# Asymptotic values of integrals

For an integral like $$D_{n}(x) \equiv \int_{0}^{x} \frac{t^{n}}{e^{t}-1} d t$$ The asymptotic values are given as

$$D_{n}(x) \simeq\left\{\begin{array}{ll} n ! \zeta(n+1)-x^{n} e^{-x}+O\left(x^{n} e^{-2 x}\right), & x \rightarrow \infty \\ x^{n} / n-x^{n+1} / 2(n+1)+O\left(x^{n+2}\right), & x \rightarrow 0 \end{array}\right.$$

Is there a way to get these expressions in Mathematica?

• For the $x=0$ series the following works Assuming[n > 0, Series[Integrate[t^n/(Exp[t] - 1), {t, 0, x}], {x, 0, 3}]]. Oct 11, 2020 at 13:20
• By the way, you should type at least something in MA format to help others who are willing to help. Oct 11, 2020 at 18:10

Substitution u=Exp[-t] respectively t=-Log[u] gives the first term of the asymptotic expansion x->Infinity of Dn[x]:

Integrate[(-Log[u])^n/(u-1),{u,0,1}]-Integrate[(-Log[u])^n/(u-1),{u,0,Exp[-x]}]


The first integral evaluates to

Integrate[(-Log[u])^n/(u-1),{u,0,1}]
(*-n Gamma[n] PolyLog[1 + n, 1]*)


the second integral (hopefully) is of order O[Exp[-x]]

• I think this is wrong: dt=-du/u so the substitution has to be (-Log[u])^n/(u(u-1))!! This stems from t[u_]:=-Log[u], D[t[u],u] is -1/u! Oct 12, 2020 at 20:33
• I think this is wrong: dt=-du/u so the substitution has to be (-Log[u])^n/(u(u-1))!! This stems from t[u_]:=-Log[u], D[t[u],u] is -1/u! The partial fraction decomposition is 1/(u(u-1))==-1/u+1/(u-1)!!! Oct 12, 2020 at 20:40
• @SteffenJaeschke Thanks for your reply. Assuming t[u]==-Log[u] it follows dt/du==-1/u and u==Exp[-t] . The first part equals dt==-du/u. Integrand transforms to (-Log[u])^n/(1/u-1)/(-u)  .That's why my answer seems to be right. Oct 13, 2020 at 6:36

As a starter:

Integrate[x^n/(-1 + E^x), {x, -\[Infinity], \[Infinity]},
Assumptions -> n \[Element] Integers && n > -1 && x \[Element] Reals]


(* Integrate[x^n/(-1 + E^x), {x, -[Infinity], [Infinity]}, Assumptions -> n [Element] Integers && n > -1 && x [Element] Reals] *)

For the solution 𝑥→∞:

IntSer[x_, n_] :=
Integrate[Series[t^n /(Exp[t] - 1), {t, Infinity, 2}], {t, 0, x}]
Table[{n, IntSer[x, n], n! Zeta[n + 1] - x E^-x}, {n, 1,
2}] // TableForm


This shows two aspects of the solution: (1) Mathematica does not expand the same way as Your sources do. (2) The condition seems to have higher validity.

The methods of Mathematica to not do the asymptotic expansion this is just the integral this way.

This result can be developed further into a asymptotic You will prefer. Reenter this with Normal:

Table[{n, Normal@IntSer[x, n], (n + 1)! Zeta[n + 1] - x E^-x}, {n, 1,
2}] // TableForm


The function freed from the condition can be expanded in Mathematica:

Series[\[Pi]^2/6 + x Log[1 - Cosh[x] + Sinh[x]] -
PolyLog[2, E^-x], {x, \[Infinity], 3}]


(* *)

Series[x^2 Log[1 - Cosh[x] + Sinh[x]] - 2 x PolyLog[2, E^-x] -
2 PolyLog[3, E^-x] + 2 Zeta[3], {x, \[Infinity], 3}]


For the solution x->0:

IntSer0[x_, n_] :=
Integrate[Series[t^n /(Exp[t] - 1), {t, 0, 3}], {t, 0, x}]
Table[{n, Normal@IntSer0[x, n], x^n/n - x^(n + 1)/(2 (n + 1))}, {n, 1,
2}] // TableForm


In the expansion 𝑥→0 this solution is expanded approximately and confirms the formula for n=1 and 2. The O can be confirmed expanding to n=2 and then drop the second-order term.

This does confirm 𝐷𝑛(𝑥)≃. I hope Mathematica is correct and I discovered a typo in Your formula.

Induction step symbolic for both:

Integrate[t^n*SeriesData[t, 0, {1, -1/2, 1/12, 0, -1/720}, -1, 4, 1], {t, 0, x}]

Integrate[t^n (
SeriesData[t, 0, {1,
Rational[-1, 2],
Rational[1, 12], 0,
Rational[-1, 720]}, -1, 4, 1]), {t, 0, x},
Assumptions -> n \[Element] Integers && n > -1 && x \[Element] Reals]


The problem stems from the term t^(n-1) according to the information given in Integrate.

Mathematica does not do this for arbitrary n in V12.0.0. Since the 𝐷𝑛(𝑥) formula is valid for all n and The approximation can be integrated for n to give n+1 up to a constant the induction step is done.

The opportunity to use a logarithmics substitution fails the same way as the direct integration:

u[t_] := Exp[-t]
D[u[t], t]


(-Exp[-t])

du/dt=-Exp[-t] => du=-udt => -du/u=dt

or

t[u_] := -Log[u]
D[t[u], u]


(* -1/u *)

dt/du=-1/u => dt=-du/u

Integrate[(-Log[u])^n/(u (u - 1)), {u, 1, Exp[x]}]


But

Solve[1/(u (u - 1)) == B/u + 1/(u - 1), B]


(* {B->-1} *)

dt = -du/ u so the substitution has to be (-Log[u])^ n/(u (u - 1))!! This stems from t[u_] := -Log[u], D[t[u], u] is - 1/u! The partial fraction decomposition is 1/(u (u - 1)) == -1/u + 1/(u - 1)!!!

Integrate[-(-Log[u])^n/u, {u, 1, Exp[x]}]

ConditionalExpression[-(((-x)^n x)/(1 + n)),
Re[x] <= 0 && Im[x] == 0 && Re[n] > -1]

Integrate[-(-Log[u])^n/(u - 1), {u, 1, Exp[x]}]


This is again open for the asymptotic expansion of 𝐷𝑛(𝑥).

• It does not solve the original question. There is a big gap between solving for $n=1,2$ and a general case. Why do you even think $n$ is integer, there is no such assumption in OP?! Your remark at the end is about integer $n$. Oct 11, 2020 at 18:09
• This is the general solution with the induction step that is not conducted by Mathematica. Now, why do I expect this to be n? This looks like Poisson distribution moment for Bose-Einstein distribution. The n! suggests this gradually, but is Complexes. Zeta induces Complexes. At most this allows to calculate from one representation of D(n) to other. Oct 11, 2020 at 19:40
• I am sorry, but what do you mean by "why do I expect this to be n?", "The n! suggests this gradually, but is Complexes. ", and what are "this" and "other" in the sentence "At most this allows to calculate from one representation of D(n) to other"? Try to be a little bit coherent in writing. Oct 11, 2020 at 19:48
• In your definition IntSer[x_, n_] := ... you interchange the order of Series and Integrate. I'm not sure wether that's allowed. Oct 11, 2020 at 19:49
• I bother too. Whether that is allowed or not? Mathematica does it and both are the very same. Oct 12, 2020 at 15:53

AsymptoticIntegrate should be the first function to try, but it seems like it kinda struggles for general values of n. This works, however, if you give it some time:

AsymptoticIntegrate[
t^2/(Exp[t] - 1),
{t, 0, x},
{x, ∞, 2}
]
`