Approximation of $\int_0^\pi \big[x(\pi-x)\csc (x)\big]^k\,dx \quad \forall k$

Question

A recent post addressed the problem of the closed form of $$I(k)=\int_0^\pi \Bigg[ \frac {x(\pi-x)} {\sin(x)}\Bigg]^k \,dx$$ When $k$ is a positive integer, they seem to be known except that they require a lot of computer ressources as soon as $k\geq 10$.

Just for the fun of it, I tried to obtain approximate values of these integrals for any exponent.

Because ot the analogy with Bhaskara I's sine approximation formula, my leading idea was to first approximate the integrand $$x(\pi-x)\csc (x) \sim a -bx(\pi-x)$$ and the coefficient $(a,b)$ where obtained minimizing the norm $$\Phi(a,b)=\int_0^\pi \Big[x(\pi-x)\csc (x)-\big[a -bx(\pi-x) \big] \Big]^2\,dx$$ which is analytic.

The optimal values are $$a=\frac{18 \left(14 \pi ^2 \zeta (3)-155 \zeta (5)\right)}{\pi ^3}\qquad \qquad b=\frac{30 \left(49 \pi ^2 \zeta (3)-558 \zeta (5)\right)}{\pi ^5}$$ $$\Phi(a,b)=\frac{6 \left(\pi ^4 \left(\pi ^2-2009 \zeta (3)\right) \zeta (3)+45570 \pi ^2 \zeta (3) \zeta (5)-259470 \zeta (5)^2\right)}{\pi ^5}=1.903\times 10^{-4}$$ Notice that these values are relatively close to $a=\frac{5 \pi ^2}{16}$ and $b=-\frac 14$ given using 2; however, the current norm is four times smaller.

All of that leads to $$J(k)=\int_ 0^\pi \big[a -bx(\pi-x) \big]^k\,dx$$ $$\color{blue}{J(k)=\pi \,\left(\frac{3(930 \zeta (5)-77 \pi ^2 \zeta (3))}{2 \pi ^3}\right)^k\, _2F_1\left(\frac{1}{2},-k;\frac{3}{2};-\frac{5 \left(49 \pi ^2 \zeta (3)-558 \zeta (5)\right)}{930 \zeta (5)-77 \pi ^2 \zeta (3)}\right)}$$

Just as for $I(k)$ (see @user64494's comment), $\lim_{k\to \infty } \, \frac{J (k+1)}{J(k)} =\pi^-$.

When $k$ is an integer, $\, _2F_1\left(\frac{1}{2},-k;\frac{3}{2};z\right)$ are quite simple polynomials with interesting patterns $$\left( \begin{array}{cc} k & \, _2F_1\left(\frac{1}{2},-k;\frac{3}{2};z\right)\\ 1 & 1-\frac{1}{3}z\\ 2 & 1-\frac{2 }{3}z+\frac{1}{5}z^2 \\ 3 & 1-\frac{3 }{3}z+\frac{3 }{5}z^2-\frac{1}{7}z^3 \\ 4 & 1-\frac{4 }{3}z+\frac{6 }{5}z^2-\frac{4}{7}z^3+\frac{1}{9}z^4 \\ 5 & 1-\frac{5 }{3}z+\frac{10 }{5} z^2-\frac{10 }{7}z^3+\frac{5}{9}z^4-\frac{1}{11}z^5 \end{array} \right)$$

This seems to lead to decent approximations $$\left( \begin{array}{ccc} k & J(k) & I(k) \\ 1 & 8.41440 & 8.41440 \\ 2 & 22.6580 & 22.6582 \\ 3 & 61.3506 & 61.3546 \\ 4 & 167.057 & 167.093 \\ 5 & 457.510 & 457.738 \\ 6 & 1260.20 & 1261.41 \\ 7 & 3491.25 & 3497.00 \\ 8 & 9743.95 & 9752.77 \\ 9 & 27314.7 & 27360.7 \\ 10 & 76794.7 & 77205.2 \end{array} \right)$$

What looks interesting (at least to me) is that this works for non integer values of $k$ $$\left( \begin{array}{ccc} k & J(k) & I(k) \\ 1.234 & 10.6042 & 10.6042 \\ 2.345 & 31.9297 & 31.9303 \\ 3.456 & 96.8040 & 96.8156 \\ 4.567 & 295.557 & 295.662 \\ 5.678 & 908.812 & 909.529 \\ 6.789 & 2814.49 & 2818.65 \\ 7.890 & 8688.35 & 8709.72 \end{array} \right)$$ and even for complex values

$$\left( \begin{array}{ccc} k & J(k) & I(k) \\ 1+i & +4.61973 +7.00583 \,i & +4.61977 +7.00576 \,i \\ 1+2 i & -3.27763 +7.65210 \,i & -3.27715 +7.65218 \,i \\ 1+3 i & -8.08205 +1.46086 \,i & -8.08159 +1.46231 \,i \\ 2+i & +12.3359 +18.9310 \,i & +12.3353 +18.9310\,i \\ 2+2i & -9.04768 +20.5009 \,i & -9.04680 +20.4992 \,i \\ 2+3i & -21.8087 +3.57177 \,i & -21.8039 +3.57352 \,i \\ 3+i & +33.1122 +51.4419 \,i & +33.1083 +51.4458 \,i \\ 3+2i & -25.1135 +55.2172 \,i & -25.1179 +55.2074 \,i \\ 3+3i & -59.1718 +8.66441 \,i & -59.1523 +8.65622 \,i \end{array} \right)$$

My questions :

• do the $\, _2F_1\left(\frac{1}{2},-k;\frac{3}{2};z\right)$ correspond to known polynomials ?
• could it be possible to develop a recurrence rekation for $\, _2F_1\left(\frac{1}{2},-k;\frac{3}{2};z\right)$
• could we find better approximations ?
• could we obtain decent and rather detailed asymptotic formulae for $Ik$ and/or $J(k)$ ?

Answer 1

I am going to derive the asymptotics for large $k$. Your integral may be re-written in the form $$ I(k) = 2\pi ^k \int_0^{\pi /2} {\exp \left( { - k\log \left[ {\frac{{\pi \sin x}}{{x(\pi - x)}}} \right]} \right)\mathrm{d}x} . $$ Performing the change of integration variables $t = \log \left[ {\frac{{\pi \sin x}}{{x(\pi - x)}}} \right]$ yields $$ I(k) = 2\pi ^k \int_0^{\log (4/\pi )} {\mathrm{e}^{ - kt} \frac{{\mathrm{d}x}}{{\mathrm{d}t}}\mathrm{d}t} . $$ By a standard series reversion argument, we have $$ \frac{{\mathrm{d}x}}{{\mathrm{d}t}} = \pi \left( {1 + \frac{{\pi ^2 - 3}}{3}t + \frac{{\pi ^4 - 6\pi ^2 + 3}}{6}t^2 + \frac{{25\pi ^6 - 219\pi ^4 + 375\pi ^2 - 45}}{{270}}t^3 + \cdots } \right) $$ in a small neighbourhood of the origin. Therefore, by Watson's lemma, $$ I(k) \sim \frac{{2\pi ^{k + 1} }}{k}\left( {1 + \frac{{\pi ^2 - 3}}{{3k}} + \frac{{\pi ^4 - 6\pi ^2 + 3}}{{3k^2 }} + \frac{{25\pi ^6 - 219\pi ^4 + 375\pi ^2 - 45}}{{45k^3 }} + \cdots } \right) $$ as $k\to +\infty$.

Answer 2

Here is the exact solution of the closed form for $I(k)$ at least for $k=2m+1$ odd. (It is quite similar for even $k=2m$, with the polynomials $\prod_{j=-m}^m(z-j)$ instead, similar situation as here.) As in my comment, define $z_j:= (2^j-1) \zeta(j)$ and take the polynomial of degree $2m$ $$p(z)=(4z^2-1^2)(4z^2-3^2)\cdots(4z^2-k^2)=\prod_{j=1-m}^m(2z+2j-1).$$ Then in umbral notation (i.e. replace $z^j$ by $z_j$), put for $n=3,5, ..., k+2$ $$p_n:= z^{2m+n}p(\frac1z).$$ (In my comment, I had forgotten to include a factor $n!!$ in the definition of $p_n$, which makes the numerators in the final explicit expression substantially smaller $-$ see the correcting comment $-$, but here I didn't include that factor, as for the general form it doesn't change anything).

Now it turns out that $I(k)$ (and BTW also $\int_{0}^{\pi} x^r (\pi-x)^s \csc^k x\, dx$ for all $r,s\ge k$) is a linear combination of these $p_n$'s, with the appropriate powers of $\pi$. In fact (using $m=\frac{k-1}2$ for convenience), $$\int_0^{ \pi }\frac{x^k(\pi-x)^k}{\sin^k x} dx= \sum_{j=0}^m \frac{(-1)^{m+j}\cdot 2^{2j}}{{ (k-1)!\cdot 2^{2k-1}}}(2k-2j)!\binom{ k} {2j}\pi^{2j}p_{k+2-2j}. $$ As the signs alternate, I have no idea whether this expression is useful to derive the asymptotics. Definitely the other answer does that in a much more elegant way. (But I like precise results, too.)

Answer 3

Answering partly my own question

The above can be improved approximating $$x(\pi -x) \csc (x) \sim a+ b \left(x-\frac{\pi }{2}\right)^2+c \left(x-\frac{\pi }{2}\right)^4=g(x)$$

Optimization of the norm $$\Phi(a,b,c)=\int_0^\pi \Big[x(\pi-x)\csc (x)-g(x) \big] \Big]^2\,dx$$

leads to $$a=\frac{105 \left(29 \pi ^4 \zeta (3)-5394 \pi ^2 \zeta (5)+51435 \zeta (7)\right)}{8 \pi ^5}$$ $$b=-\frac{105 \left(175 \pi ^4 \zeta (3)-27342 \pi ^2 \zeta (5)+257175 \zeta (7)\right)}{\pi ^7}$$ $$c=\frac{1890 \left(49 \pi ^4 \zeta (3)-7130 \pi ^2 \zeta (5)+66675 \zeta (7)\right)}{\pi ^9}$$ which give $\Phi=2.080\times 10^{-7}$ (that is to say $915$ times lower than before).

$$\color{blue}{J_k(x)=\int \big[g(x)\big]^k\,dx}$$ $$\color{blue}{J_k(x)=\pi\, a^k\,F_1\left(\frac{1}{2};-k,-k;\frac{3}{2};-\frac{\pi ^2 c}{2 \left(\sqrt{b^2-4 a c}+b\right)},\frac{\pi ^2 c}{2 \left(\sqrt{b^2-4 a c}-b\right)}\right)}$$ where appears the Appell hypergeometric function of two variables.

Computing for a few integer values of $k$

$$\left( \begin{array}{ccc} k & J(k) & I(k) \\ 1 & 8.41440 & 8.41440 \\ 2 & 22.6582 & 22.6582 \\ 3 & 61.3546 & 61.3546 \\ 4 & 167.093 & 167.093 \\ 5 & 457.737 & 457.738 \\ 6 & 1261.41 & 1261.41 \\ 7 & 3496.98 & 3497.00 \\ 8 & 9752.63 & 9752.77 \\ 9 & 27360.0 & 27360.7 \\ 10 & 77202.4 & 77205.2 \end{array} \right)$$ which is significantly better.

Update

Using series around $x=\frac \pi 2$, we have,as approximations of $x(\pi -x) \csc (x)$

$$f_1(x)=\frac{\pi ^2}{4}+\left(\frac{\pi ^2}{8}-1\right) \left(x-\frac{\pi }{2}\right)^2+\left(\frac{5 \pi ^2}{96}-\frac{1}{2}\right) \left(x-\frac{\pi }{2}\right)^4+O\left(\left(x-\frac{\pi }{2}\right)^6\right)$$ while the simplest Padé approximant is $$f_2(x)=\frac{\frac{\pi ^2}{4}+\frac{384-48 \pi ^2+\pi ^4}{48 \left(\pi ^2-8\right)}\left(x-\frac{\pi }{2}\right)^2 } {1 +\frac{48-5 \pi ^2}{12 \left(\pi ^2-8\right)}\left(x-\frac{\pi }{2}\right)^2}$$ $$\Phi_1=\int_0^\pi \Big[x(\pi-x)\csc (x)-f_1(x) \big] \Big]^2\,dx=3.40\times 10^{-5}$$ $$\Phi_2=\int_0^\pi \Big[x(\pi-x)\csc (x)-f_2(x) \big] \Big]^2\,dx=1.74\times 10^{-6}$$ that is to say $\Phi_2 \sim \frac 1 {20} \Phi_1$.

So, more than likely a form like $$x(\pi -x) \csc (x) \sim \frac {a+b\left(x-\frac{\pi }{2}\right)^2} {1+c\left(x-\frac{\pi }{2}\right)^2 }$$ would be better.

The problem is that I am not able to compute $$\int_0^\pi \Bigg[x(\pi -x) \csc (x)-\frac {a+b\left(x-\frac{\pi }{2}\right)^2} {1+c\left(x-\frac{\pi }{2}\right)^2 }\Bigg]^2\,dx$$ which makes, at least for the time being, impossible to know the optimum $(a,b,c)$.

This would give $$\int_0^\pi \Bigg[\frac {a+b\left(x-\frac{\pi }{2}\right)^2} {1+c\left(x-\frac{\pi }{2}\right)^2 }\Bigg]^k\,dx=\pi\, a^k\,\,F_1\left(\frac{1}{2};-k,k;\frac{3}{2};-\frac{\pi ^2 b}{4 a},-\frac{\pi ^2 c}{4}\right)$$ more pleasant than the previous one since it only contains real arguments.

For lack of anything better, using $f_2$ the first values are $(8.41557,22.6652,61.3880)$ while, using $f_1$, they are $(8.40913,22.6269,61.2144)$

Integrated numerically, the minimum values of $\Phi_1$ and $\Phi_2$ are respectively $2.080\times 10^{-7}$ and $6.893\times 10^{-9}$ that is to say $\Phi_2 \sim \frac 1 {30} \Phi_1$.

Answer 4

Here we derive an alternative exact formula for

$$I(k)=\int_{0}^{\pi} \left(\frac{x(\pi-x)}{\sin(x)} \right)^k\;dx \tag{1}$$

valid for any positive integer $k$, as a finite sum, similar to that of @Wolfgang.

Starting from https://math.stackexchange.com/a/4437901/198592 we find that we can write

$$I(k) = 2 \Im \int_{0}^{\infty}\frac{t^k (\pi+i t)^k}{\text{sinh}(t)^k}\;dt \tag{2}$$

Expanding $\frac{1}{\text{sinh}(t)^k}$ in a power series in terms of $e^{-t}$ and expanding the imaginary part in the numerator (giving rise to the selecting factor $\sin(m \pi/2)$) we get:

$$I(k) = 2^{k+1} \sum_{m=0}^{k} \sin(m \pi/2) \pi^{k-m}\binom{k}{m}(k+m)! g(k,m)\tag{3}$$

Here

$$g(k,m) = \sum_{n\ge 0}\binom{n+k-1}{k-1} \frac{1}{(k+2n)^{1+k+m}}\tag{4}$$

In this function a $\zeta$-function structure appears.

The binomial coefficient in $g(k,m)$ can be conveniently generated using an auxiliary variable $z$ so that we can do the $n$-sum and write

$$\begin{align} g(k,m) &= \frac{1}{(k-1)!}(\frac{d}{dz})^{k-1} \sum_{n\ge 0}z^{n+k-1} \frac{1}{(k+2n)^{1+k+m}}|_{z\to 1}\\ &=\frac{1}{2^{k+m+1}}\frac{1}{(k-1)!}(\frac{d}{dz})^{k-1} ( z^{k-1}\Phi(z,1+k+m,\frac{k}{2})|_{z\to 1} \end{align}\tag{5}$$

Here we have $\Phi(z,s,a)=\sum_{k=0}^{\infty} z^k (k+a)^{-s}$ is the Hurwitz-Lerch transcendent (https://mathworld.wolfram.com/LerchTranscendent.html).

Noicing that

$$\frac{d}{dz}\Phi(z, a, s) = \frac{1}{z}\left(\Phi(z, -1 + a, s) - s \Phi(z, a, s) \right)$$

we get for the first 4 values of $k$

$$\begin{align} &{g(1,m)= \Phi(1, 2 + m, 1/2)}\\ &{g(2,m)= \text{Li}(2 + m, 1)}\\ &{g(3,m) = 1/8 (4 \Phi(1, 2 + m, 3/2) - \Phi(1, 4 + m, 3/2))}\\ &{g(4,m) = 1/6 (\Phi(1, 2 + m, 2) - \Phi(1, 4 + m, 2))} \end{align} $$

Inserting these expressions into $(3)$ and observing that $\Phi$ reduces to a linear combination of $\zeta$-functions for odd $m$ we get for the 4 first values of $k$

$$\begin{align} &I(1)= 7 \zeta(3)\\ &I(2)=6 \pi \zeta(3)\\ &I(3) = 126 \pi^2 \zeta(3) - 1395 \zeta(5) - \frac{279}{2} \pi^2 \zeta(5) + \frac{5715}{4} \zeta(7)\\ &I(4) = 40 \pi^3 (\zeta(3) - \zeta(5)) - 420 \pi (\zeta(5) - \zeta(7)) \end{align}$$