Solving an equation with integral with one variable

y2=4;
Ly2=y2*exp(y2^2)/2-.5*1j*(pi)^.5*erf(sym(1j*y2))
Ly2 =
149084195602433/8388608 - (erf(4i)*3991211251234741i)/4503599627370496
   y1=-4;
  Ly1=y1*exp(y1^2)/2-.5*1j*(pi)^.5*erf(sym(1j*y1))
Ly1 =
(erf(4i)*3991211251234741i)/4503599627370496 - 149084195602433/8388608
the value of the integral is
   L= Ly2-Ly1
  L =
  149084195602433/4194304 - (erf(4i)*3991211251234741i)/2251799813685248

way around erf() only working for real inputs

L= double(Ly2-Ly1)
L =
   3.7843e+07

therefore

    C=1-L
    C =
      -3.7843e+07

Repeating with

format double

the result is

L =
     3.784324335121135e+07

Comment:

When attempting direct integration with command sum

    x=[-4:.001:4];L=sum(exp(x.^2) .* x.^2)
    L =
       3.4537e+10
    x=[-4:.00001:4];L=sum(exp(x.^2) .* x.^2)
    L =
       3.4396e+12
    x=[-4:.0000001:4];L=sum(exp(x.^2) .* x.^2)
    L =
       3.4395e+14

The slopes when approaching -4 and 4 are too sharp for command sum.

Grazie tante per la sua attenzione.

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thanks in advance

John BG

<mailto:jgb2012@sky.com jgb2012@sky.com>

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Answer 2

Star Strider 2017-8-10

2
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https://ww2.mathworks.cn/matlabcentral/answers/352313-solving-an-equation-with-integral-with-one-variable#answer_277564

编辑：Star Strider 2017-8-10

在 MATLAB Online 中打开

If I understand correctly what you want to do, this works:

f = @(x) exp(x.^2) .* x.^2;
Cs = fzero(@(C) C + integral(f, -4, 4) - 1, 1)
Cs =
  -3.4395e+07

Or more simply, since ‘C’ is a constant:

C = 1 - integral(f, -4, 4)
C =
  -3.4395e+07

If you are using R2011b or earlier, use quad instead of integral:

C = 1 - quad(f, -4, 4)

The result is the same:

C =
         -34395040.4474417

here using format long g.

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Sergio Manzetti 2017-8-11

Hi Star Strider, I was thinking of something real simple, like on Wolfram Alpha, where you type in "C*integral of f(x) + integral g(x) = 1" . Then you get the output, for the numerical integration of the two integrands, and you get the value of C automatically. Is this feasible in Matlab?