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Integral to sum

  1. May 12, 2008 #1


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    1. The problem statement, all variables and given/known data

    Show that

    [tex]\int_{0}^{t}\frac{\lambda^{k}u^{k-1}e^{-\lambda u}}{(k-1)!}\ du[/tex]


    [tex]\sum_{j=k}^{\infty}\frac{(\lambda\cdot t)^{j}e^{-\lambda t}}{j!}[/tex]

    Hint: Use a taylor series to express e^(-lambda*u)

    2. Relevant equations

    3. The attempt at a solution

    I used the taylor series to rewrite the e^ term and then I interchanges the integral sign with the summation and integrated the term which leaves me with

    [tex]\sum_{j=0}^{\infty}\left[\frac{(\lambda\cdot t)^{j}}{j!}\cdot\frac{(\lambda\cdot t)^{k}}{(k+j)(k-1)!}\right][/tex]

    However, I have no idea how to rewrite the left-erm to e^{-\lambda t} especially since we don't sum over k (I suspect it has something to do with shifting the index to j=k but I don't really know how that can be done without creating a second summation)
  2. jcsd
  3. May 12, 2008 #2
    The Taylor series of [itex]e^{-\lambda u}[/itex] will be an alternating series. Your result is not an alternating series.

    But in any case, I suspect there may be a typo in the answer. You can shift the index by setting i = j + k and change the summand to work with i.
  4. May 12, 2008 #3


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    yeah you are right the (-1)^n is definitely missing.

    Nevertheless, I tried your method of index shifting and I could even simplify the expression to some degree; however, I'm still not able to get a e^... in the answert. To get that I would need a taylor series or some expoential function I could rewrite in terms of e^... but how to do that is a complete mystery to me.

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