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Help with ingetral

  1. Sep 11, 2008 #1
    So i dont really know where to paste this.

    I'm reading a example in my modern physics book and i dont understand what they did.

    [tex]\int[/tex] (2(pi)hc2)/([tex]\lambda[/tex]5(ehc/[tex]\lambda[/tex]kT-1))d[tex]\lambda[/tex]

    if we make the change of variable x=hc/[tex]\lambda[/tex]kT

    (2(pi)k4T4)/(c2h3)[tex]\int[/tex]((x3)/(ex-1))dx

    so how did they get to this:

    (2(pi)k4T4)/(c2h3)[tex]\int[/tex]((x3)/(ex-1))dx

    when I solve for [tex]\lambda[/tex] in x=hc/[tex]\lambda[/tex]kT and plug it into the first ingetral I get:

    (2(pi)x5k5T5)/(h4c3)

    basically, i'm off by 1 power for everything except for x (that i'm off by 2).

    can someone explain this?
     
  2. jcsd
  3. Sep 11, 2008 #2

    HallsofIvy

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    Science Advisor

    That's
    [tex]\int \frac{2\pi hc^2}{\lambda^5 e^{\frac{hc}{\lambda}kT^{-1}-1} d\lambda[/tex]?

    Did you forget to replace "[itex]d\lambda[/itex]" as well?
    [tex]x= \frac{hc}{kT}\lambda^{-1}[/tex]
    so [tex]dx= -\frac{hc}{kT}\lambda^{-2}d\lambda[/tex]
    Solving that for [itex]d\lambda[/itex],
    [tex]d\lambda= -\frac{kT}{hc}\lambda^2 dx[/itex]
    and since [itex]\lambda= (hc)/(kT) x^{-1}[/itex], [itex]\lambda^2= (h^2c^2)/(k^2T^2) x^{-2}[/itex] giving
    [tex]d\lambda= \frac{hc}{kT}x^{-2}dx[/tex].
     
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