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Maximizing the fraction of two integrals

  1. Nov 7, 2013 #1
    EDIT:
    I left out something of major importance, I want to maximize with respect to a!

    1. The problem statement, all variables and given/known data
    My problem is rather complex, but in the end it boils down to maximizing the fraction
    integrals.jpg


    2. Relevant equations
    With the calculus I know, I can't evaluate these integrals, so I have no applicable formula's


    3. The attempt at a solution
    So I am pretty stuck at this point. I want to maximize this fraction for positive a, but I simply don't know how. I tried using mathematica, and although it can evaluate the integrals, it cannot maximize the fraction. So instead I figured I should use matlab (which I also have to my disposal) to do so instead, but I simply don't know how.
    Could anyone help me out? Either there exists a clever way of evaluating them, or simply by maybe giving me an idea of how to maximize this in matlab.

    I don't get much further than defining the integrals, as
    fun1 = @(x) x.^2./(exp(x)-1);
    fun2 = @(x) x.^3./(exp(x)-1);

    Kind regards
     
    Last edited: Nov 7, 2013
  2. jcsd
  3. Nov 7, 2013 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    In Maple, if I change variables to z = exp(y) and then help the program by assuming a < 0, I get
    that the numerator N and denominator D are:
    [tex]N= -a^2 \ln(1-e^a)-2 a \,\text{polylog}(2,e^a)+2\, \text{polylog}(3,e^a)+\frac{1}{3} a^3 \\
    D = -a^3 \ln(1-e^a)-3a^2\, \text{polylog}(2,e^a)
    +6a\, \text{polylog}(3,e^a)-6\, \text{polylog}(4,e^a)+\frac{1}{4} a^4+\frac{2}{15} \pi^4
    [/tex]
    According to Maple,
    [tex] \text{polylog}(a,z) = \sum_{n=1}^{\infty} \frac{z^n}{n^a}[/tex] for
    ##|z| < 1##, and is defined by analytic continuation otherwise. The index ##a## can be any complex number. If ##\Re(a) \leq 1## the point ##z = 1## is a singularity. For all indices ##a## the point ##z=1## is a branch point, and in Maple the branch cut is taken to be the interval ##(1,\infty)##.

    We see from this that for a<0 the formulas above are unambiguous; furthermore, the function ##\ln(1 - e^a)## is also real and unambiguous. For a > 0 both the log and the polylog functions become complex, but when evaluated in Maple, the imaginary parts seem to go away and the result remains well defined and real. When plotted, the function N/D has its maximum at a = 0.
     
  4. Nov 7, 2013 #3
    I noticed you didn't say, "maximize symbolically" or algebraic which mean numerically would do so you can't just plot them and look for a maximum?

    Suppose I define the integral functions:

    $$f(a)=\int_a^{\infty}\frac{y^3}{e^y-1}$$

    $$g(a)=\int_a^{\infty}\frac{y^2}{e^y-1}$$

    Ok then, solve them numerically and then just plot

    Plot[f(a)/g(a),{a,0,100}]

    or so.

    First need to set them up correctly in Mathematica. I'll do the first one:

    Code (Text):

    f[a_?NumericQ] := NIntegrate[y^3/(Exp[y] - 1), {y, a, \[Infinity]}]
     
    Now you do the other and try the plot routine to see what happens. I haven't so I'm curious what you find. After you do so numerically, you'll have an empirical grounding and you can then more comfortably if you wish attempt to do so symbolically.
     
    Last edited: Nov 7, 2013
  5. Nov 7, 2013 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    I would expect the maximum at a=0.

    At a=0, both integrals are positive. A smaller a will reduce the numerator and increase the denominator -> bad
    Between a=0 and a=1, the fraction is smaller than 1. Increasing a will reduce the numerator more than the denominator -> bad
    For a>1, the fraction is smaller than 1/a. Increasing a will reduce the numerator by 1/a more than the denominator -> bad

    Therefore, the fraction has its maximum at a, and it is decreasing towards both sides.

    This is just a result of a plot of both functions, no integration is necessary.
     
  6. Nov 7, 2013 #5
    I request a plot. Verdict, it's your thread. Code too.
     
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