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Homework Help: Integrating double exponential terms in a function

  1. Sep 16, 2011 #1
    Hi, this is not really a homework problem, but something that came up during my research, upon trying to integrate some empirical function. This function consists of many terms, but specifically, there is a term containing double exponential functions which is giving me some trouble.

    If anyone could please assist in helping me with this problem my appreciation would approach infinity :)

    1. The problem statement, all variables and given/known data
    I want to compute the integral of an empirical relation with respect to x, which I will refer to as y. Here here a, b and c are constant and x>0.

    2. Relevant equations
    \begin{equation} y = \int_{x_1}^{x_2} \frac{e^{-ax}}{e^{b+e^{-cx}}} dx \end{equation}

    3. The attempt at a solution
    By using the properties of exponential functions, I simplify the relation:
    \begin{equation} y = \int_{x_1}^{x_2} e^{-ax}e^{-b} e^{-e^{-cx}} dx \end{equation}

    I have become aware that if a = c, i could make the substitution:
    \begin{equation} u = e^{-ax} \end{equation}
    and it would be possible to obtain an analytical solution, however this is not the case and I have become stuck in trying to solve this problem. I have also used Mathematica & Matlab to try and obtain symbolic solutions, however both these packages fail (I realize this is hardly an substitution to using analysis, but I was examining different avenues).

    Is it perhaps that an analytical solution does not exists? I have plotted the function and it is continuous and reasonably I expected an analytical integral to be possible. Please assist.
  2. jcsd
  3. Sep 16, 2011 #2


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    Since e-b is a constant, pull it out of the integral:

    \begin{equation} y = e^{-b}\int_{x_1}^{x_2} e^{-ax} e^{-e^{-cx}} dx . \end{equation}

    The exponential function in the exponent is the nuisance, but we can deal with it. Make the substitution [itex]u = e^{-cx} [/itex] and continue from there: what is du, and how would you express [itex]e^{-ax} dx[/itex] in terms of u and du , using properties of exponents?
  4. Sep 16, 2011 #3
    Thank you very much dynamicsolo, I appreciate your help, from the substitution [itex]u = e^{-cx} [/itex] the problem progresses:
    du & = -c e^{-cx} dx \\
    y & = -\frac{e^{-b}}{c} \int_{u_1}^{u_2} e^{-ax} e^{cx} e^{-u} du
    where it follows:
    x & = - \frac{\ln u}{c} \\
    y & = -\frac{e^{-b}}{c} \int_{u_1}^{u_2} e^{\frac{a \ln u}{c}} e^{-\ln u} e^{-u} du \\
    & = -\frac{e^{-b}}{c} \int_{u_1}^{u_2} e^{\ln u ( \frac{a}{c} - 1 )} e^{- u} du \\
    & = -\frac{e^{-b}}{c} \int_{u_1}^{u_2} u^{(\frac{a}{c} - 1)} e^{ - u} du
    Now I hit my next wall, using integration by parts I run into a loop....
    y & = -\frac{e^{-b}}{c} \int_{u_1}^{u_2} u^{(\frac{a}{c} - 1)} e^{ - u} du = -\frac{e^{-b}}{c} \bigg( \frac{c}{a} u^{\frac{a}{c}} e^{-u} |_{u_1}^{u_2} - \frac{c}{a} \int_{u_1}^{u_2} u^{\frac{a}{c}} (-e^{-u}) du \bigg)
    However, it is now possible to use a symbolic toolbox to integrate from here, which results in the use of gamma functions. I do wonder if it would be possible to go further without the symbolic toolbox. Or perhaps somewhere I am missing something. Thanks to all following & to dynamicsolo.
    Last edited: Sep 16, 2011
  5. Sep 16, 2011 #4


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    I haven't worked this out completely, but I believe there are a few cases. If a/c is a positive integer, the remaining integral is recursive, but the "chain" terminates. If a/c is a negative integer, the recursion also terminates, but the final integral [itex]\int \frac{e^{-u} du}{u} [/itex] leads to a power series. If a/c is not an integer, I think the "chain" doesn't terminate. Also, you're only going to have gamma functions if the integral is improper (type I), aren't you?

    I'll need to look at this further. Perhaps someone else has more familiarity with this type of integral and could comment...
  6. Sep 17, 2011 #5

    Ray Vickson

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    Your integral of u^(r-1)*exp(-u) can be written in terms of the so-called incomplete Gamma function. You are right that when r is a positive integer the integral can be done completely in terms of "elementary" functions, but for something like r = 1000 why would you want to do it? If you have a package like Maple or Mathematica, leaving it in terms of the incomplete Gamma function is the best way to deal with it.

  7. Sep 19, 2011 #6
    Thanks dynamicsolo. I must say with sadness that is where my math knowledge ends and I am not familiar with how to terminate the recursive integral. Just a remark on the problem, the constants, a & c, are always positive real numbers, however, the integral is to be performed for different cases such that, a & c will not always be the same. If I understand correctly, different values of a & c will result in a different final integral expression, such that it may not be the best way for the application. The original integration limits i.e. x_1 and x_2 are positive real numbers such that u_1 and u_2 will also be positive real numbers and it therefore follows that the integral is proper. Does this mean that an exact solution must exist if the function being integrated is continuous?

    Thanks RGV for your help and pointing out that it follows the form of the incomplete Gamma function. I was not familiar with this function, except knowing it pops up now and again! This does seem the best way to leave the problem for this application, which has to be applied for different cases of a & c. For completeness I write out the lower incomplete Gamma function for one of the limits:
    \Gamma(\frac{a}{c},u_2) = \int_0^{u_2} e^{-u} u^{\frac{a}{c} - 1} du
    it follows
    y = -\frac{e^{-b}}{c} \int_{u_1}^{u_2} u^{(\frac{a}{c} - 1)} e^{ - u} du = -\frac{e^{-b}}{c} \bigg( \Gamma(\frac{a}{c},u_2) - \Gamma(\frac{a}{c},u_1) \bigg)

    I hate to bring another problem, but I have also come across another form of y for a similar empirical model (I will call y_2). After following the same analysis as for y above, I arrive at:
    y_2 = \int_{u_1}^{u_2} e^{\frac{-a}{b+c u}} u^{\frac{g}{d} - 1}du
    (a, b, c, d & g are real number constants)
    The key difference being the term inside the exponential function. This does no longer follow the form of the gamma function. Does anyone have some advice for dealing with this form?

    Thanks to dynamicsolo & RGV for helping me with this problem, I really appreciate your insights.
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