# How to get the result from a RootSum?

While working on an integral, I got this scary-looking expression. I tried to use // Normal // ToRadicals // FullSimplify to get the answer, but it is taking lot of time. I don't know how to do it in a simple way.Please help me in this regard, thanks in advance.

2 \[Pi] -
RootSum[1 + 2 #1^2 + 6 n #1^2 - 6 n^2 #1^2 + 2 n^3 #1^2 - 23 #1^4 +
50 n #1^4 - 25 n^2 #1^4 - 12 n^3 #1^4 + 15 n^4 #1^4 - 6 n^5 #1^4 +
n^6 #1^4 + 36 #1^6 - 168 n #1^6 + 316 n^2 #1^6 - 304 n^3 #1^6 +
156 n^4 #1^6 - 40 n^5 #1^6 + 4 n^6 #1^6 &, Log[-#1]/#1 &]
• That is the answer. It won't simplify further. If you want a numerical value you need to provide a value for n. If you choose n=1 you get $4\pi$ : FullSimplify[thething /. n -> 1] – flinty Jun 11 '20 at 12:29
• I have the numerical values, I need a closed-form or a nice output from that expression. Thanks. – Shivam K Jun 11 '20 at 12:34
• There are some n that work e.g FullSimplify[ToRadicals[thething/.n->4]] gives you $2 \left(4+\sqrt{7}\right) \pi$, n=3 gives you $8\pi$, and n=1 gives you $4\pi$, but most others e.g n=6 do not simplify to anything nice. – flinty Jun 11 '20 at 12:41
• What do you need the closed form for? What integral did this come from? – flinty Jun 11 '20 at 12:53
• Integrate[Log[16*(n - 1)^4*x^6 + (4*(n - 3)*(n - 1)^2*x^4 + 1 + ((n - 1)^3 + 4)*x^2)^2]/x^2, {x, 0, Infinity}] This is the integral. 'n' is a positive integer. – Shivam K Jun 11 '20 at 13:06

RootSum[] is necessary for reasons similar to why Root[] is needed: namely, as a way to represent sums of a function over the roots of a polynomial. In brief, if you have the expression

r == RootSum[C[0] + C[1] # + ... + C[n] #^n &, f[#] &]

this is equivalent to

$$r=\sum_{k=1}^n f(x_k),\qquad c_n x_k^n+\cdots+c_1 x_k+c_0=0,\; k=1,\dots,n$$

In the case of the expression you have, note that the generating polynomial (i.e. the first argument of RootSum[]) has the following structure:

Collect[1 + 2 #1^2 + 6 n #1^2 - 6 n^2 #1^2 + 2 n^3 #1^2 - 23 #1^4 + 50 n #1^4 -
25 n^2 #1^4 - 12 n^3 #1^4 + 15 n^4 #1^4 - 6 n^5 #1^4 + n^6 #1^4 + 36 #1^6 -
168 n #1^6 + 316 n^2 #1^6 - 304 n^3 #1^6 + 156 n^4 #1^6 - 40 n^5 #1^6 +
4 n^6 #1^6 &[x], x]
1 + (2 + 6 n - 6 n^2 + 2 n^3) x^2 +
(-23 + 50 n - 25 n^2 - 12 n^3 + 15 n^4 - 6 n^5 + n^6) x^4 +
(36 - 168 n + 316 n^2 - 304 n^3 + 156 n^4 - 40 n^5 + 4 n^6) x^6

and you should immediately observe that all the powers involved are even. This means that if we let $$u=x^2$$, we get a polynomial that is cubic in $$u$$, which lends itself to using things like Cardano's formula to express the roots.

I won't go that far. (The sufficiently masoch- ...er, motivated reader is free to go further than me.) Instead, the obervation of even powers is why you get the following result after applying Normal[]:

and you should now notice that you obtain a result of the form $$\sum_k\frac{\log(-x_k)}{x_k}$$, where the $$x_k$$ come in $$\pm\sqrt{\cdot}$$ pairs, corresponding to the roots of the cubic polynomial I mentioned earlier. You could try using ToRadicals[] to get the Cardano-like radical form, but it won't be pretty.

Good luck!

• Thanks, sir, I will try in this way. – Shivam K Jun 11 '20 at 13:05