# integral, explicit form

I want to estimate the ratio of integrals:

$$\frac{\int \frac{4 a T^3}{\frac{4 a T^4}{3}+\Lambda_1 \left(\frac{4}{T^3}+1\right)}}{\int \frac{4 a T^3}{\frac{4 a T^4}{3}+\Lambda_0 \left(\frac{4}{T^3}+1\right)}}$$

where $\Lambda_1$=$\Lambda$ (constant) and $\Lambda_1$=0 for some temperature $T_0$ (arbitrary constants are set to zero).

When I try to integrate the function:

f[T_] := (4 a T^3)/(4/3 a T^4 + Λ (1 + 4/T^3))
Integrate[ f[T], T]


I get the result like this (with # and &) :

12a RootSum[ 12 Λ + 3 Λ #1^3 + 4 a #1^7 &, (Log[T - #1] #1^4)/(9 Λ + 28 a #1^4) &]


What am I doing wrong? How do I get an explicit expression for the integral? How do I solve this problem ?

• (1) The documentation provides some helpful hints. (2) Typically there are seven discrete poles of the integrand in the Complex plane, so the best you can hope for--except for some special values of $a$ and $\Lambda$--is precisely a sum over those poles, which is what RootSum accomplishes. What, then, would a more "explicit expression" look like? – whuber Jan 22 '13 at 19:24
• How can I, at least, approximate this function for $T\rightarrow\infty$? – molkee Jan 22 '13 at 19:51
• You can do Series[f[bigT], {bigT, Infinity, 10}]. – b.gates.you.know.what Jan 22 '13 at 19:53
• The integral diverges logarithmically at $\infty$, because eventually the $4aT^4/3$ term in the denominator overwhelms the other term, giving an integral proportional to $\int dT/T$. – whuber Jan 22 '13 at 20:14
• A comment on the mathematical problem itself: an indefinite integral is defined only up to an arbitrary constant, so for the ratio to be well defined you need additional conditions. – Szabolcs Jan 22 '13 at 20:37

It would be more appropriate to use this definition :

f[T_, a_, Λ_] := (4 a T^3)/(4/3 a T^4 + Λ (1 + 4/T^3))


In general you cannot get an explicit expresion (i.e. in terms of radicals) when roots of higher order polynomials are taken into account. There is a fundamental mathematical barrier see e.g. Abel's Impossibility Theorem, Galois's Theorem. Only for special values of a and Λ you could get it. Root and RootSum objects are symbolic representations of certain well defined mathematical concepts. So for example you can evaluate them with arbitrary numerical accuracy. For a relation between RootSum and Root you can use Normal, e.g. Normal @ Integrate[ f[T, a, Λ], T]. More on Root objects you can find here How do I work with Root objects ?

You can find the limit of your ratio when T goes to Infinity :

Limit[  Integrate[ f[T, a, Λ1], T, Assumptions :> {a > 0, Λ1 > 0}]/
Integrate[ f[T, a, Λ2], T, Assumptions :> {a > 0, Λ2 > 0}] ,
T -> Infinity]

 1


If you have values a, Λ1 and Λ2 you can compute the ratio for any temerature T :

ratio[ T_?NumericQ, a_?NumericQ, Λ1_?NumericQ, Λ2_?NumericQ] :=
NIntegrate[f[x, a, Λ1], {x, 0, T}]/  NIntegrate[f[x, a, Λ2], {x, 0, T}]
ratio[300, 10, 20, 15]

0.990688

Plot[ ratio[T, 10, 20, 15], {T, 0, 300}] One can see that the ratio is monotonic and tends to 1 quite rapidly.

• Thnx for your help, guys! – molkee Jan 23 '13 at 2:24

Can set it up as below. You need not look at the intermediate results, hence can pretend you never ran into those RootSum things.

f[t_, lam_] := (4 a*t^3)/(4/3 a*t^4 + lam*(1 + 4/t^3));
f2[t_, lam_] := Integrate[f[t, lam], t];

quot = f2[t, lam0]/f2[t, lam1];


Here is the asymptotic behavior of the quotient as t --> infinity.

Limit[quot, t -> Infinity, Assumptions -> Element[a, Reals]]

(* Out= 1 *)