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Integrals and gamma functions manipulation

  1. Dec 29, 2014 #1
    1. The problem statement, all variables and given/known data
    I am working through some maths to deepen my understanding of a topic we have learnt about. However I am not sure what the author has done and I have copied below the chunk I am stuck on. I would be extremely grateful if someone could just briefly explain what is going on i.e how to get from one step to another and why.

    2. Relevant equations

    $\int^\infty_y P(2y;2\nu,2k) dK$

    where $P(2y;2\nu,2k)$ =$ \frac{1}{2} (\frac{x}{\lambda})^{\frac{\nu-2}{4}} I_{\frac{\nu-2}{2}}(\lambda x)^{\frac{1}{2}} e^{\frac{-(\lambda+x)}{2}}$
    Note that I is the bessel function of the first kind of order K and is defined as $ (\frac{1}{2}Z) \sum\limits_{j=0}^\infty \frac{(\frac{z^2}{4})^j}{{j!\gamma(k+j+1)}}$
    $$=\int^\infty_y e^{-z-k} (\frac{z}{k})^{\tau-1} (kz)^{\frac{\tau-1}{2}} \sum\limits_{n=0}^\infty \frac{(zk)^n}{n!\gamma(n+\nu-1+1)}dK$$
    $$=\int^\infty_y \frac{e^{-z} z^{n+v-1}}{\gamma(n+\nu)} \int^\infty_y \frac{e^{-k} k^{n}}{\gamma(n+1)}dK$$
    $$=\sum\limits_{n=0}^\infty g(n+v,z)G(n+1,y)$$

    Please also note that $\gamma$ is meant to be that symbol that looks like a T when working with gamma funtions but I do not know what it is.

    3. The attempt at a solution

    The only thing I know is that it has something to do with gamma functions and integration by parts.
    By the way this is not assignment related but something I really want to understand.
  2. jcsd
  3. Dec 29, 2014 #2
    Does anyone know why latex is not working on the first part of equation?
  4. Dec 29, 2014 #3

    Ray Vickson

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    See remarks following the re-write of your post, which I have edited to:

    [tex]J = \int_y^\infty P(2y;2\nu,2k) dk[/tex]

    [tex]P(2y;2\nu,2k) = \frac{1}{2} (\frac{x}{\lambda})^{\frac{\nu-2}{4}} I_{\frac{\nu-2}{2}}(\lambda x)^{\frac{1}{2}} e^{\frac{-(\lambda+x)}{2}} \leftarrow \text{makes no sense!}[/tex]
    Note that ##I## is the bessel function of the first kind of order ##k## and is defined as
    [tex] \frac{z}{2} \sum_{j=0}^\infty \frac{(\frac{z^2}{4})^j}{{j!\gamma(k+j+1)}}[/tex]
    Thus, we have
    [tex] J = \int_y^\infty e^{-z-k} (\frac{z}{k})^{\tau-1} (kz)^{\frac{\tau-1}{2}} \sum_{n=0}^\infty \frac{(zk)^n}{n!\gamma(n+\nu-1+1)}dk \\
    \;\;=\int_y^\infty \frac{e^{-z} z^{n+v-1}}{\gamma(n+\nu)} \int_y^\infty \frac{e^{-k} k^{n}}{\gamma(n+1)}dk \\
    \;\; =\sum_{n=0}^\infty g(n+v,z)G(n+1,y)[/tex]

    I did not understand what your \gamma is, so I have not addressed that issue.

    Several points:
    (1) Your definition of P makes no sense because you have y on one side and x on the other, and do not say what the relationship is between x and y.
    (2) Best to avoid "$" in LaTeX; just use "# # 'material' # # (remove blanks between the #s) to put 'material' in an in-line equation, or use [t e x] 'material' [/t e x] (remove spaces) to put it as a displayed equation.
    (3) You can say "\sum_a^b"; no need to say "sum_limits_a^b", so that is what I have done in the above.
    (4) Same as (2) but for integrals. It is customary (and good practice if you want to exchange documents with others) to say "int_a^b" rather than "int^b_a", so I changed those.
    (5) Decide if you mean k or K and use it consistently; I changed dK to dk to fix it up.
    (6) In a multi-line equation you can just do what I did above: start by using "[t e x]" (no spaces), use "\\" to end a line, then start the next line with an "=" or whatever. Keep going like that until you run out of lines, then end the thing by "[/ t e x]" (no spaces)
    (7) The way you wrote it was confusing and misleading. You need a break after the definition of ##I ## and before the remaining material. I did that by defining a symbol (J) for the thing you want and then saying "Thus, we have J = ..."
  5. Dec 29, 2014 #4
    Thanks for your reply and thanks for having a look. Gamma was meant to be $$\Gamma$$ (sorry about that)
    I have copied what is above directly from the book.
    Do you know how to get from j to the next step ?
  6. Dec 29, 2014 #5

    Ray Vickson

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    I cannot hope to make any headway on the problem until you tell me:
    (1) In the definition of P, you have 2y on the left and x on the right. Is x = y, or what? Also, your definition of P has ##\lambda## in it, but there is no indication of how ##\lambda## is related to ##\nu## or ##k## or whatever.
    (2) In the second-last line you have written two integrations (and no sum), but have only one integration variable "dk". Is that a typo? If not, what is the correct expression?
  7. Dec 30, 2014 #6


    Staff: Mentor

    Use pairs of dollar signs, not single ones.
    $$\int^\infty_y P(2y;2\nu,2k) dk$$
  8. Dec 30, 2014 #7
    Hey mate its ok, I worked it out.
    Thanks anyway
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