Reduce/nf/urad: Reducing radicals [RootOf(_Z^2+1,index = 1)] at time 157.889 evala/Reduce/nf/urad: Reducing radicals [RootOf(_Z^2+1,index = 1)] at time 157.889 int/risch/DEratpoly: solving Risch d.e. y' + f y = g where f,g are: 2 RootOf(_Z + 1, index = 1) 2 1 (2 RootOf(_Z + 1, index = 1) x~ + a + 1), ---- x~ int/risch/exppoly: Risch d.e. has no solution int/indef1: first-stage indefinite integration int/indef1: first-stage indefinite integration int/indef2: second-stage indefinite integration int/exp: case of integrand containing exp int/indef1: first-stage indefinite integration int/indef2: second-stage indefinite integration int/exp: case of integrand containing exp combine: combining with respect to exp int/indef1: first-stage indefinite integration int/indef2: second-stage indefinite integration int/exp: case of integrand containing exp int/prpexp: case ratpoly*exp(arg) int/rischnorm: enter Risch-Norman integrator radfield: computing a basis for {RootOf(_Z^2+1,index = 1)} radfield: field is {RootOf(_Z^2+1,index = 1)} radfield: extension degree is 2 radfield: substitutions are [`radnormal/Radical`(-1,2,0,`radnormal/notation/RootOf`) = RootOf(_Z^2+1,index = 1)] radfield: computing a basis for {RootOf(_Z^2+1,index = 1)} radfield: field is {RootOf(_Z^2+1,index = 1)} radfield: extension degree is 2 radfield: substitutions are [`radnormal/Radical`(-1,2,0,`radnormal/notation/RootOf`) = RootOf(_Z^2+1,index = 1)] solve/linear: # equations 3 solve/linear/sparse/polyalg: # equations is: 2 int/rischnorm: exit Risch-Norman integrator int/indef1: first-stage indefinite integration int/indef1: first-stage indefinite integration int/indef2: second-stage indefinite integration int/exp: case of integrand containing exp int/indef1: first-stage indefinite integration int/indef2: second-stage indefinite integration int/exp: case of integrand containing exp combine: combining with respect to exp int/indef1: first-stage indefinite integration int/indef1: first-stage indefinite integration int/indef2: second-stage indefinite integration int/exp: case of integrand containing exp int/prpexp: case ratpoly*exp(arg) int/rischnorm: enter Risch-Norman integrator radfield: computing a basis for {RootOf(_Z^2+1,index = 1)} radfield: field is {RootOf(_Z^2+1,index = 1)} radfield: extension degree is 2 radfield: substitutions are [`radnormal/Radical`(-1,2,0,`radnormal/notation/RootOf`) = RootOf(_Z^2+1,index = 1)] radfield: computing a basis for {RootOf(_Z^2+1,index = 1)} radfield: field is {RootOf(_Z^2+1,index = 1)} radfield: extension degree is 2 radfield: substitutions are [`radnormal/Radical`(-1,2,0,`radnormal/notation/RootOf`) = RootOf(_Z^2+1,index = 1)] solve/linear: # equations 3 solve/linear/sparse/polyalg: # equations is: 2 int/rischnorm: exit Risch-Norman integrator int/risch: exit Risch integration int/definite/contour: contour integration infinity / 2 | sin(a x + x) exp(-x ) | --------------------- dx | x / 0 Constrast that with the factored form. The following output is brief. Maple finds the correct approach very quickly. > restart; > infolevel[all]:= 5: > Int(sin((a+1)*x)*exp(-x^2)/x, x= 0..infinity); infinity / 2 | sin((a + 1) x) exp(-x ) | ----------------------- dx | x / 0 > value(%); simplify/do: applying simplify/exp function to expression simplify/do: applying simplify/exp function to expression simplify/do: applying commonpow function to expression simplify/do: applying power function to expression simplify/do: applying simplify/exp function to expression int/indef1: first-stage indefinite integration int/indef1: first-stage indefinite integration int/indef2: second-stage indefinite integration int/indef2: applying algebraic substitution int/indef1: first-stage indefinite integration int/indef1: first-stage indefinite integration int/indef2: second-stage indefinite integration int/indef2: applying change of variables int/indef1: first-stage indefinite integration int/indef1: first-stage indefinite integration int/indef2: second-stage indefinite integration int/exp: case of integrand containing exp int/prpexp: case ratpoly*exp(arg) 1/2 Pi erf(1/2 a + 1/2) The next step is to turn trace on and see where the differences in the problems start. But I don't feel like it now. Let's see what happens with a simple problem. > restart; > infolevel[all]:= 5: > int(sin(a*x+x), x); int/indef1: first-stage indefinite integration int/indef2: second-stage indefinite integration int/indef2: applying change of variables int/indef1: first-stage indefinite integration int/indef1: first-stage indefinite integration cos((a + 1) x) - -------------- a + 1 Note that the answer is returned in argument-factored form.

Previous by date: [MUG] Bug in LinearAlgebra:-GenerateMatrix?, Bernhard Roider
Next by date: [MUG] large program on the web, modules vs. worksheets?,  Dr David Levitt
Previous thread: [MUG] Is the any `maple-compiler`?,  Andrey Siver
Next thread: [MUG] large program on the web, modules vs. worksheets?,  Dr David Levitt