# $A_{mn} = \frac{1}{\pi}\int_0^{\pi}d\theta\, \sin(2m\theta)\, \frac{1-\cos^{2n}(\theta)}{\tan(\theta)} =$ ? $m$, $n$ integers > 0

The integral $$A_{mn} = \frac{1}{\pi}\int_0^{\pi}d\theta\, \sin(2m\theta)\, \frac{1-\cos^{2n}(\theta)}{\tan(\theta)}$$ popped up when I was playing around with the integral representation of the harmonic numbers $H_n$.

I find that Mathematica can give a result for specific values of $m$ and $n$ but I can't get a general formula. Some observations that appear to be true, based on Mathematica results for specific cases:

1) $A_{mn} > 0$.

2) When $m = n$, $A_{nn} = \frac{1}{4^n}$.

3) When $m > n$, $A_{mn} = 0$. (I think this should follow from expanding the $\cos^{2n}(\theta)$ part out in a finite sum of terms like $\sin$ or $\cos$ of integers times $\theta$.)

I tried doing this via contour integration by extending the integral (which is symmetric around $\pi$) to $[0,2\pi)$, and making the substitution $z = e^{i\theta}$. Unless I've made a mistake, this leads to: $$A_{mn} = \frac{1}{2\pi i}\oint dz\, \frac{\left(z^{2m} - z^{-2m}\right)\left(z+z^{-1}\right)\left[1 - \frac{1}{2^{2n}}{\left(z + z^{-1}\right)}^{2n}\right]}{z^2 - 1}\, ,$$ where the integral is over the counter-clockwise unit circle. This doesn't seem to work: The zeroes appear to cancel on the top and bottom at $z = \pm 1$, so the residues are zero.

Have I made a mistake in the contour integration approach? Is there a better way to evaluate this whole thing?

• The question started with "this integral" but didn't not say what the integral was or what $A_{mn}$ was. Don't assume that someone reading the post has read the title; the post itself should include all the necessary information. – Carl Mummert Jul 12 '16 at 12:04
• @CarlMummert OK, thanks. – John Barber Jul 12 '16 at 16:52

Using the identities $$\frac{\sin (2m \theta)}{\sin (\theta)} = 2 \sum_{k=0}^{m-1} \cos[(2k+1)\theta]$$

and

\begin{align} A_{mn} &= \frac{1}{\pi} \int_{0}^{\pi} \frac{\sin (2 m \theta)}{\sin (\theta)}\left(\cos (\theta) - \cos^{2n+1}(\theta) \right) \, d \theta \\ &= \frac{2}{\pi} \sum_{k=0}^{m-1} \int_{0}^{\pi} \cos[(2k+1)\theta] \cos(\theta) \, d \theta \\ &- \frac{2}{\pi} \, \frac{1}{4^{n}}\sum_{k=0}^{m-1} \sum_{j=0}^{n} \binom{2n+1}{j}\int_{0}^{\pi} \cos[(2k+1)\theta ] \cos[(2n+1-2j) \theta] \, d \theta. \end{align}

First assume that $m <n$.

Then using the fact that

$$\int_{0}^{\pi} \cos(mx) \cos(nx) \, dx = \begin{cases} \frac{\pi}{2} & m = n \\ 0 & \text{otherwise} \end{cases}$$ we get

$$A_{mn} = \frac{2}{\pi} \left(\frac{\pi}{2} \right) - \frac{2}{\pi} \, \frac{1}{4^{n}} \sum_{j=n-m+1}^{n} \binom{2n+1}{j} \frac{\pi}{2} = 1 - \frac{1}{4^{n}} \sum_{j=n-m+1}^{n} \binom{2n+1}{j}.$$

Now if $m> n$, then

$$A_{mn} = \frac{2}{\pi} \left(\frac{\pi}{2} \right)- \frac{2}{\pi} \, \frac{1}{4^{n}} \sum_{j=0}^{n} \binom{2n+1}{j} \frac{\pi}{2} = 1 - \frac{1}{4^{n}} \left(4^{n}\right) = 0. \tag{1}$$

And if $m=n$, $$A_{mn} = 1- \frac{1}{4^{n}} \sum_{j=1}^{n} \binom{2n+1}{j} = 1 - \frac{1}{4^{n}} \left(4^{n}-1\right) = \frac{1}{4^{n}}.$$

• @joriki I added an extra line to make it more clear. – Random Variable Jul 12 '16 at 7:10


The Question: $\ds{A_{mn} \equiv {1 \over \pi}\int_{0}^{\pi}\sin\pars{2m\theta}\, {1 - \cos^{2n\,}\pars{\theta} \over \tan\pars{\theta}}\,\dd\theta\,,\qquad m,n \in \braces{1,2,3,\ldots}}$

\begin{align} \color{#f00}{A_{mn}} & = {1 \over 2\pi}\,\Im\int_{-\pi}^{\pi}\expo{2m\ic\theta}\, {1 - \cos^{2n\,}\pars{\theta} \over \tan\pars{\theta}}\,\dd\theta \\[3mm] & = {1 \over 2\pi}\,\Im\oint_{\verts{z} = 1}z^{2m}\, {1 \over -\ic\pars{z^{2} - 1}/\pars{z^{2} + 1}}\, \bracks{1 - \pars{z^{2} + 1 \over 2z}^{2n}}\,{\dd z \over \ic z} \\[3mm] & = {1 \over 2\pi}\,{1 \over 2^{2n}}\Im\oint_{\verts{z} = 1}z^{2m - 2n - 1}\,\,\, {z^{2} + 1 \over z^{2} - 1}\, \bracks{2^{2n}z^{2n} - \pars{z^{2} + 1}^{2n}}\,\dd z\tag{1} \end{align}

Indeed, as the OP already claimed there isn't any poles at $\ds{z = \pm 1}$. So, we concentrate our efforts with poles $\ds{\pars{~\mbox{it depends on the particular values of}\ m\ \mbox{and}\ n ~}}$ at $\ds{z = 0}$. It amounts to rewrite $\ds{\pars{1}}$ in a convenient way. Namely, \begin{align} \color{#f00}{A_{mn}} & = {1 \over 2\pi}\,{1 \over 2^{2n}}\Im\oint_{\verts{z} = 1^{-}} \pars{z^{2m + 1}\ +\ z^{2m - 1}\ } \sum_{\ell = 0}^{\infty}z^{2\ell}\ \bracks{\sum_{\ell' = 0}^{2n}{2n \choose \ell'}z^{2\pars{\ell' - n}} - 2^{2n}} \,\dd z \\[8mm] & = {1 \over 2\pi}\,{1 \over 2^{2n}}\Im\bracks{% \sum_{\ell = 0}^{\infty}\,\,\sum_{\ell' = 0}^{2n}{2n \choose \ell'} \oint_{\verts{z} = 1^{-}}\,\frac{\dd z}{z^{2n - 2\ell' - 2\ell - 2m - 1}}\,\,} \\[3mm] & + {1 \over 2\pi}\,{1 \over 2^{2n}}\Im\bracks{% \sum_{\ell = 0}^{\infty}\,\,\sum_{\ell' = 0}^{2n}{2n \choose \ell'} \oint_{\verts{z} = 1^{-}}\,\frac{\dd z}{z^{2n - 2\ell' - 2\ell - 2m + 1}}\,\,} \\[3mm] & - {1 \over 2\pi}\,{1 \over 2^{2n}}\Im\bracks{% \sum_{\ell = 0}^{\infty}\underbrace{% \oint_{\verts{z} = 1^{-}}\,\frac{\dd z}{z^{-2n - 2\ell - 2m - 1}}\,\,} _{\ds{= 0}}} - {1 \over 2\pi}\,{1 \over 2^{2n}}\Im\bracks{% \sum_{\ell = 0}^{\infty}\underbrace{% \oint_{\verts{z} = 1^{-}}\,\frac{\dd z}{z^{-2n - 2\ell - 2m + 1}}\,\,} _{\ds{=\ 0}}} \\[8mm] & = {1 \over 2^{2n}}\sum_{\ell' = 0}^{2n}{2n \choose \ell'}\sum_{\ell = 0}^{\infty} \delta_{\ell,n - \ell' - m - 1}\,\,\,\, + {1 \over 2^{2n}}\sum_{\ell' = 0}^{2n}{2n \choose \ell'}\sum_{\ell = 0}^{\infty} \delta_{\ell,n - \ell' - m} \\[3mm] & = \left.{1 \over 2^{2n}}\sum_{\ell' = 0}^{2n}{2n \choose \ell'} \right\vert_{\ \ell'\ \leq\ n - m - 1} + \left.{1 \over 2^{2n}}\sum_{\ell' = 0}^{2n}{2n \choose \ell'} \right\vert_{\ \ell'\ \leq\ n - m} \\[3mm] & = \color{#f00}{\left\lbrace\begin{array}{ccrcl} \ds{{1 \over 4^{n}}\bracks{-{2n \choose n - m} + 2\sum_{\ell' = 0}^{n - m}{2n \choose \ell'}}} & \mbox{if} & \ds{m} & \ds{<} & \ds{n} \\[2mm] \ds{1 \over 4^{n}} & \mbox{if} & \ds{m} & \ds{=} & \ds{n} \\[2mm] \ds{0} & \mbox{if} & \ds{m} & \ds{>} & \ds{n} \end{array}\right.} \end{align}