When answering this question I came across the integrals

$$ I_n \equiv \int \limits_0^\infty [1-x^n \operatorname{arccot}^n(x)] \, \mathrm{d} x \, , ~n \in \mathbb{N} \, . $$

I needed $I_1 = \frac{\pi}{4}$ , $I_2 = \frac{\pi}{6}[2 \ln(2)+1]$ and $I_3 = \frac{\pi}{32}[32 \ln(2)+ 4 -\pi^2]$ in my answers. They can be evaluated by writing $$ I_n = \lim_{r \to \infty} \left[r - \int \limits_0^r x^n \operatorname{arccot}^n (x) \, \mathrm{d} x \right] $$ and using repeated integration by parts to reduce the remaining integral to a few terms that cancel the $r$ in the limit and some well-known integrals.

This calculation should work for any $n \in \mathbb{N}$, but it gets more tedious for larger values of course. Mathematica gives reasonably nice expressions for the integrals in terms of $\pi$ , $\ln(2)$ and values of the zeta function (for example $I_4 = \frac{\pi}{40} [4 + 80 \ln(2) - \pi^2 (3 + 4 \ln(2)) + 18 \zeta(3)]$), so it might be possible to evaluate the integrals in terms of suitable special functions in general.

However, I have not yet found a way to transform them into such an expression. The obvious substitution $x = \cot (t)$ leads to $$ I_n = \int \limits_0^{\pi/2} \frac{\sin^n (t) - t^n \cos^n (t)}{\sin^{n+2}(t)} \, \mathrm{d} t \, ,$$ which does not seem to help much. We can use $$ 1 - a^n = (1-a) \sum \limits_{k=0}^{n-1} a^k = \sum \limits_{k=0}^{n-1} \sum \limits_{l=0}^k (-1)^l {k \choose l} (1-a)^{l+1} $$ for $n \in \mathbb{N}$ and $a \in \mathbb{R}$ to obtain $$ I_n = \sum \limits_{k=0}^{n-1} \sum \limits_{l=0}^k (-1)^l {k \choose l} J_{l+1} $$ in terms of $$ J_n \equiv \int \limits_0^\infty [1-x \operatorname{arccot}(x)]^n \, \mathrm{d} x \, , ~n \in \mathbb{N} \, . $$ Interchanging the sums, using $\sum_{k=l}^{n-1} {k \choose l} = {n \choose l+1}$ (which apparently is known as the hockey-stick identity) and defining $I_0 = J_0 = 0$ , we find that the two sequences are binomial transforms of each other (except for a minus sign): $$I_n = - \sum \limits_{m=0}^n (-1)^m {n \choose m} J_m \, , \, n \in \mathbb{N}_0 \, . $$ I do not know which of the two families of integrals is easier to evaluate though. Note that the same method enables us to compute $$ \int \limits_0^\infty [1-x \operatorname{arccot}(x)] P[x \operatorname{arccot}(x)] \, \mathrm{d} x $$ for any polynomial $P$ once we know $(I_n)_{n \in \mathbb{N}}$ or $(J_n)_{n \in \mathbb{N}}$.

I am also interested in the asymptotic behaviour of the integrals. Numerical calculations and plots suggest that we have \begin{align} I_n &\sim \sqrt{\frac{\pi n}{3}} \, , \, n \to \infty \, , \\ J_n &\sim \frac{2}{\pi n} \, , \, n \to \infty \, , \end{align} but I have no idea how to prove that.

Therefore I am left with the following two questions:

  1. How can we find a closed-form expression for $I_n$ or $J_n$ , $n \in \mathbb{N}$ ?
  2. What can we say about the asymptotics of $I_n$ or $J_n$ as $n \to \infty$ ?

Any hints or (partial) solutions to either of them would be greatly appreciated.

  • 1
    $\begingroup$ $$I_n=\int_0^{\frac{\pi }{2}} \csc ^2(y) \left(1-y^n \cot ^n(y)\right) \, dy$$ $$I_n=-\int_0^{\frac{\pi }{2}} y^n \cot ^n(y) \, dy+\int_0^{\frac{\pi }{2}} \cot ^2(y) \left(1-y^n \cot ^n(y)\right) \, dy+\frac{\pi }{2}$$ $\endgroup$ – James Arathoon Jul 29 '18 at 13:54
  • 1
    $\begingroup$ $$I_n=n \int_0^{\frac{\pi }{2}} y^n \cot ^n(y) \left(\csc ^2(y)-\frac{\cot (y)}{y}\right) \, dy$$ $\endgroup$ – James Arathoon Jul 29 '18 at 16:32
  • $\begingroup$ @JamesArathoon Thank you! I like the last form, in which the 'problematic' factor with the cancelling poles does not contain $n$-th powers anymore. That definitely makes an approach based on Laurent series seem less intimidating. $\endgroup$ – ComplexYetTrivial Jul 29 '18 at 20:05

Cool problem. Both proposed asymptotic forms are correct. I'm only going to do a first order calc for both, and for $J_n$ I'll say where the proof needs some work. For $I_n$ use the expression given by Arathoon's first note. Observe that $t\cot(t)$ is unimodal decreasing and at $t=0$ has the value 1. Thus by raising it to a high power we expect it to become more sharply peaked. In fact it is gaussian because $$ t\cot(t) = 1-t^2/3-t^4/45... \sim \exp(-t^2/3)(1-7/90t^4 + ...) $$ Therefore $$I_n = \int_0^{\pi/2} \frac{dt}{\sin^2t} \big(1-(t\cot(t))^n \big) \sim \int_0^{\pi/2} \frac{dt}{\sin^2t} \big(1-\exp{(-n\,t^2/3)} \big).$$ Because $n \to \infty$ and because of the $\csc^2t,$ it is seen that the most important region is near $t=0.$ The trick now is to recognize that $\tan^2t = t^2 + O(t^4).$ Replace $t^2$ in the exponential with $\tan^2t$ because it just so happens (Mathematica knows it too) that $$ \int_0^{\pi/2} \frac{dt}{\sin^2t} \big(1-\exp{(-a\,\tan^2t)} \big)=\sqrt{a \pi}.$$ With $a=n/3,$ we indeed have $I_n \sim \sqrt{n \pi /3}.$

For $J_n$ I used a technique call $\textit{depoissonization}.$ Make an exponential power series, $$\sum_{n=0}^\infty \frac{y^n}{n!}J_n = \sum_{n=0}^\infty\frac{y^n}{n!} \int_0^{\infty}(1-t\cot^{-1}(t))^n\, dt = e^y\int_0^{\infty}\exp{(-y\,t\,\cot^{-1}{t} )} \, dt. $$ Since $t\cot^{-1}t$ is monotonically increasing, all we need is the behavior near $0$ to asymptotically estimate the integral. In fact, $t\cot^{-1}t = \pi\,t/2 +O(t^2).$ Using the first term we therefore find $$J(y) := e^{-y}\sum_{n=0}^\infty \frac{y^n}{n!}J_n \sim \int_0^{\infty} \exp{(-\pi\,y\,t/2)} \, dt = \frac{2}{\pi\,y}.$$ By depoissonization, $J_n \sim J(n) =2/(\pi n)$ as long as the next term in the sequence is smaller than this first term. I don't intend to show that. One thing that makes this problem interesting is that in fact it can be shown that $I_n = \sum_{k=1}^n(-1)^{k+1}\binom{n}{k}J_k$ so that $I_n$ and $J_n$ are binomial transforms. Putting in the first order asymptotic for for either $I_n$ or $J_n$ will $\textit{not}$ give you the asymptotics of its transform.

  • $\begingroup$ Great answer with very interesting methods! I also added the observation on the binomial transform, which I had overlooked when posting the question. Thanks a lot! $\endgroup$ – ComplexYetTrivial Jul 30 '18 at 7:54

The following partial answer gives the integrals $(I_n)_{n\in\mathbb{N}}$ in terms of $$K_n^{(m)} \equiv \int \limits_0^{\pi/2} t^n \cot^m (t) \, \mathrm{d} t \, , \, n \in \mathbb{N}_0 \, , \, 0 \leq m \leq n \, .$$ A closed-form expression for these integrals is discussed in this question, but in principle they can be evaluated in terms of $$K_n^{(0)} = \int \limits_0^{\pi/2} t^n \, \mathrm{d} t = \frac{1}{n+1} \left(\frac{\pi}{2}\right)^{n+1} \, , \, n \in \mathbb{N}_0 $$ and \begin{align} K_n^{(1)} &= \int \limits_0^{\pi/2} t^n \cot (t) \, \mathrm{d} t = -n \int \limits_0^{\pi/2} t^{n-1} \ln(\sin(t)) \, \mathrm{d} t \\ &= \frac{n!}{2^n} \left[\sum \limits_{l=0}^{\left \lfloor \frac{n-1}{2} \right \rfloor} (-1)^l \frac{\pi^{n-2l}}{(n-2l)!} \eta (2l+1) + \operatorname{1}_{2 \mathbb{N}} (n) (-1)^{\left \lfloor \frac{n+1}{2} \right \rfloor} [\zeta (n+1) + \eta (n+1)]\right] \end{align} using the recurrence relation $$K_n^{(m)} = \frac{n}{m-1} K_{n-1}^{(m-1)} - K_n^{(m-2)} \, , \, n \geq m \geq 2 \, .$$

We start from the representation $$I_n = n \int \limits_0^{\pi/2} t^n \cot^n (t) \left[\csc^2 (t) - \frac{\cot(t)}{t}\right] ^\, \mathrm{d} t \, , \, n \in \mathbb{N} \, , $$ given by James Arathoon in the comments. Using $\csc^2(t) = 1+\cot^2(t)$ and integrating by parts we find \begin{align} I_n &= n K_n^{(n)} + n \int \limits_0^{\pi/2} t^{n-1} \left[t \cot^{n+2}(t) -\cot^{n+1}(t)\right] \, \mathrm{d} t \\ &= n K_n^{(n)} + \int \limits_0^{\pi/2} \left[(n+2) t^{n+1} \cot^{n+1} (t) \csc^2(t) - t^n \cot^{n+2} (t) - (n+1) t^n \cot^n (t) \csc^2(t)\right] \, \mathrm{d} t \\ &= - K_n^{(n)} + (n+2) \left[K_{n+1}^{(n+1)} + \int \limits_0^{\pi/2} t^n \left[t \cot^{n+3} (t) - \cot^{n+2} (t)\right] \, \mathrm{d} t \right] \\ &= - K_n^{(n)} + \frac{n+2}{n+1} I_{n+1} \, . \end{align} This yields the recurrence relation $$ I_{n+1} = \frac{n+1}{n+2} (I_n + K_n^{(n)}) \, , \, n \in \mathbb{N}_0 \, ,$$ with the initial condition $I_0 = 0$ , which has the solution $$ I_n = \frac{1}{n+1} \sum \limits_{l=0}^{n-1} (l+1) K_l^{(l)}$$ for $n \in \mathbb{N}$ .


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.