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Integration problem

  1. Jul 23, 2014 #1
    1. The problem statement, all variables and given/known data
    I'm trying to show that the definite integral:

    [itex] \int_0^{\infty} \frac 1{\sqrt{2 \pi}} \sqrt {y} e^{\frac {-y}2} dy [/itex],

    equals 1.

    2. Relevant equations

    it's already known that [itex] \int_0^{\infty}\frac 1{\sqrt{2 \pi}} y^{\frac {-1}2} e^{\frac {-y}2} dy = 1 [/itex], since f(x) is a probability density function.

    3. The attempt at a solution
    I've tried integration by parts, but that hasn't helped.

    [itex] \int_0^{\infty} \frac 1{\sqrt{2 \pi}} \sqrt {y} e^{\frac {-y}2} dy = \frac 1{\sqrt{2 \pi}} (-2 \sqrt {y} e^{\frac {-y}2})|_0^{\infty} + \int_0^{\infty}\frac 1{\sqrt{2 \pi}} y^{\frac {-1}2} e^{\frac {-y}2} dy [/itex],

    but [itex]lim_{t \to \infty} -2 \sqrt {y} e^{\frac {-y}2})|_0^{\infty} = 0 [/itex], so that's not very helpful.

    Any ideas to evaluate the integral?
     
  2. jcsd
  3. Jul 23, 2014 #2

    SammyS

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    If you know that [itex] \int_0^{\infty}\frac 1{\sqrt{2 \pi}} y^{\frac {-1}2} e^{\frac {-y}2} dy = 1 \ ,\ [/itex] then it seems to me that you have all but arrived at the result.
     
  4. Jul 23, 2014 #3
    I believe Mogarr knows that it equals 1, though doesn't analytically know how to demonstrate that with standard integration techniques.
     
  5. Jul 23, 2014 #4
    That's right
     
  6. Jul 23, 2014 #5

    SammyS

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    So, which part of this problem are you having difficulty with.

    Please be specific.
     
  7. Jul 24, 2014 #6
    I'm having a problem finding a method to evaluate:

    [itex] \int_0^{\infty} y \frac 1{\sqrt {2\pi}} \cdot y^{\frac {-1}2} e^{\frac {-y}2} dy =\int_0^{\infty} \frac 1{\sqrt {2\pi}} \cdot y^{\frac 12} e^{\frac {-y}2} dy [/itex].

    Here's some background. I know that the integral should evaluate to 1. In a previous part of the problem, I saw that for this standard normal function,

    [itex] f_X(x) = (\frac 1{\sqrt {2\pi}}) e^{\frac {-x}2} [/itex],

    the expected value of the random variable squared, [itex] E X^2 [/itex], is equal to 1. I had to look at the answer to solve this problem. The integral I am trying to solve is the expected value of Y, [itex] E Y [/itex], where [itex] Y = X^2 [/itex]. So this is a transformation of the random variable, but the result should be the same as [itex] E X^2 [/itex] (this was stated in the problem description).

    The conundrum I have is, after seeing the answer, I don't believe it to be true. Here is the answer given (hopefully without typos) in equations and italics:

    [itex] E Y = \int_0^{\infty} \frac {Y}{\sqrt {2\pi y}}e^{\frac {-y}2} dy = \frac 1{2\pi} (-2y^{\frac 12}e^{\frac {-y}2}|_0^{\infty} + \int_0^{\infty} y^{\frac {-1}2}e^{\frac {-y}2}dy) = \frac 1{\sqrt {2\pi}} \cdot \sqrt {2\pi} = 1[/itex].

    This was obtained using integration by parts with [itex]u=2y^{\frac 12}[/itex] and [itex]dv= \frac 12 e^{\frac {-y}2} [/itex], and the fact that [itex] f_Y(y) [/itex] integrates to 1.

    I don't think this answer is right though.

    Isn't [itex] -2y^{\frac 12} e^{\frac {-y}2} |_0^{\infty} = 0[/itex] ? And if I am right, then I am fresh out of ideas to solve this problem.
     
  8. Jul 24, 2014 #7
    If we make a simple substitution, [itex] u = \sqrt{y} [/itex], then we can see that $$\int \frac{1}{2\sqrt{y}}e^{-y/2}dy = \int e^{-u^2/2}du $$, what does that tell you about the integral itself?
     
  9. Jul 24, 2014 #8
    I think I'm getting really close, but thus far...

    [itex] \int_0^{\infty} \sqrt {y} e^{\frac {-y}2} dy = 2\sqrt {y} (-e^{\frac {-y}2})|_0^{\infty} + \int_0^{\infty} \frac 1{\sqrt {y}} e^{\frac {-y}2} dy [/itex]

    where I did integration by parts with [itex] u = 2\sqrt {y}[/itex] and [itex] dv = \frac 12e^{\frac {-y}2} dy [/itex].

    The first part of the sum is 0. So continuing on...

    [itex] \int_0^{\infty} \frac 1{\sqrt {y}} e^{\frac {-y}2} dy [/itex] can be simplified with a u-substitution. If I let [itex] u = \sqrt {y} [/itex], then [itex] du = \frac 12 y^{\frac {-1}2}dy [/itex], and the limits of integration won't change.

    So [itex] \int_0^{\infty} \frac 1{\sqrt {y}} e^{\frac {-y}2} dy = 2 \int_0^{\infty} e^{-y/2} \frac 12 \frac 1{\sqrt {y}} dy = 2\int_0^{\infty} e^{\frac {-(u)^2}2}du [/itex],

    but [itex] 2\int_0^{\infty} e^{\frac {-(u)^2}2}du = 2 \cdot \sqrt {2\pi}[/itex], where I need it to be [itex] \sqrt {2\pi} [/itex] so the whole thing can evaluate to 1.

    I think a mistake was made. Don't know where though.
     
    Last edited: Jul 24, 2014
  10. Jul 24, 2014 #9

    Ray Vickson

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    You can do it in two ways:
    (1) recognize that ##\sqrt{y} \exp(-y/2) = y^{3/2 - 1} \exp(-y/2)## is proportional to the density function of the Gamma random variable G(3/2,1/2); see, eg., http://en.wikipedia.org/wiki/Gamma_distribution . You can then reduce the integral to a something involving ##\Gamma(3/2) = (1/2) \Gamma(1/2)## and use the known value of ##\Gamma(1/2)## (or work it out yourself).
    (2) Use the change of variable ##y = x^2## to get ##c \int_0^{\infty} x^2 \exp(-x^2/2) \, dx## for a constant ##c##; then you need to know that the standard normal random variable has variance = 1, as you have already mentioned. The easiest way to get that is to use integration by parts:
    [tex] \int x^2 e^{-x^2/2} \, dx = \int u dv, \\
    u = x, \: dv = x e^{-x^2/2} \, dx = d \left(- e^{-x^2/2} \right)[/tex]
     
  11. Jul 24, 2014 #10
    I found my mistake. I found this from Wikipedia:

    [itex] \int_0^{\infty} e^{-ax^{b}}dx = \frac 1{b} a^{\frac {-1}{b}} \Gamma (\frac 1{b})[/itex]

    So after doing looking back at the u-substitution, I completed the problem. Thanks.
     
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