3

I'd like to prove, using a partial fraction decomposition (I don't want to use residue calculus), that $$\int_0^{+\infty} \frac{t^{m-1}}{1+t^{2n}}{\rm d}t=\frac{\pi}{2n\sin\frac{m\pi}{2n}}$$ where $1\le m<2n$.

  1. I find that $$\frac{X^{m-1}}{1+X^{2n}} = \sum_{k=-n}^{n-1} \frac{\alpha_k}{X-\omega_k}$$ where $\omega_k={\rm e}^{i\theta_k}$, $\theta_k=\frac{(2k+1)\pi}{2n}$ and $\alpha_k=-\frac{\omega_k^m}{2n}$.
  2. Next, with $A>0$, $$\int_0^A \frac{{\rm d}t}{t-\omega_k} = \ln A+\frac12\ln(1-2/A\cos\theta_k+1/A^2) + {\rm i}\left(\arctan\frac{A-\cos\theta_k}{\sin\theta_k} + \arctan\frac{\cos\theta_k}{\sin\theta_k}\right)$$
  3. When $A\to+\infty$, the sum of all the logarithms tends to $0$, because the sum of the $\alpha_k$ is zero.
  4. To find the limit of the sum of the $\arctan$, I group the term in $k$ and the term in $-k-1$, with $0\le k\le n-1$, because $\theta_{-k-1}=-\theta_k$, and $\sin\theta_k>0$ for $0\le k\le n-1$ and $\sin\theta_k<0$ for $-n\le k\le -1$. I find $${\rm i}(\alpha_k-\overline{\alpha_k})\left(\frac\pi2+\arctan\frac{\cos\theta_k}{\sin\theta_k}\right)$$
  5. Now I find that whether $\cos\theta_k$ is positive or negative, this limit can be written $\frac{1}{n}\sin(m\theta_k)(\pi-\theta_k)$.
  6. So all that is left is to find a closed form of the sum $$\frac1n\sum_{k=0}^{n-1} \sin(m\theta_k)(\pi-\theta_k)$$

If you had the patience to follow me until now, can you see where I made a mistake ? I have done this computations four or five times now, discovered a few mistakes, and I'm not closer to the result than at the beginning :-(

If you could help, it would be greatly appreciated :-)

\bye

Nicolas FRANCOIS
  • 4,846

2 Answers2

1

A first manipulation to reduce the integration range to $(0,1)$:

$$ I(m,n)=\int_{0}^{+\infty}\frac{t^{m-1}}{1+t^{2n}}\,dt=\int_{0}^{1}\frac{t^{m-1}}{1+t^{2n}}\,dt+\int_{1}^{+\infty}\frac{t^{m-1}}{1+t^{2n}}\,dt= \int_{0}^{1}\frac{t^{m-1}+t^{2n-m-1}}{1+t^{2n}}\,dt.$$ It leads to $$ I(m,n)=\int_{0}^{1}\frac{t^{m-1}+t^{2n-m-1}-t^{2n+m-1}-t^{4n-m-1}}{1-t^{4n}}\,dt $$

$$ I(m,n)=\sum_{k\geq 0}\left(\frac{1}{4kn+m}+\frac{1}{4kn+2n-m}-\frac{1}{4kn+2n+m}-\frac{1}{4kn+4n-m}\right) $$

so $I(m,n)$ can be computed by applying a discrete Fourier transform (based on the $4n$-th roots of unity) to $\sum_{s\geq 1}\frac{x^s}{s}=-\log(1-x)$. We have (also due to the reflection formula for the $\psi=\Gamma'/\Gamma$ function) $$\sum_{k\geq 0}\left(\frac{1}{4kn+m}-\frac{1}{4kn+4n-m}\right) = \frac{\pi}{4n}\cot\left(\frac{\pi m}{4n}\right) $$ and $$ \sum_{k\geq 0}\left(\frac{1}{4kn+2n-m}-\frac{1}{4kn+2n+m}\right) = \frac{\pi}{4n}\tan\left(\frac{\pi m}{4n}\right) $$ immediately leading to the claim.

Jack D'Aurizio
  • 353,855
  • 1
    Thanks, but that's not what I was asking (and I can't use your formulas about function $\Gamma$). I'll post another question with the result I'm looking for. – Nicolas FRANCOIS Jun 14 '22 at 10:33
  • The reflection formula for the $\psi$ function is not strictly needed - you just have to evaluate $\log(1-x)$ at some roots of unity, i.e. to exploit the factorization of $x^{4n}-1$. – Jack D'Aurizio Jun 14 '22 at 10:47
1

  1. … …

  2. So all that is left is to find a closed form of the sum $$S=\frac1n\sum_{k=0}^{n-1} \sin(m\theta_k)(\pi-\theta_k)$$

  3. With $\theta_k=\frac{(2k+1)\pi}{2n}$, proceed as follows \begin{align} S=&\ \frac1{2n}\sum_{k=0}^{2n-1} \sin(m\theta_k)(\pi-\theta_k) = -\frac1{2n}\sum_{k=0}^{2n-1} \sin(m\theta_k)\theta_k\\ =& -\frac {\pi}{2n^2}\sum_{k=0}^{2n-1} k\sin\frac{\pi(2k+1)m}{2n} = \frac 1{2n}\frac{d}{dx}\bigg(\sum_{k=0}^{2n-1} \cos\frac{\pi(2k+1)x}{2n}\bigg)_{x=m}\\ =& \ \frac1{2n}\frac d{dx}\bigg(\frac{\sin 2\pi x}{2\sin\frac{\pi x}n} \bigg)_{x=m} =\frac{\pi}{2n} \csc \frac{m\pi}{2n} \end{align}

Quanto
  • 97,352