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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg83874] Re: The integrand has evaluated to non-numerical values for all
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
  • Date: Mon, 3 Dec 2007 05:46:57 -0500 (EST)
  • Organization: The Open University, Milton Keynes, UK
  • References: <fitstp$69s$1@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


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