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I did the integral $$I_{k,n}=\int_0^\infty \frac{\log^k(x)}{x^n+1}\,\mathrm dx, k\in\mathbb{N}, n\in\mathbb{N}, n\geq 2$$ via a wedge contour, which gives a recursive definition for $I_{k,n}$ in terms of all previous $I_{j,n}$. Using induction it can be proved that $$I_{2k,n}={\left(\frac{\pi}{n}\right)}^{2k+1}P_{2k}\left(\csc\left(\frac{\pi}{n}\right)\right)$$ and $$I_{2k+1,n}=-{\left(\frac{\pi}{n}\right)}^{2k+2}P_{2k+1}\left(\csc\left(\frac{\pi}{n}\right)\right)\cot\left(\frac{\pi}{n}\right)$$ where $P_k$ is the polynomial sequence of interest. Applying this gives a recursive definition for the sequence: $$P_k={(-1)}^\frac{k}{2}x+x^2\sum_{m=0}^{\frac{k}{2}-1}{k\choose 2m+1}{(-4)}^{\frac{k}{2}-m-1}P_{2m+1}+\sum_{m=0}^{\frac{k}{2}-1}{(-4)}^{\frac{k}{2}-1}\left(2{k\choose 2m}P_{2m}-{k\choose 2m+1}P_{2m+1}\right)$$ for even $k$ and $$P_k=-\sum_{m=1}^{\frac{k-1}{2}}{k\choose 2m-1}2^{k-2m}{(-1)}^{\frac{k-1}{2}-m}P_{2m-1}+\sum_{m=0}^{\frac{k-1}{2}}{k\choose 2m}2^{k-2m-1}{(-1)}^{\frac{k-1}{2}-m}P_{2m}$$ for odd $k$, assuming I have made no mistakes in copying things over. Of interest to me is two things:

  1. Showing that the imaginary part of the recursive definition cancels out for all $k$. I have omitted the full recursive definition, but clearly since it is a real integral the imaginary part must cancel, though I am uncertain about how to show it. Perhaps there are some interesting binomial coefficient properties which would help.
  2. Whether or not these polynomials show up elsewhere or have a definition which is more simple. The highest degree coefficient is always $k!$, but other than that I can’t see anything. The first few polynomials are: \begin{align} P_0&=x\\ P_1&=x\\ P_2&=2x^3-x\\ P_3&=6x^3-5x\\ P_4&=24x^5-36x^3+17x\\ P_5&=120x^5-220x^3+169x \end{align}

If anyone recognizes these I would be grateful.

Sparkle-Lin
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Nile Waldal
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1 Answers1

7

This answer evaluates the integral $I_{k,n}$ as defined above.

Let us start with the integral

$$J_n(s):=\int_{0}^{\infty}\frac{x^s}{x^n+1}\,\mathrm{d}x,$$

which can be evaluated using the substitution $u=x^n$:

$$\begin{aligned}J_n(s)&=\int_{0}^{\infty}\frac{u^{s/n}}{u+1}\cdot\frac{1}{n}u^{1/n-1}\,\mathrm{d}{u}=\frac{1}{n}\int_{0}^{\infty}\frac{u^{(s+1)/n-1}}{u+1}\,\mathrm{d}u=\frac{\pi}{n}\csc\left(\pi\frac{s+1}{n}\right)\\ &=\frac{\pi}{n}\sec\left(\frac{\pi}{2}-\frac{\pi(s+1)}{n}\right).\end{aligned}$$

We can relate $J_n(s)$ to $I_{k,n}$ by noting that

$$\frac{\mathrm{d}^k}{\mathrm{d}s^k}\Bigg|_{s=0}J_n(s)=\int_{0}^{\infty}\frac{x^s(\log x)^k}{x^n+1}\,\mathrm{d}x\Bigg|_{s=0}=I_{k,n}.$$

The $k$-th derivative of $\sec(x)$ has the following closed-form expression (for example, see the comment by @Mariusz Iwaniuk in this post):

$$\begin{aligned} \frac{\partial^k}{\partial x^k}\sec(x)=&\frac{k!}{(2\pi)^{k+1}}\left\{\zeta\left(1+k,\frac{\pi-2x}{4\pi}\right)+(-1)^k\zeta\left(1+k,\frac{\pi+2x}{4\pi}\right)\right.\\&\quad\qquad\qquad\left.-(-1)^k\zeta\left(1+k,\frac{3\pi+2x}{4\pi}\right)-\zeta\left(1+k,\frac{3\pi-2x}{4\pi}\right)\right\}, \end{aligned}$$

where $\zeta(s,a)$ denotes the Hurwitz zeta function.

Using this, we compute

$$\begin{aligned} \frac{\partial^k}{\partial x^k}\sec(x)\Bigg|_{x=\frac{\pi}{2}-\frac{\pi(s+1)}{n}}&=\frac{k!}{(2\pi)^{k+1}}\left\{\zeta\left(1+k,\frac{s+1}{2n}\right)+(-1)^k\zeta\left(1+k,\frac{1}{2}-\frac{s+1}{2n}\right)\right.\\&\qquad\qquad\qquad\left.-(-1)^k\zeta\left(1+k,1-\frac{s+1}{2n}\right)-\zeta\left(1+k,\frac{1}{2}+\frac{s+1}{2n}\right)\right\}. \end{aligned}$$

Therefore, we obtain

$$\begin{aligned} I_{k,n}&=\frac{\pi}{n}\left(-\frac{\pi}{n}\right)^k\left\{\frac{\partial^k}{\partial x^k}\sec(x)\Bigg|_{x=\frac{\pi}{2}-\frac{\pi(s+1)}{n}}\right\}\Bigg|_{s=0}\\ &=(-1)^k\cdot\frac{\pi^{k+1}}{n^{k+1}}\cdot\frac{k!}{(2\pi)^{k+1}}\left\{\zeta\left(1+k,\frac{1}{2n}\right)+(-1)^k\zeta\left(1+k,\frac{n-1}{2n}\right)\right.\\ &\qquad\qquad\qquad\qquad\qquad\qquad\left.-(-1)^k\zeta\left(1+k,\frac{2n-1}{2n}\right)-\zeta\left(1+k,\frac{n+1}{2n}\right)\right\}\\ &=\frac{k!}{(2n)^{k+1}}\left\{\zeta\left(1+k,\frac{n-1}{2n}\right)+(-1)^k\zeta\left(1+k,\frac{1}{2n}\right)\right.\\ &\left.\qquad\qquad\qquad+(-1)^{k+1}\zeta\left(1+k,\frac{n+1}{2n}\right)-\zeta\left(1+k,\frac{2n-1}{2n}\right)\right\}. \end{aligned}$$

Sparkle-Lin
  • 1,061