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Integrals/Non-Elementary Antiderivatives

  1. Mar 26, 2014 #1
    Hey everyone,

    I'm wondering how to solve the following definite integral,

    [itex]\int^\infty_{-\infty}{x^4e^{-x^2}dx}[/itex].

    I know the answer is ##\frac{3 \sqrt{\pi}}{4}##, but I'm not positive how to get there.

    I understand how to evaluate the definite "Gaussian" integral $$\int^\infty_{-\infty}{e^{-x^2}}=\sqrt \pi$$ using a switch to polar coordinates and a u sub, but not sure if/how that applies here.

    Extending the question, I'm also wondering about an integral like $$\int^\infty_{-\infty}{e^{ikx}e^{-x^2}}$$.

    Thanks!
     
  2. jcsd
  3. Mar 26, 2014 #2

    micromass

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    Try integration by parts for the first one. Try completing the square for the second one.
     
  4. Mar 26, 2014 #3
    So I tried parts for the first one, and assuming I didn't mess up somewhere, if I choose u to be the exponential term then I end up with $$e^{-x^2}\frac{x^5}{5} - \int^{\infty}_{-\infty}\frac{x^5}{5}(-2xe^{-x^2})$$, with the first term being evaluated from minus infinity to infinity as well. Wouldn't I just have to keep doing partial fractions, this way, and it would just keep going indefinitely?

    Conversely, if I chose u to be the x term, then I think I'd get $$x^4 \sqrt{\pi} - \int^{\infty}_{-\infty}\sqrt{\pi}4x^3$$, with the first term being evaluated from x is minus infinity to positive infinity and the ##\sqrt{\pi}## coming from the exponential term being integrated over all space. Doing this, then wouldn't you just get 0?
     
  5. Mar 26, 2014 #4

    micromass

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    What happens if you use integration by parts on

    [tex]\int_{-\infty}^{+\infty} e^{-x^2} dx[/tex]
     
  6. Mar 26, 2014 #5
    I do not like the idea of integration by parts here. Step by step

    1. Instead solve a slightly more difficult but related problem definite integral (x squared exp - a times x squared).
    2. Call this definite integral I(a).
    3. Take the derivative with respect to a (not x) of the integrand inside the integral. Here you may note you end up multiplying the integrand by a times x squared.
    4. Equate this integral with dI(a) / da.
    5. do step 3 to the previous integral expression to get a x fourth power in the integrand.
    6 equate this integral to d2I / da2.
    7. Plug in a = 1.
    8 This will give you the answer.
     
  7. Mar 26, 2014 #6
    Correction: Step 3 you end up multiplying the integrand by -x squared not a times x squared, but you get the idea.

    Step 1 can be obtained with the change of variables x = sqrt(a) * y. then when you get the result later substitute in x for y to preserve the notation in terms of x not y. (this is optional)
     
  8. Mar 26, 2014 #7
    Sorry step 1 is still wrong. Start with definite int( exp-x squared ) = sqrt(pi). Change variables y = sqrt(a) times x to eventually get I(a) = definte int( -x squared times exp (- x sqared)).

    Step 4 should be calculate dI(a)/da and equate coefficient (take the derivative)
    Step 6 is similar.

    In the long run you change variables and take two derivatives. You do not even have to evaluate an integral (i.e. you already have all you need as long as you do not consider changing variables in step 1 to be evaluating an integral)
     
  9. Mar 26, 2014 #8
    Sorry step 1 is still wrong. Start with definite int( exp-x squared ) = sqrt(pi). Change variables y = sqrt(a) times x to eventually get I(a) = definte int( -x squared times exp (- a times x sqared)).
     
  10. Mar 27, 2014 #9
    Integration by parts also gives the answer.

    Notice that the integral can be written as

    $$\int_{-\infty}^{\infty} x^3(xe^{-x^2})\,dx$$
    Can you see how to use by parts here? :)
     
  11. Mar 27, 2014 #10
    To: Pranav-Arora

    Very good. It still takes several steps but it can be integrated by parts without the dodge of derivating inside the integral with respect to a and introducing the a in the exponent.
     
  12. Mar 27, 2014 #11
    Better idea is to write your integral: Def integral x (x cubed times x squared exp(- x squared) ). Then substitute y in for x squared.
     
  13. Mar 27, 2014 #12
    Internet kicked me off.

    Correction:

    Better idea is to write your integral: Def integral x times (x cubed times exp(- x squared) ). Then substitute y in for x squared.
     
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