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Integral using the gamma function

  1. Apr 17, 2017 #1
    1. The problem statement, all variables and given/known data
    I am trying to evaluate the following integral: ##\displaystyle \int^{\infty}_0 (1 - e^{-x}) x^{-\frac{3}{2}} \, dx##

    2. Relevant equations


    3. The attempt at a solution
    When I split the above integral, I get the following ##\int^{\infty}_0 x^{-\frac{3}{2}} \, dx - \Gamma (-\frac{1}{2})## = ##\int^{\infty}_0 x^{-\frac{3}{2}} \, dx + 2 \sqrt{\pi}##, but that integral doesn't converge, so I'm not sure now what I should do.
     
  2. jcsd
  3. Apr 17, 2017 #2

    fresh_42

    Staff: Mentor

    I haven't done it, but what about taking the power series for ##e^{-x}##, splitting ##x^{-\frac{3}{2}}=x^{-1}\cdot x^{-\frac{1}{2}}## and multiplying ##(1-e^{-x})x^{-1}##?
     
  4. Apr 17, 2017 #3
    If I use the power series of ##e^{-x}##, then what I get is that ##(1-e^{-x})x^{-1} = 1 - \frac{x}{2!} + \frac{x^2}{3!} - \frac{x^3}{4!} + \cdots##, which I don't see to be useful.

    Also, one question, if we find that ##\int^{\infty}_0 x^{-\frac{3}{2}} \, dx## doesn't converge, why should ##\int^{\infty}_0 (1 - e^{-x}) x^{-\frac{3}{2}} \, dx## converge at all?
     
  5. Apr 17, 2017 #4

    fresh_42

    Staff: Mentor

    Yes, you're right. So back to your calculation. Can you show how you ended up at ##\int x^{-\frac{3}{2}}dx -\Gamma(-\frac{1}{2})##?
     
  6. Apr 18, 2017 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The integral ##\int^{\infty}_0 e^{-x} x^{-3/2} \, dx## is not ## \Gamma (-\frac{1}{2})##. It's divergent. The defining relation ##\int^{\infty}_0 e^{-x} x^{n-1} dx = \Gamma(n)## only holds for ##Re(n) \gt 0##.
     
    Last edited: Apr 18, 2017
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