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This Question asked on math over flow

I tried to find the indefinite integral $$ f_n(x)=\int \prod_{k=1}^n \cos^k(kx)dx$$ by using Euler's formula and put $x=\frac{\ln y}{2i}$ I got $$ f_n(x)=-i2^{-\frac{n(n+1)}{2}-1}\int y^{-\frac{n(n+1)(2n+1)}{12}-1} \prod_{k=1}^n (y^k+1)^k dy$$ now lets define $a(n,k)$ as the coefficient of $x^k$ in the expression $\prod_{p=1}^n (x^p+1)^p$ then $$ \prod_{k=1}^n (y^k+1)^k =\sum_{k=0}^{\frac{n(n+1)(2n+1)}{6}} a(n,k) y^k$$

So $$ f_n(x)=2^{-\frac{n(n+1)}{2}-1}\sum_{k=0}^{\frac{n(n+1)(2n+1)}{6}} \frac{a(n,k)}{k-\frac{n(n+1)(2n+1)}{12}} (-i)\exp\left(-2x\left(k-\frac{n(n+1)(2n+1)}{12}\right) i\right)+c $$ and where $f_n(x)$ is real So we will take the real part of the result and get $$ f_n(x)=2^{-\frac{n(n+1)}{2}-1}\sum_{k=0}^{\frac{n(n+1)(2n+1)}{6}} \frac{a(n,k)}{k-\frac{n(n+1)(2n+1)}{12}}\sin\left(2x\left(k-\frac{n(n+1)(2n+1)}{12}\right)\right)+c $$ and if $k=\frac{n(n+1)(2n+1)}{12}$ then take limit to get $\frac{\sin(2ax)}{a}=2x , a\to0$

finally if we know $$ a(n,k)=a\left(n,\frac{n(n+1)(2n+1)}{6}-k\right)$$ then $$ f_n(x)=2^{-\frac{n(n+1)}{2}} a\left(n,\frac{N}{2}\right) x+2^{-\frac{n(n+1)}{2}-1}\sum_{k=1}^{\frac{N}{2}} \frac{a\left(n,\frac{N}{2}-k\right)}{k} \sin\left(2kx\right)+c ,\text{if } N \text{ is even}$$ and $$ f_n(x)=2^{-\frac{n(n+1)}{2}+1}a\left(n,\frac{N-1}{2}\right)\sin\left(x\right)+2^{-\frac{n(n+1)}{2}}\sum_{k=1}^{\frac{N-1}{2}} \frac{a\left(n,\frac{N-1}{2}-k\right)}{2k+1} \sin\left((2k+1)x\right)+c ,\text{if } N \text{ is odd}$$ where $N=\frac{n(n+1)(2n+1)}{6} $

now my QUESTIONS

How to calculate $a(n,k)$ or even what is the recurrence relation?

also How to prove that $a(n,k)=a\left(n,\frac{n(n+1)(2n+1)}{6}-k\right)$?

and when we took the real part if we took the imaginary part it will be zero So How to prove $$\sum_{k=0}^{\frac{n(n+1)(2n+1)}{6}} \frac{a(n,k)}{k-\frac{n(n+1)(2n+1)}{12}}\cos\left(2x\left(k-\frac{n(n+1)(2n+1)}{12}\right)\right)=c $$

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  • $\begingroup$ Have you considered the log of the original expression and use the Taylor expansion of $\log(1+u)$? $\endgroup$ Commented Feb 12 at 21:20
  • $\begingroup$ @StefanLafon No..please show how can you get it by that way $\endgroup$
    – Faoler
    Commented Feb 12 at 21:35
  • $\begingroup$ Hmm, on second thoughts, that was not one of my brightest ideas :) I don't think it helps are all. $\endgroup$ Commented Feb 12 at 23:12
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    $\begingroup$ Related post $\endgroup$
    – Sil
    Commented Feb 12 at 23:39
  • $\begingroup$ If one takes $n\to\infty$, the resulting infinite series is tabulated on OEIS as A026007. This entry does not include an explicit formula for the $n$th term, so I'm not sure one should hope for such in that case (let alone the case for finite $n$). $\endgroup$ Commented Feb 13 at 19:01

1 Answer 1

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I got it...

firstly the degree of $(x^p+1)^p$ is $p^2$ So the degree of $\prod_{p=1}^n (x^p+1)^p$ is $$N=1+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$$ now we have $$\prod_{p=1}^n (x^p+1)^p=\sum_{p=1}^N a(n,p)x^p $$ by taking kth derivative and put $x\to0$ we get $$\lim_{x\to0} \frac{d^k}{dx^k} \prod_{p=1}^n (x^p+1)^p=\lim_{x\to0} \frac{d^k}{dx^k}\sum_{p=1}^N a(n,p)x^p $$ But for natural $k,p$ $$\lim_{x\to0} \frac{d^k}{dx^k} x^p=0 ,p\ne k$$ So $$\lim_{x\to0} \frac{d^k}{dx^k} a(n,k)x^k=\lim_{x\to0} \frac{d^k}{dx^k} \prod_{p=1}^n (x^p+1)^p $$ then $$ a(n,k)=\frac{1}{k!}\lim_{x\to0} \frac{d^k}{dx^k} \prod_{p=1}^n (x^p+1)^p $$ now to find the kth derivative we need to use General Leibniz rule and get $$ \frac{1}{k!}\lim_{x\to0} \frac{d^k}{dx^k} \prod_{p=1}^n (x^p+1)^p=\frac{1}{k!}\sum_{k_1+k_2+...+k_n=k} \binom{k}{k_1,k_2,...,k_n} \prod_{j=1}^n \lim_{x\to0} \frac{d^{k_j}}{dx^{k_j}} (x^j+1)^j$$ where $$ \lim_{x\to0} \frac{d^{k_j}}{dx^{k_j}} (x^j+1)^j=\sum_{p=0}^j \binom{j}{p} \lim_{x\to0} \frac{d^{k_j}}{dx^{k_j}} x^{pj} $$ So it must be $k_j=pj$ which mean $\frac{k_j}{j}\in N$ or its value is zero then $$ \lim_{x\to0} \frac{d^{k_j}}{dx^{k_j}} (x^j+1)^j=\binom{j}{\frac{k_j}{j}} k_j!f\left(\frac{k_j}{j}\right) , 0 \leq\frac{k_j}{j}\leq j $$ where $f(x)=1$ if $x\in N$ and $f(x)=0$ if $x \notin N$

back to the formula we have $$ a(n,k)=\frac{1}{k!}\sum_{k_1+k_2+...+k_n=k} \binom{k}{k_1,k_2,...,k_n} \prod_{j=1}^n \lim_{x\to0} \frac{d^{k_j}}{dx^{k_j}} (x^j+1)^j$$ $$ =\sum_{k_1+k_2+...+k_n=k} \frac{1}{k_1!k_2!...k_n!} \prod_{j=1}^n \binom{j}{\frac{k_j}{j}} k_j!f\left(\frac{k_j}{j}\right)$$ So $$ a(n,k)=\sum_{k_1+k_2+...+k_n=k}\prod_{j=1}^n \binom{j}{\frac{k_j}{j}} f\left(\frac{k_j}{j}\right)$$ now put $g_j=\frac{k_j}{j}$ So $$ a(n,k)=\sum_{g_1+2g_2+...+ng_n=k}\prod_{j=1}^n \binom{j}{g_j} f\left(g_j\right)$$ where $g_1+2g_2+...+ng_n=k$ with $0\leq g_j \leq j$ which mean $g_j\in\{0,1,2,...,j\}$ So $f(g_j)=1$

finally I got $$ a(n,k)=\sum_{\substack{\sum_{j=1}^n j g_j=k \\ g_j\in\{0,1,..,j\}}}\prod_{j=1}^n \binom{j}{g_j}$$

and because of $g_j\in\{0,1,..,j\}$ then we can put $g_j\to j-g_j$ then $$ \sum_{j=1}^n j (j-g_j)=k \to \sum_{j=1}^n j g_j=N-k $$ which mean $a(n,k)=a(N-k)$

and for the last question to prove the given series is constant function for $x$ lets define $f(x)$ and rewrite $\cos x$ as real part of $e^{ix}$ So $$ f(x)=\Re\left(\sum_{\substack{k=0 \\ k\ne \frac{N}{2}}}^N \frac{a(n,k)}{k-\frac{N}{2}} \exp\left(2ix\left(k-\frac{N}{2} \right) \right) \right)$$ note that the case $k=\frac{N}{2}$ is real valued by using limit : $\frac{\sin(ax)}{a} , a\to 0$ , then by derivative $$f'(x)=\Re\left(2i\sum_{\substack{k=0 \\ k\ne \frac{N}{2}}}^N a(n,k) \exp\left(2ix\left(k-\frac{N}{2} \right) \right) \right) $$ $$=-2\Im\left(e^{-iNx}\sum_{k=0}^N a(n,k) e^{2ikx}-a\left(n,\frac{N}{2}\right) \right)=-2\Im\left(e^{-iNx}\prod_{k=1}^n \left(e^{2ikx}+1\right)^k\right) $$ $$=-2\Im\left(\prod_{k=1}^n e^{-ik^2x} \left(e^{2ikx}+1\right)^k\right)=-\Im\left(\prod_{k=1}^n \left(2 \cos (kx)\right)^k \right)=0 $$ therefore $f'(x)=0$ which mean $f(x)$ is constant for $x$

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