Re: The integrand has evaluated to non-numerical values for all

*To*: mathgroup at smc.vnet.net*Subject*: [mg83955] Re: The integrand has evaluated to non-numerical values for all*From*: "Steve Luttrell" <steve at _removemefirst_luttrell.org.uk>*Date*: Wed, 5 Dec 2007 07:16:00 -0500 (EST)*References*: <fitstp$69s$1@smc.vnet.net> <fj0n9k$j2v$1@smc.vnet.net>

"Jean-Marc Gulliet" <jeanmarc.gulliet at gmail.com> wrote in message news:fj0n9k$j2v$1 at smc.vnet.net... > Kreig Hucson wrote: > >> Being given the function: >> >> f[x_,y_]:= Sin[x*y]*Exp[-(y-1)^2]/(x*y) , >> >> I want to integrate it with respect to y, from 0 to Infinity, and to >> obtain a function of x only, F[x]. >> >> I typed the command: >> >> F[x_]:= Assuming[x>0, Integrate[f[x,y],{y,0,Infinity}]] >> >> but the integral remained unevaluated by Mathematica 6. >> >> After this, I typed the command: >> >> F[x_]:= Assuming[x>0, NIntegrate[f[x,y],{y,0,Infinity}]] >> >> but I received the error message: "The integrand f[x,y] has evaluated to >> non-numerical values for all sampling points in the region with >> boundaries {{Infinity,0.}}". >> I looked at the explanations of the error message and there it was >> suggested to give a particular value to x and to compute the integral for >> that particular value. I gived one particular value and I obtained the >> expected result. Anyway I need to obtain an analytical expression for all >> x>0. >> >> My question is, how can be performed the integral and to obtain the >> analytical function F[x]? > > You could change the form of the original expression to be written as > exponentials only, then do the integration. For instance, > > In[1]:= expr = Sin[x*y]*Exp[-(y - 1)^2]/(x*y) > > Out[1]= > Sin[x y] > ---------------- > 2 > (-1 + y) > E (x y) > > In[2]:= f = TrigToExp@expr > > Out[2]= > 2 2 > -(-1 + y) - I x y -(-1 + y) + I x y > I E I E > --------------------- - --------------------- > 2 x y 2 x y > > In[3]:= F[x_] = Assuming[x > 0, Integrate[f, {y, 0, Infinity}]] > > Out[3]= > 1 3 > --------------- ((-2 - I x) (2 I + x) > 2 > 24 E x (4 + x ) > > 5 1 2 > HypergeometricPFQ[{1, 1}, {2, -}, -(-) (-2 I + x) ] + > 2 4 > > 2 > x /2 3 > (E (-2 + I x) (-2 I + x) > > 5 1 2 > HypergeometricPFQ[{1, 1}, {2, -}, -(-) (2 I + x) ] - > 2 4 > > 2 > 1 - I x + x /4 I x > 6 (2 E Sqrt[Pi] (-2 I + x) Erf[1 - ---] + > 2 > > 2 > 2 + 1/4 (2 I + x) I x > 2 E Sqrt[Pi] (2 I + x) Erf[1 + ---] + > 2 > > 2 > x /2 2 I x I x > I E (4 + x ) (Pi Erfi[1 - ---] - Pi Erfi[1 + ---] - > 2 2 > > 2 > 2 Log[-2 - I x] + 2 Log[-2 + I x] + Log[-(-2 I + x) ] - > > 2 > 2 x /2 > Log[-(2 I + x) ]))) / E ) > > In[4]:= F[2] > > Out[4]= > 1 5 > ----- (64 HypergeometricPFQ[{1, 1}, {2, -}, 2 I] + > 384 E 2 > > 1 2 5 > -- (64 E HypergeometricPFQ[{1, 1}, {2, -}, -2 I] - > 2 2 > E > > 2 - 2 I > 6 ((4 - 4 I) E Sqrt[Pi] Erf[1 - I] + > > 2 + 2 I > (4 + 4 I) E Sqrt[Pi] Erf[1 + I] + > > 2 > 8 I E (I Pi + Pi Erfi[1 - I] - Pi Erfi[1 + I] - > > 2 Log[-2 - 2 I] + 2 Log[-2 + 2 I])))) > > In[5]:= % // N // Chop > > Out[5]= 0.453078 > > Regards, > -- > Jean-Marc > I tried comparing my series solution (see my parallel posting to yours) with your closed-form solution. When I use my most accurate series approximation Last[fapprox] (1/(3628800 x))\[ExponentialE]^(-(x^2/4)) (x (-20 \[ExponentialE]^(x^2/4) (-37512+2252 x^2-54 x^4+x^6)+Sqrt[\[Pi]] (1149120-110880 x^2+4032 x^4-72 x^6+x^8))+665280 \[ExponentialE]^(x^2/4) \[Pi] Erf[x/2]+10 Sqrt[\[Pi]] (196224-41664 x^2+2352 x^4-56 x^6+x^8) Erfi[x/2]) from my posting I get F[2]=0.636495, which disagrees with your 0.453078 (which I can reproduce). Curious! I then tried evaluating the integral numerically using With[{x = 2.}, NIntegrate[(\[ExponentialE]^-(-1 + y)^2 Sin[x y])/(x y), {y, 0, \[Infinity]}]] which gives 0.63577 in close agreement with my series approximation, but in disagreement with your result. I wonder what is going on here? Steve Luttrell West Malvern, UK