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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
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