Maximizing the fraction of two integrals

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EDIT:
I left out something of major importance, I want to maximize with respect to a!

Homework Statement


My problem is rather complex, but in the end it boils down to maximizing the fraction
integrals.jpg

Homework Equations


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

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
 
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Verdict said:
EDIT:
I left out something of major importance, I want to maximize with respect to a!

Homework Statement


My problem is rather complex, but in the end it boils down to maximizing the fraction
integrals.jpg



Homework Equations


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


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

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:
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 \\<br /> D = -a^3 \ln(1-e^a)-3a^2\, \text{polylog}(2,e^a)<br /> +6a\, \text{polylog}(3,e^a)-6\, \text{polylog}(4,e^a)+\frac{1}{4} a^4+\frac{2}{15} \pi^4<br />
According to Maple,
\text{polylog}(a,z) = \sum_{n=1}^{\infty} \frac{z^n}{n^a} 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.
 
Verdict said:
EDIT:
I left out something of major importance, I want to maximize with respect to a!

Homework Statement


My problem is rather complex, but in the end it boils down to maximizing the fraction
integrals.jpg

Homework Equations


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

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

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:
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.
 
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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.
 
I request a plot. Verdict, it's your thread. Code too.
 
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