Who wanna get some fun? (Improper Integral)

Discussion in 'Homework' started by JustNrik, Aug 28, 2016.

  1. JustNrik

    JustNrik Well-Known Member

    [​IMG]

    Find m and n.
     
  2. Eutychius

    Eutychius Moderator

    From my calculations and assuming m and n are not constants:

    m=-x and n=-1/(2x)

    If m=-x then it cancels out the x^3 at the beginning. The n=-1/(2x) will get the other x out of the way.

    The integral 1/(x^2 + x + 1) has a nice solution of inverse tangent which means that at the limits of minus infinity and plus infinity it approaches π/2 with different signs. With the subtraction of the finite integral in mind (and some other numbers somewhere in there), you get 2π/sqrt(3).
    But since n=-1/(2x), the 2 from above vanishes and the final result is negative. Hence -π/sqrt(3).

    There's another solution that works the exact same way under the same conditions:

    m=-1/(2x^2) and n=-x^2

    If m and n are in fact constants, then I can't see any way that the integral doesn't diverge (hence can't be a finite solution).
     
    Last edited: Aug 29, 2016
  3. JustNrik

    JustNrik Well-Known Member

    Sadly, they are constants unless the problem is stated wrong
     
  4. Eutychius

    Eutychius Moderator

    I don't see how it will ever work with constants.

    If you solve the indefinite integral equivalent for any random real numbers m and n, you'll see that the product has "uncounterable" factors of x that will blow up at infinity, so it diverges.

    You can tell that the x^3 is a big problem there when we talk integrals of polynomials. My only solution for a finite number solution is to let m and n be variables.
     
  5. JustNrik

    JustNrik Well-Known Member

    Definitely the problem was stated wrong, solving the integral step by step I noticed that there's an x remaining there making the integral to diverge no matter what (assuming m and n must be constants) so yes, the correct answer in this case must allow variables or there's simply no answer as the integral will diverge no matter what value takes m and n.