I'm new to Mathematica. I have written a code for performing a convolution integral (as follows) but it seems to be giving out error messages:

My code is:

a[x_?NumericQ] := PDF[NormalDistribution[40, 2], x]
b[k_?NumericQ, x_?NumericQ] := 0.0026*Sin[1.27*k/x]^2
c[k_?NumericQ, x_?NumericQ] := {a[x]*b[k, x]}
d[k_?NumericQ] := NIntegrate[c[k, x], {x, 0, Infinity}]
Plot [d[k], {k, 0, 350}]

It gives the following error message multiple times:

Integrand c[0.00715,x] is not numerical at {x} = {124.67}

And the plot dosen't show up..! Does anyone have any suggestions on how to fix it ?

  • 1
    $\begingroup$ You can remove curly brackets from the definition of cto make it numeric instead of a list. $\endgroup$ – Shadowray May 20 '17 at 6:48
  • $\begingroup$ It takes a while, but it's doable: Plot[0.0026 NIntegrate[Sin[1.27 k/x]^2 Exp[-(x - 40)^2/8], {x, 0, ∞}]/(2 Sqrt[2 π]), {k, 0, 350}] $\endgroup$ – J. M. is in limbo May 20 '17 at 6:52
  • $\begingroup$ If you use Method -> "GaussKronrodRule" it's about ten times faster. $\endgroup$ – Michael E2 May 20 '17 at 13:04
  • $\begingroup$ Hmm, a related bug: Integrate[(0.0026 E^(-(1/8) (-40 + x)^2) Sin[(1.27 k)/x]^2)/(2 Sqrt[2 Pi]), {x, 0, Infinity}] returns a constant 0.0013 (independent of k) which does not agree with NIntegrate. It seems to be the limit as k approaches infinity. $\endgroup$ – Michael E2 May 20 '17 at 19:22
  • $\begingroup$ Thank you all...! @Shadowray, This process works for the Normal Distribution function, but yields an error for the function a[x_] := 0.00003*(x^2) (1 - x/79.2).....this time it's an NIntegrate error..saying that the integral fails to converge. Changing the limit of integration from infinity to 90 helps, but is there a way to do the infinity integral? $\endgroup$ – Epari shalini May 21 '17 at 4:52

The only function definition that needs to have its argument restricted to numeric values is that for d since that is the only function that uses numeric techniques.

Clear[a, b, c, d]

a[x_] = PDF[NormalDistribution[40, 2], x];

b[k_, x_] = 0.0026*Sin[1.27*k/x]^2 // Rationalize;

As @Shadowray pointed out in the comments, the output of c should be a scalar rather than a list.

c[k_, x_] = a[x]*b[k, x]

(*  (13*Sin[(127*k)/(100*x)]^2)/
   (E^((1/8)*(-40 + x)^2)*
      (10000*Sqrt[2*Pi]))  *)

Using suggestion by @MichaelE2 to use Method -> "GaussKronrodRule"

d[k_?NumericQ] := NIntegrate[c[k, x], {x, 0, Infinity},
  Method -> "GaussKronrodRule"]

Plot[d[k], {k, 0, 350}]

enter image description here

  • $\begingroup$ Thank you for your inputs! I tried to follow the same procedure for a different function: a[x_] := 0.00003*(x^2) (1 - x/79.2). The NIntegrate shows error: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small. "NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in x near {x} = {8.16907*10^224}. NIntegrate obtained -1.815272304467489*10^55882 and 1.815272304467489`15.954589770191005*^55882 for the integral and error estimates" $\endgroup$ – Epari shalini May 21 '17 at 4:48
  • 2
    $\begingroup$ @Eparishalini - for the a given in your comment, the integral does not converge on {0, Infinity}. $\endgroup$ – Bob Hanlon May 21 '17 at 5:09

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