9

The integral \begin{equation} I_m=\int_0^1 \frac{x^m-x}{\sin \pi x} \:dx \end{equation} is expressed as a finite sum of terms of the shape $r_n \zeta(2n+1) \pi^{-(2n+1)}$, where $r_n$ is a rational number that depends on $n$.

In this question Evaluating $\int_0^1 \frac{x^2-x}{\sin \pi x} dx = - \frac{7 \zeta(3)}{\pi^3}.$ The user "user90369" provided a answer for $I_2$. So I followed his steps and generalized the integral.

Define: \begin{align} &f(a) = \int_0^1 x e^{ax} dx= \frac{1+e^a(a-1)}{a^2} \\ &g(a) = \int_0^1 x^m e^{ax} dx = \frac{e^a}{a^{m+1}} \left( (-1)^mm! + \sum_{n=1}^{m} (-1)^{m+n} \frac {a^n}{n!} \right) - \frac{(-1)^mm!}{a^{m+1}} \end{align}
Rewrite $I_m$ as:

\begin{align} I_m &= \int_0^1 \frac{x^m-x}{\sin \pi x} = 2i \int_0^1 \frac{x^m-x}{ e^{i\pi x } - e^{-i \pi x} }\nonumber\\ &= 2i \sum_{k=0}^{\infty} \int_0^1 (x^m-x)e^{-i\pi x(2k+1) } dx \\ &= 2i \sum_{k=0}^{\infty} g(-i\pi(2k+1)) - f(-i\pi (2k+1)) \nonumber \end{align}

From that I arrived at the following formulas for $I_m$.

When $m$ is even and $\geq 2$:

\begin{align*} I_m = \frac{4m!i^m}{\pi^{m+1}} \left( 1-\frac{1}{2^{m+1}}\right) \zeta(m+1) + 2 m! \sum_{n\:odd \geq 0}^{m-2} \frac{i\: \zeta (m-n)}{(n+1)!(i\pi)^{m-n}} \left( 1-\frac{1}{2^{m-n}} \right) \end{align*}

When $m$ is odd and $\geq 3$:

\begin{align*} I_m = 2 m! \sum_{n\:even \geq 0}^{m-2} \frac{i\: \zeta (m-n)}{(n+1)!(i\pi)^{m-n}} \left( 1-\frac{1}{2^{m-n}} \right). \end{align*}

A few values for $I_m$:

\begin{align*} & I_2 = - \frac{7\zeta(3)}{\pi^3} \\ & I_3 = -\frac{21 \zeta(3)}{2\pi^3}\\ & I_4 = \frac{93\zeta(5)}{\pi^5}-\frac{21\zeta(3)}{\pi^3} \\ & I_5 = \frac{465\zeta(5)}{2\pi^5} - \frac{35\zeta(3)}{\pi^3} \\ & I_6 = -\frac{1}{2} \left( \frac{5715\zeta(7)}{\pi^7} - \frac{1395\zeta(5)}{\pi^5} + \frac{105\zeta(3)}{\pi^3} \right) \\ & I_7 = -\frac{1}{2} \left( \frac{40005\zeta(7)}{2\pi^7} - \frac{3255\zeta(5)}{\pi^5} + \frac{147\zeta(3)}{\pi^3} \right) \\ &I_8 = \frac{160965\zeta(9)}{\pi^9} - \frac{40005\zeta(7)}{\pi^7} + \frac{3255\zeta(5)}{\pi^5}-\frac{98\zeta(3)}{\pi^3} \end{align*}

It's also possible to integrate by parts $I_m$. Anyway, I found this integral interesting and wanted to share. Perphaps someone can do something cool with it, or not.

Sorry for the weird/sloppy english.

Quanto
  • 97,352
Pinteco
  • 2,631

2 Answers2

7

Not an answer but too long for a comment.

This is for sure a very interesting integral.

Just by curiosity, I wondered if the approximation $$\sin(t) \simeq \frac{16 (\pi -t) t}{5 \pi ^2-4 (\pi -t) t}\qquad (0\leq t\leq\pi)$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician, could lead to something simple.

Changing variable $x=\frac y \pi$ and considering $$\int y^m \csc (y)\,dy \approx \frac{y^m \left(5 m y \, _2F_1\left(1,m+1;m+2;\frac{y}{\pi }\right)-4 m y+5 \pi (m+1)\right)}{16 m (m+1)}$$ we can arrive to the incredibly simple result $$I_m \approx \frac{2 m^2+3 m+5}{16\,m\, (m+1)}-\frac 5 {16} H_m $$

The table below reproduces some values (the so-called exact values being obtained by numerical integration) $$\left( \begin{array}{ccc} m & \text{approximation} & \text{exact} \\ 2 & -0.270833 & -0.271377 \\ 3 & -0.406250 & -0.407066 \\ 4 & -0.497917 & -0.499007 \\ 5 & -0.567708 & -0.569075 \\ 6 & -0.624256 & -0.625896 \\ 7 & -0.671875 & -0.673782 \\ 8 & -0.713046 & -0.715212 \\ 9 & -0.749330 & -0.751749 \\ 10 & -0.781780 & -0.784444 \\ 20 & -0.995573 & -1.000305 \\ 30 & -1.121081 & -1.127370 \\ 40 & -1.210330 & -1.217854 \\ 50 & -1.279654 & -1.288200 \\ 60 & -1.336350 & -1.345767 \\ 70 & -1.384318 & -1.394494 \\ 80 & -1.425892 & -1.436739 \\ 90 & -1.462578 & -1.474027 \\ 100 & -1.495406 & -1.507399 \end{array} \right)$$

Edit

More amazing (at least to me) would be $$I_{m+1}-I_m\approx-\frac{5}{16 m}+\frac{1}{4 (m+1)}-\frac{1}{4 (m+2)}$$

Claude Leibovici
  • 260,315
3

Substitute $y=e^{i\pi x}$ \begin{align} &\int_0^1 \frac{x^m-x}{\sin \pi x} \ dx\\ =& \ \frac2\pi \int_{-1}^1 \bigg[ \bigg(\frac{\ln y}{i\pi}\bigg)^m-\frac{\ln y}{i\pi}\bigg]\frac{dy}{1-y^2}\\ = & \ \frac2\pi \ \Re \int_{0}^1 \bigg[{\bigg(1+\frac{\ln y}{i\pi}\bigg)^m}-1 +\frac{\ln^m y}{(i\pi)^m}\bigg]\frac{dy}{1-y^2}\\ = & \ \frac2\pi \ \Re \int_{0}^1 \bigg[\sum_{k=1}^{m} \binom mk \frac{\ln^k y}{(i\pi)^k} +\frac{\ln^m y}{(i\pi)^m}\bigg]\frac{dy}{1-y^2}\\ = & \ \frac{4 \Re (-i)^m}{\pi^{m+1}} J_m +\sum_{k=2}^{m-1}\binom mk\frac{2\Re (-i)^k}{\pi^{k+1}}J_k \end{align} where $ J_k=\int_0^1 \frac{\ln^k y}{1-y^2}dy=(1-2^{-k-1})k!\zeta(k+1)\\ $. Note that only the even terms in the sum is non-zero due to $\Re (-i)^k$.

Quanto
  • 97,352