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Mathematics Calculus Wolfram Language

Problem in integrating Erfc function

Seb Fig

Posted 4 years ago

Hello folks, I'm having some troubles on the analytical integration over the variable z of this function using Mathematica: z^6 Erfc[b \[Rho] - d \[Rho] - a n \[Rho] - c z \[Rho] - a m z \[Rho]] With the following assumptions: Assumptions -> \[Rho] > 0 && a \[Element] Reals && b \[Element] Reals && c \[Element] Reals && d \[Element] Reals && m \[Element] Reals && n \[Element] Reals] Thus, I use the following command: Integrate[ z^6 Erfc[ b \[Rho] - d \[Rho] - a n \[Rho] - c z \[Rho] - a m z \[Rho]], z, Assumptions -> \[Rho] > 0 && a \[Element] Reals && b \[Element] Reals && c \[Element] Reals && d \[Element] Reals && m \[Element] Reals && n \[Element] Reals] I can't understand why Mathematica cannot integrate this function, while if I lower the degree of *z^6* the integration is performed without any problems. For instance, this function gets integrated fine and smoothly over z and using the same assumptions: z^5 Erfc[b \[Rho] - d \[Rho] - a n \[Rho] - c z \[Rho] - a m z \[Rho]] Thank you for the time spent on reading this. Looking forward to your help! S

Hello folks,

I'm having some troubles on the analytical integration over the variable z of this function using Mathematica:

     z^6 Erfc[b \[Rho] - d \[Rho] - a n \[Rho] - c z \[Rho] - a m z \[Rho]]

With the following assumptions:

Assumptions -> \[Rho] > 0 && a \[Element] Reals && b \[Element] Reals && c \[Element] Reals && d \[Element] Reals && m \[Element] Reals && n \[Element] Reals]

Thus, I use the following command:

Integrate[
 z^6 Erfc[
   b \[Rho] - d \[Rho] - a n \[Rho] - c z \[Rho] - 
    a m z \[Rho]], z, 
 Assumptions -> \[Rho] > 0 && a \[Element] Reals && 
   b \[Element] Reals && c \[Element] Reals && d \[Element] Reals && 
   m \[Element] Reals && n \[Element] Reals]

I can't understand why Mathematica cannot integrate this function, while if I lower the degree of z^6 the integration is performed without any problems. For instance, this function gets integrated fine and smoothly over z and using the same assumptions:

z^5 Erfc[b \[Rho] - d \[Rho] - a n \[Rho] - c z \[Rho] - a m z \[Rho]]

Thank you for the time spent on reading this. Looking forward to your help!

POSTED BY: Seb Fig

3 Replies

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Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 4 years ago

POSTED BY: Mariusz Iwaniuk

Seb Fig

Posted 4 years ago

Thanks a lot for your reply! The integral gives indeed a closed form. I just miss the part where the variable "*cov*" seems being squared before the integral. Why is that? cov^2 Integrate[(( b \[Rho] - d \[Rho] - a n \[Rho] - \[Phi])/((c + a m) \[Rho]))^6 Erf[\[Phi]], \[Phi]] Since: cov=-(1/((c + a m) \[Rho])) Inspired by your answer, I would have set the integral this way: z=((b \[Rho] - d \[Rho] - a n \[Rho] - \[Phi])/((c + a m) \[Rho])) cov = dz/d\[Phi] = -(1/((c + a m) \[Rho])) Integrate[((b \[Rho] - d \[Rho] - a n \[Rho] - \[Phi])/((c + a m) \[Rho]))^6 Erf[\[Phi]] (-(1/((c + a m) \[Rho]))), \[Phi]] Am I right or am I missing something?

Thanks a lot for your reply! The integral gives indeed a closed form.

I just miss the part where the variable "cov" seems being squared before the integral. Why is that?

cov^2 Integrate[((
     b \[Rho] - d \[Rho] - 
      a n \[Rho] - \[Phi])/((c + a m) \[Rho]))^6 Erf[\[Phi]], \[Phi]]

Since:

cov=-(1/((c + a m) \[Rho]))

Inspired by your answer, I would have set the integral this way:

z=((b \[Rho] - d \[Rho] - a n \[Rho] - \[Phi])/((c + a m) \[Rho]))
cov = dz/d\[Phi] = -(1/((c + a m) \[Rho]))

Integrate[((b \[Rho] - d \[Rho] - a n \[Rho] - \[Phi])/((c + a m) \[Rho]))^6 Erf[\[Phi]] (-(1/((c + a m) \[Rho]))), \[Phi]]

Am I right or am I missing something?

POSTED BY: Seb Fig

W. Craig Carter

W. Craig Carter, MIT

Posted 4 years ago

The WL language may miss some integrals. It happens. For this one, you could try: sol = SolveValues[ b \[Rho] - d \[Rho] - a n \[Rho] - c z \[Rho] - a m z \[Rho] == \[Phi], z][[1]] cov = D[sol, \[Phi]] integral = cov (-(1/((c + a m) \[Rho]))) Integrate[(( b \[Rho] - d \[Rho] - a n \[Rho] - \[Phi])/((c + a m) \[Rho]))^6 Erf[\[Phi]], \[Phi]] gives a closed form. checking: dintegral = D[integral /. \[Phi] -> b \[Rho] - d \[Rho] - a n \[Rho] - c z \[Rho] - a m z \[Rho], z]; Simplify[dintegral] (z^6 Erf[(b - d - a n - c z - a m z) \[Rho]])

The WL language may miss some integrals. It happens. For this one, you could try:

sol = SolveValues[
   b \[Rho] - d \[Rho] - a n \[Rho] - c z \[Rho] - 
     a m z \[Rho] == \[Phi], z][[1]]

cov = D[sol, \[Phi]]

integral = cov (-(1/((c + a m) \[Rho]))) Integrate[((
     b \[Rho] - d \[Rho] - 
      a n \[Rho] - \[Phi])/((c + a m) \[Rho]))^6 Erf[\[Phi]], \[Phi]]

gives a closed form.

checking:

dintegral = 
  D[integral /. \[Phi] -> 
     b \[Rho] - d \[Rho] - a n \[Rho] - c z \[Rho] - a m z \[Rho], z];

Simplify[dintegral]  (*z^6 Erf[(b - d - a n - c z - a m z) \[Rho]]*)

POSTED BY: W. Craig Carter

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