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Homework Help: Higher dimension Integrals

  1. Jun 6, 2010 #1
    1. The problem statement, all variables and given/known data
    Calculate the following integrals:

    (a) [tex]I(n,\alpha) = \int_{0}^{\infty} e^{-\alpha x^2}x^n dx[/tex] for n whole integers and [tex]n \ge 0[/tex]

    Calculate all results till n=5.

    Tip: First calculate [tex]I^2(0,\alpha)[/tex] and [tex]I(1,\alpha)[/tex] and then use this to calculate n>1.

    (b) [tex]I(n)=\int_{0}^{\infty} e^{-x}x^n dx[/tex] for n whole and half integer where [tex]n\ge -1/2[/tex]

    Calculate all results till n=5

    Tip: Calculate I(n) using I(0) and I(-1/2).

    2. Relevant equations



    3. The attempt at a solution
    (а) I managed to do the first part using polar coordinates and substitution [tex]I(0,\alpha) = \sqrt{\frac{\pi}{\alpha}}[/tex] but I keep getting 0 for [tex]n \ge 1[/tex]

    For example with 1: [tex]I(1,\alpha) = \int_0^{\infty} e^{-\alpha x^2} x^1 dx = x \sqrt{\frac{\pi}{\alpha}} -\int_0^{\infty} \sqrt{\frac{\pi}{\alpha}} 2x dx = x^2 \sqrt{\frac{\pi}{\alpha}} - x^2 \sqrt{\frac{\pi}{\alpha}} = 0[/tex]

    (b) The first part here is ok: [tex]I(0) = \int_0^{\infty} e^{-x} dx = -e^{-x}[/tex] but I tried a thousand times to do I(-1/2) and keep going in circle with integration by parts.

    Any ideas would be greatly appreciated!
     
  2. jcsd
  3. Jun 6, 2010 #2

    vela

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    What happened to the exponential when you integrated by parts?
    You can use the result from part (a) to solve (b), so figure out (a) first.
     
  4. Jun 6, 2010 #3
    Well, we know from the first part that [tex]\int_{0}^{\infty} e^{-\alpha x^2} dx = \sqrt{\frac{\pi}{\alpha}}[/tex] or am I missing something?
     
  5. Jun 6, 2010 #4

    vela

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    You're doing things out of order. You can't plug the limits in until the end. Consider a simpler example:

    [tex]\int_0^\infty xe^{-x}\,dx[/tex]

    Let f=x and dg=e-x dx. Then df=dx and g=-e-x, and you should get

    [tex]\int_0^\infty xe^{-x}\,dx = \left[-xe^{-x}\right]_0^\infty + \int_0^\infty e^{-x}\,dx[/tex]

    But what you're doing is saying

    [tex]g = \int_0^\infty dg = \left[-e^x\right]_0^\infty = 1[/tex]

    so that

    [tex]\int_0^\infty xe^{-x}\,dx = -x\cdot 1 + \int_0^\infty 1\,dx[/tex]

    See the difference?

    Anyway, if you attempt integration by parts properly, you're stuck trying to integrate exp(-ax2) without the limits, which you can't do. Instead, for the n=1 case, try the substitution u=-ax2.
     
    Last edited: Jun 6, 2010
  6. Jun 7, 2010 #5
    Well, I did what you said and I got [tex]- \frac{1}{2}[/tex] by substitution in [tex]I(1,\alpha)= \int_0^{\infty} e^{-\alpha x^2} x^1 dx[/tex] but now I`m stuck on all the other expressions [tex]n \ge 2 [/tex]
    The Tip says to get solutions for [tex]I^2(0,\alpha)[/tex] and [tex]I(1,\alpha)[/tex] and use this information to solve the others to n=5 but I don`t see how this helps me any further.
     
    Last edited: Jun 7, 2010
  7. Jun 7, 2010 #6

    vela

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    Recheck your work. The integrand is positive over the interval of integration, so the integral should be positive. Also, the α is missing from the answer.

    Now use integration by parts to find I(n,α) for n≥2.
     
  8. Jun 7, 2010 #7
    I corrected it, there was a mistake in the substitution. So, now I have [tex]\scriptsize I(1,\alpha) = \frac{1}{2\alpha}[/tex]
    I`m having real trouble with the integration by parts of I(n,a) for n>1, though.

    If I choose [tex]\scriptsize u=x^2[/tex] and [tex]\scriptsize \dot{v} = e^{-\alpha x^2}[/tex] I`m stuck because I cannot integrate [tex]\scriptsize \dot{v}[/tex] without transformation to polar coordinates.
    If I choose the other way around, I get stuck again because I just get x with a growing exponent in the integral.
     
  9. Jun 7, 2010 #8

    vela

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    Keep one power of x with the exponential so you can integrate it with a substitution.
     
  10. Jun 7, 2010 #9
    I`m not sure I understand what you mean by that. I cannot get to [tex]\scriptsize x^1[/tex].

    I tried integration by parts that way:

    [tex]\int_0^{\infty} e^{- \alpha x^2} x^2 dx =[/tex]
    [tex]= \biggl [\frac{x^3}{3} e^{- \alpha x^2} \biggr ]_0^{\infty} + \int_0^{\infty} 2 \alpha x e^{- \alpha x^2} \frac{x^3}{3} dx = [/tex]

    Now, if I substitute [tex]u=-\alpha x^2[/tex] and [tex]dx=\frac{du}{-2 \alpha x}[/tex] I cannot manage to get rid of that nasty [tex]\scriptsize x^3[/tex].

    In the end, whatever I try I just get [tex]\scriptsize - \frac{x^3}{3} + 0[/tex] from the integral.
     
  11. Jun 7, 2010 #10

    vela

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    What I'm saying is you have choices other than u=xn and dv=exp(-ax2)dx. You don't have to put all of the x's from the xn factor into u and just the exponential into dv.
     
  12. Jun 7, 2010 #11
    That way I got an answer 1/2a for I(2,a), too. Could this possibly be right?
     
  13. Jun 7, 2010 #12

    vela

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    No, it's not right. You should find I(n+2) is proportional to I(n).

    By the way, your answer for I(0,α) is missing a factor of 1/2.
     
  14. Jun 7, 2010 #13
    I`ve got I(0,a) from my "Mathematical methods in physics" textbook, I thought it was supposed to be correct.

    [PLAIN]http://img267.imageshack.us/img267/426/answer.jpg [Broken]

    Where do I get the 1/2 factor from?
     
    Last edited by a moderator: May 4, 2017
  15. Jun 7, 2010 #14

    vela

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    The lower limit of the integral in the book is [itex]-\infty[/itex]. The lower limit on your integral is 0.
     
  16. Jun 7, 2010 #15
    I see the logic of that but I still cannot mathematically derive this factor of 1/2...
     
  17. Jun 7, 2010 #16

    vela

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    The easy way to show this is to note that the integrand is an even function, so you'd have

    [tex]\int_0^\infty e^{-\alpha x^2} dx = \frac{1}{2}\int_{-\infty}^\infty e^{-\alpha x^2} dx[/tex]

    You could also follow the derivation in the book with slight modifications. Both the x and y integrals would go from 0 to ∞ instead of from -∞ to ∞. The region of integration would therefore be only the first quadrant rather than the entire xy plane, so when you change to polar coordinates, the angle φ would go from 0 to π/2 instead of 0 to 2π. The end result is a factor of 1/2 in the final answer.
     
  18. Jun 7, 2010 #17
    Thank you for your time, this really helped. At least now I have a small part from the first part of the first problem. Too bad that there is no time to finish it, I have to submit the homework in 20 minutes.
    I`ll still look further into that later today, although it will be too late for points from the homework.
     
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