I derived equation of sum from the following problem,

$\int_{a}^{b}\sum_{n=0}^{\infty} cos^n(x)dx$.

Using the following definitions,

$cos(x)=\frac{e^{ix}+e^{-ix}}{2}$ and $(a+b)^n=\sum_{m=0}^{n}\binom {n}m a^{n-m}b^m$,

my solution to the above is,

$\sum_{n=0}^{\infty}\sum_{m=0}^{n} \binom {n}m (-i)\frac{e^{i(n-2m)}}{n-2m} $.

However, Mathematica cannot evaluate this summation because of singularities. Either my derivation is incorrect although steps are valid or I must add conditionals to this summation evaluation (if it is valid way to do). If my derivation is correct, how do I evaluate this summation,


Is there a way to avoid singularities? Like $n \neq 2m$

  • 1
    $\begingroup$ Your sum is a geometric series that is convergent for most real xand easily summable. I would compute that and then integrate. $\endgroup$
    – mikado
    Commented Jun 19, 2021 at 7:04
  • $\begingroup$ Sum[Cos[x]^n, {n, 0, \[Infinity]}, Assumptions -> x \[Element] Reals] produces 1/(1 - Cos[x]). If $\cos(x)=\pm1$, the series diverges. $\endgroup$
    – user64494
    Commented Jun 19, 2021 at 7:28

1 Answer 1


A direct approach to your problem works

sum = Sum[Cos[x]^n, {n, 0, ∞}]
(* 1/(1 - Cos[x]) *)

Integrate[sum, x]
(* -Cot[x/2] *)

Of course, the answer is only valid if the interval doesn't span a singularity (where Cos[x]==1)

  • $\begingroup$ If Cos[x]==-1, the series diverges so such points should be excluded too, when integrating. $\endgroup$
    – user64494
    Commented Jun 19, 2021 at 12:52
  • $\begingroup$ Think of sum = Sum[Cos[x]^n, {n, 0, \[Infinity]}, GenerateConditions -> True] which results in ConditionalExpression[1/(1 - Cos[x]), Abs[Cos[x]] < 1]. $\endgroup$
    – user64494
    Commented Jun 19, 2021 at 12:59

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