# Evaluating $\sum_{k=1}^{\infty}\frac{\sin\left(k\theta\right)}{k^{2u+1}}$ with multiple integrals

I am trying to evaluate $$\sum_{k=1}^{\infty}\frac{\sin\left(k\theta\right)}{k^{2u+1}}\tag{u\in\mathbb{N}}$$ using some results I've got. I know that $$\sum_{k=1}^{\infty}\frac{\sin\left(k\theta_0\right)}{k}=\tan^{-1}\left(\cot\left(\theta_0/2\right)\right)$$ which equals $\displaystyle\frac{1}{2}\left(\pi-\theta_0\right)$ whenever $0\leq\theta_0<2\pi.$ I realized that if I integrate both sides from $0$ to $\theta_1$, I get $$\sum_{k=1}^{\infty}\frac{\cos\left(k\theta_1\right)-1}{k^{2}}=-\frac{\pi\theta_1}{2}+\frac{\theta_1^{2}}{4}\\\implies\sum_{k=1}^{\infty}\frac{\cos\left(k\theta_1\right)}{k^{2}}=\zeta\left(2\right)-\frac{\pi\theta_1}{2}+\frac{\theta_1^{2}}{4}$$ and integrating one more time from $0$ to $\theta_2$ show me that $$\sum_{k=1}^{\infty}\frac{\sin\left(k\theta_2\right)}{k^{3}}=\frac{\theta_2^{3}}{12}-\frac{\pi\theta_2^{2}}{4}+\zeta\left(2\right)\theta_2.$$ I considered integrating the first equation a $n$ number of times, so I could write $$\sum_{k=1}^{\infty}\left(\int_{0}^{\theta_{n}}\cdots\left(\int_{0}^{\theta_{2}}\left(\int_{0}^{\theta_{1}}\frac{\sin\left(k\theta_{0}\right)}{k}\,\mathrm{d}\theta_{0}\right)\,\mathrm{d}\theta_{1}\right)\cdots\,\mathrm{d}\theta_{n-1}\right)\\=\frac{1}{2}\left(\int_{0}^{\theta_{n}}\cdots\left(\int_{0}^{\theta_{2}}\left(\int_{0}^{\theta_{1}}\pi-\theta_{0}\,\mathrm{d}\theta_{0}\right)\,\mathrm{d}\theta_{1}\right)\cdots\,\mathrm{d}\theta_{n-1}\right)\\=\frac{1}{2}\left(\frac{\pi\theta_{n}^{n}}{n!}-\frac{\theta_{n}^{n+1}}{\left(n+1\right)!}\right)$$ using the well-established zeta function of a even number and somehow extracting the trigonometric sum apart from that mess. Looking at some attempts like $$\int_0^{\theta_1}\frac{\sin(k\theta_0)}{k}\,\mathrm d\theta_0=\frac{-\cos(k\theta_1)}{k^2}+\frac{1}{0!k^2}$$ $$\int_0^{\theta_2}\frac{-\cos(k\theta_1)}{k^2}+\frac{1}{k^2}\,\mathrm d\theta_1=\frac{-\sin(k\theta_2)}{k^3}+\frac{\theta_2}{1!k^2}$$ $$\int_0^{\theta_3}\frac{-\sin(k\theta_2)}{k^3}+\frac{\theta_2}{1!k^2}\,\mathrm d\theta_2=\frac{\cos(k\theta_3)}{k^4}+\frac{\theta_3^2}{2!k^2}-\frac{1}{0!k^4}$$ $$\int_0^{\theta_4}\frac{\cos(k\theta_3)}{k^4}+\frac{\theta_3^2}{2!k^2}-\frac{1}{0!k^4}\,\mathrm d\theta_3=\frac{\sin(k\theta_4)}{k^5}+\frac{\theta_4^3}{3!k^2}-\frac{\theta_4}{1!k^4}$$ $$\int_0^{\theta_5}\frac{\sin(k\theta_4)}{k^5}+\frac{\theta_4^3}{3!k^2}-\frac{\theta_4}{1!k^4}\,\mathrm d\theta=\frac{-\cos(k\theta_5)}{k^6}+\frac{\theta_5^4}{4!k^2}-\frac{\theta_5^2}{2!k^4}+\frac{1}{0!k^6}$$ show some pattern. Which pattern? I conjectured that $$\sum_{k=1}^{\infty}\left(\int_{0}^{\theta_{n}}\cdots\left(\int_{0}^{\theta_{2}}\left(\int_{0}^{\theta_{1}}\frac{\sin\left(k\theta_{0}\right)}{k}\,\mathrm{d}\theta_{0}\right)\,\mathrm{d}\theta_{1}\right)\cdots\,\mathrm{d}\theta_{n-1}\right)\\=\left(-1\right)^{n}\sum_{k=1}^{\infty}\frac{\sin\left(n\pi/2+k\theta_{n}\right)}{k^{n+1}}+\sum_{i=1}^{\lfloor\frac{n+1}{2}\rfloor}\frac{\left(-1\right)^{i+1}\theta^{n+1-2i}_{n}\zeta\left(2i\right)}{(n+1-2i)!}$$ which seems quite plausible (otherwise, it wouldn't be a conjecture). This implies $$\sum_{k=1}^{\infty}\frac{\sin\left(n\pi/2+k\theta_{n}\right)}{k^{n+1}}=\frac{\left(-1\right)^{n}}{2}\left(\frac{\pi\theta_{n}^{n}}{n!}-\frac{\theta_{n}^{n+1}}{\left(n+1\right)!}\right)+\sum_{i=1}^{\lfloor\frac{n+1}{2}\rfloor}\frac{\left(-1\right)^{n+i}\theta_{n}^{n+1-2i}\zeta\left(2i\right)}{(n+1-2i)!}$$ and, because I am interested in the case $n=2u$, this turns out to be $$\sum_{k=1}^{\infty}\frac{\sin\left(u\pi+k\theta_{n}\right)}{k^{2u+1}}=\left(-1\right)^{u}\sum_{k=1}^{\infty}\frac{\sin\left(k\theta_{}\right)}{k^{2u+1}}\\=\frac{1}{2}\left(\frac{\pi\theta_{2u}^{2u}}{(2u+1)!}-\frac{\theta_{2u}^{2u+1}}{\left(2u+1\right)!}\right)+\sum_{i=1}^{\lfloor\frac{2u+1}{2}\rfloor}\frac{\left(-1\right)^{i}\theta^{2u+1-2i}\zeta\left(2i\right)}{(2u+1-2i)!}$$ and, therefore $$\sum_{k=1}^{\infty}\frac{\sin\left(k\theta_{}\right)}{k^{2u+1}}=\frac{\left(-1\right)^u}{2}\left(\frac{\pi\theta^{2u}}{(2u)!}-\frac{\theta^{2u+1}}{\left(2u+1\right)!}\right)+\sum_{i=1}^{\lfloor\frac{2u+1}{2}\rfloor}\frac{\left(-1\right)^{u+i}\theta^{2u+1-2i}\zeta\left(2i\right)}{(2u+1-2i)!}.$$ Now, some questions: how do I prove whether my conjecture is right? Is there any cool tricks I can use when dealing with multiple integrations like these? Is there another approach with this series?

EDIT

Maybe induction can help me. My own attempt: Assume $$\sum_{k=1}^{\infty}\frac{\sin\left(k\theta_{}\right)}{k^{2u+1}}=\frac{\left(-1\right)^u}{2}\left(\frac{\pi\theta^{2u}}{(2u)!}-\frac{\theta^{2u+1}}{\left(2u+1\right)!}\right)+\sum_{i=1}^{\lfloor\frac{2u+1}{2}\rfloor}\frac{\left(-1\right)^{u+i}\theta^{2u+1-2i}\zeta\left(2i\right)}{(2u+1-2i)!}$$ $$=\frac{\left(-1\right)^u}{2}\left(\frac{\pi\theta^{2u}}{(2u)!}-\frac{\theta^{2u+1}}{\left(2u+1\right)!}\right)+\sum_{i=1}^{u}\frac{\left(-1\right)^{u+i}\theta^{2u+1-2i}\zeta\left(2i\right)}{(2u+1-2i)!}.$$ Base case $u=1:$ $$\sum_{k=1}^{\infty}\frac{\sin\left(k\theta_{}\right)}{k^{3}}=-\frac{1}{2}\left(\frac{\pi\theta^2}{2}-\frac{\theta^3}{6}\right)+\frac{\theta\zeta(2)}{1}=-\frac{\pi\theta^2}{4}+\frac{\theta^3}{12}+\theta\,\zeta(2)$$ Surprisingly, it is right! Now, I will try to work out the case $u=t+1.$

SECOND EDIT As my answer shows, induction can prove that my conjecture was right. Problem solved.

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Holy cow! That looks like anyone's worst nightmare...! What about using the exponential function $\,\sin k\theta=\frac{1}{2}\left(e^{ik\theta}-e^{-ik\theta}\right)\,$ ? I don't know yet what happens, but it looks less dreadful, and since you get the zeta function into the mess I guess you already know some complex analysis, don't you? – DonAntonio Jan 16 '13 at 20:39
@DonAntonio I have tried it. I don't think it will work because the similar sum $$\sum_{k=1}^{\infty}\frac{\cos\left(k\theta\right)}{k}=-\log(2) -\log(1-\cos(\theta))$$ is awfully more complicated to integrate. – Ian Mateus Jan 16 '13 at 20:42
In fact, if I could easily integrate this sum, I should be able to "easily" find a formula for $\zeta\left(2n+1\right)$ simply plugging in $\theta=0$. You know what it means. – Ian Mateus Jan 16 '13 at 20:48

My own attempt by induction: Assume $$\sum_{k=1}^{\infty}\frac{\sin\left(k\theta_{}\right)}{k^{2u+1}}=\frac{\left(-1\right)^u}{2}\left(\frac{\pi\theta^{2u}}{(2u)!}-\frac{\theta^{2u+1}}{\left(2u+1\right)!}\right)+\sum_{i=1}^{\lfloor\frac{2u+1}{2}\rfloor}\frac{\left(-1\right)^{u+i}\theta^{2u+1-2i}\zeta\left(2i\right)}{(2u+1-2i)!}$$ $$=\frac{\left(-1\right)^u}{2}\left(\frac{\pi\theta^{2u}}{(2u)!}-\frac{\theta^{2u+1}}{\left(2u+1\right)!}\right)+\sum_{i=1}^{u}\frac{\left(-1\right)^{u+i}\theta^{2u+1-2i}\zeta\left(2i\right)}{(2u+1-2i)!}.$$ Base case $u=1$: $$\sum_{k=1}^{\infty}\frac{\sin\left(k\theta_{}\right)}{k^{3}}=-\frac{1}{2}\left(\frac{\pi\theta^2}{2}-\frac{\theta^3}{6}\right)+\frac{\theta\zeta(2)}{1}=-\frac{\pi\theta^2}{4}+\frac{\theta^3}{12}+\theta\,\zeta(2)$$

Now, I will integrate both sides twice. The left hand side:

$$\int_{0}^{\theta_2}\int_{0}^{\theta_1}\sum_{k=1}^{\infty}\frac{\sin\left(k\theta_{0}\right)}{k^{2t+1}}\,\mathrm{d}\theta_{0}\,\mathrm{d}\theta_{1}=\sum_{k=1}^{\infty}\int_{0}^{\theta_2}-\frac{\cos(k\theta_1)}{k^{2t+2}}+\frac{1}{k^{2t+2}}\,\mathrm{d}\theta_1\\=-\sum_{k=1}^{\infty}\frac{\sin(k\theta_2)}{k^{2(t+1)+1}}+\theta_2\,\zeta(2(t+1))$$

The right hand side: $$\int_{0}^{\theta_2}\int_{0}^{\theta_1}\frac{\left(-1\right)^u}{2}\left(\frac{\pi\theta^{2u}}{(2u)!}-\frac{\theta^{2u+1}}{\left(2u+1\right)!}\right)+\sum_{i=1}^{u}\frac{\left(-1\right)^{u+i}\theta^{2u+1-2i}\zeta\left(2i\right)}{(2u+1-2i)!}\,\mathrm{d}\theta\,\mathrm{d}\theta_{1}\\=\int_{0}^{\theta_2}\int_{0}^{\theta_1}\frac{\left(-1\right)^u}{2}\left(\frac{\pi\theta^{2u}}{(2u)!}-\frac{\theta^{2u+1}}{\left(2u+1\right)!}\right)\,\mathrm{d}\theta\,\mathrm{d}\theta_{1}+\int_{0}^{\theta_2}\int_{0}^{\theta_1}\sum_{i=1}^{u}\frac{\left(-1\right)^{u+i}\theta^{2u+1-2i}\zeta\left(2i\right)}{(2u+1-2i)!}\,\mathrm{d}\theta\,\mathrm{d}\theta_{1}$$ The first double integral: $$\frac{\left(-1\right)^u}{2}\int_{0}^{\theta_2}\int_{0}^{\theta_1}\left(\frac{\pi\theta^{2u}}{(2u)!}-\frac{\theta^{2u+1}}{\left(2u+1\right)!}\right)\,\mathrm{d}\theta\,\mathrm{d}\theta_{1}=\frac{\left(-1\right)^u}{2}\left(\frac{\pi\theta_{2}^{2(u+1)}}{(2(u+1))!}-\frac{\theta_{2}^{2(u+1)+1}}{\left(2(u+1)+1\right)!}\right)$$

The second double integral: $$\int_{0}^{\theta_2}\int_{0}^{\theta_1}\sum_{i=1}^{u}\frac{\left(-1\right)^{u+i}\theta^{2u+1-2i}\zeta\left(2i\right)}{(2u+1-2i)!}\,\mathrm{d}\theta\,\mathrm{d}\theta_{1}=\sum_{i=1}^{u}\frac{\left(-1\right)^{u+i}\theta^{2(u+1)+1-2i}\zeta\left(2i\right)}{(2(u+1)+1-2i)!}$$

Putting everything together: $$-\sum_{k=1}^{\infty}\frac{\sin(k\theta_2)}{k^{2(u+1)+1}}+\theta_2\,\zeta(2(u+1))=\frac{\left(-1\right)^u}{2}\left(\frac{\pi\theta_{2}^{2(u+1)}}{(2(u+1))!}-\frac{\theta_{2}^{2(u+1)+1}}{\left(2(u+1)+1\right)!}\right)\\+\sum_{i=1}^{u}\frac{\left(-1\right)^{u+i}\theta_{2}^{2(u+1)+1-2i}\zeta\left(2i\right)}{(2(u+1)+1-2i)!}\\\implies\sum_{k=1}^{\infty}\frac{\sin(k\theta_2)}{k^{2(u+1)+1}}=\frac{\left(-1\right)^{u+1}}{2}\left(\frac{\pi\theta_{2}^{2(u+1)}}{(2(u+1))!}-\frac{\theta_{2}^{2(u+1)+1}}{\left(2(u+1)+1\right)!}\right)\\+\sum_{i=1}^{u}\frac{\left(-1\right)^{(u+1)+i}\theta_{2}^{2(u+1)+1-2i}\zeta\left(2i\right)}{(2(u+1)+1-2i)!}+\theta_2\,\zeta(2(u+1))\\\implies\sum_{k=1}^{\infty}\frac{\sin(k\theta_2)}{k^{2(u+1)+1}}=\frac{\left(-1\right)^{u+1}}{2}\left(\frac{\pi\theta_{2}^{2(u+1)}}{(2(u+1))!}-\frac{\theta_{2}^{2(u+1)+1}}{\left(2(u+1)+1\right)!}\right)\\+\sum_{i=1}^{u+1}\frac{\left(-1\right)^{(u+1)+i}\theta_{2}^{2(u+1)+1-2i}\zeta\left(2i\right)}{(2(u+1)+1-2i)!}$$ Success, the conjecture is true. Then $$\sum_{k=1}^{\infty}\frac{\sin(k\theta_2)}{k^{2t+1}}=\frac{\left(-1\right)^{t}}{2}\left(\frac{\pi\theta_{2}^{2t}}{(2t)!}-\frac{\theta_{2}^{2t+1}}{\left(2t+1\right)!}\right)+\sum_{i=1}^{t}\frac{\left(-1\right)^{t+i}\theta_{2}^{2t+1-2i}\zeta\left(2i\right)}{(2t+1-2i)!}$$

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It seems to have escaped attention that these sums may be evaluated using harmonic summation techniques. In fact we can reduce the amount of computational effort quite considerably.

To see this, introduce the sum $$S(x) = \sum_{k\ge 1} \frac{\sin(kx)}{k^{2p+1}}$$ with $p$ a positive integer and $x$ a real number.

This sum converges absolutely and it is harmonic and may be evaluated by inverting its Mellin transform.

Recall the harmonic sum identity $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$

In the present case we have $$\lambda_k = \frac{1}{k^{2p+1}}, \quad \mu_k = k \quad \text{and} \quad g(x) = \sin x.$$

We need the Mellin transform $g^*(s)$ of $g(x)$ which is $$\int_0^\infty \sin x \times x^{s-1} dx.$$

Compute the fundamental strip of this transform. In a neighborhood of zero, $\sin x \sim x$ and we need $\Re(s) > -1$ so the integral converges there. At infinity $\sin x$ is bounded which gives convergence for $\Re(s-1)<-1$ or $\Re(s) < 0.$ Therefore the fundamental strip is $\langle -1, 0 \rangle.$

The following series of MSE posts explains in great detail using a variety of methods why $$\int_0^\infty e^{\pm ix} x^{s-1} dx = e^{\pm \pi i s/2} \Gamma(s).$$ This implies that $$g^*(s) = \sin\left(\frac{\pi s}{2}\right)\Gamma(s).$$

It follows that the Mellin transform $Q(s)$ of the harmonic sum $S(x)$ is given by

$$Q(s) = \sin\left(\frac{\pi s}{2}\right)\Gamma(s) \zeta(s+2p+1) \\ \text{because}\quad \sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \sum_{k\ge 1}\frac{1}{k^{2p+1}} \frac{1}{k^s} = \zeta(s+2p+1)$$ for $\Re(s) > -2p.$

The Mellin inversion integral here is $$\frac{1}{2\pi i} \int_{-1/2-i\infty}^{-1/2+i\infty} Q(s)/x^s ds$$ which we evaluate by shifting it to the left for an expansion about zero. The choice of line was determined by the fundamental strip of $g^*(s)$ which is completely embedded in the half-plane of convergence $\Re(s)>-2p$ of the sum term.

We now proceed to collect the contribution in residues from the poles. The sine zerm cancels all poles at even integers of the gamma function. The trivial zeros of the zeta function term cancel the poles $s+2p+1 = -2q$ where $q\ge 1$ i.e. $-2(q+p)-1$ or the poles at odd integers strictly less than $-2p-1.$ This leaves the odd integers between $-2p-1$ and $-1.$

First, the pole at $s=-2p-1:$ $$\mathrm{Res}\left(Q(s)/x^s; s=-2p-1\right) = (-1)^{p+1} \frac{(-1)^{2p+1}}{(2p+1)!}\times -\frac{1}{2} \times x^{2p+1} \\ = \frac{1}{2} \frac{(-1)^{p+1}}{(2p+1)!} \times x^{2p+1}.$$

Second, the pole at $s=-2p$ which is even, the one exception. This is a double pole which is why it did not get canceled. Combining the expansion of the zeta term with the sine term about $s=-2p$
leaves us with a factor of $(-1)^p\pi/2$ which we combine with the gamma function term to obtain $$\mathrm{Res}\left(Q(s)/x^s; s=-2p\right) = \frac{\pi}{2} (-1)^p \frac{(-1)^{2p}}{(2p)!} x^{2p} = \frac{\pi}{2} \frac{(-1)^p}{(2p)!} x^{2p}.$$

The contibution from the remaining poles is easy and given by $$\sum_{q=0}^{p-1} \mathrm{Res}\left(Q(s)/x^s; s=-(2q+1)\right) = \sum_{q=0}^{p-1} (-1)^{q+1} \frac{(-1)^{2q+1}}{(2q+1)!} \zeta(2p-2q) x^{2q+1}\\ = \sum_{q=0}^{p-1} (-1)^q \frac{\zeta(2p-2q)}{(2q+1)!} x^{2q+1}.$$

Collecting everything into one sum we obtain $$\frac{1}{2} (-1)^{p+1} \left(\frac{x^{2p+1}}{(2p+1)!} - \pi \frac{x^{2p}}{(2p)!}\right) + \sum_{q=0}^{p-1} (-1)^q \frac{\zeta(2p-2q)}{(2q+1)!} x^{2q+1}.$$

This is indeed the same expansion that we saw earlier in the other posts. It holds for $x\in[0,2\pi).$

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Here is another approach. You can write the sum in terms of the polylogarithm function $$\sum_{k=1}^{\infty}\frac{\sin\left(k\theta\right)}{k^{2u+1}}=\frac{1}{2i}\sum_{k=1}^{\infty}\frac{e^\left(ik\theta\right)}{k^{2u+1}}-\frac{1}{2i}\sum_{k=1}^{\infty}\frac{e^\left(-ik\theta\right)}{k^{2u+1}}=\frac{1}{2i}(\operatorname{Li}_{2u+1}(e^{i\theta})- \operatorname{Li}_{2u+1}(e^{-i\theta}) )$$

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Thanks, but I have already seen this form. My objective here is to make it more explicit with the "multiple integrals" idea. – Ian Mateus Jan 16 '13 at 20:56
@IanMateus: You said if there was another approach and this is one. You are welcome. – Mhenni Benghorbal Jan 16 '13 at 21:01
@Argon: Thank you for the edit. – Mhenni Benghorbal Jan 17 '13 at 12:27

Ian, multiple integrals are hard. For example for Apéry's constant there is the formula $$\zeta(3) = \int \limits_0^1 \int \limits_0^1 \int \limits_0^1 \frac{\mathrm{d}x \mathrm{d}y \mathrm{d}z}{1 - xyz}$$ source.

It looks nice, but it still is hard.

A close variant of this series was investigated by Abel and he developed his uniform convergence test in this context. It is usable to prove the euler formula for the even zeta values.

If you are careful with your signs then if you plug in $\theta=0$ you find the odd values, not of the Riemann zeta, but of the Dirichlet eta function! It turns out that the odd values of the eta function as easy - but the even values are hard!

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$$\mbox{Lets}\quad{\cal J}_{u}\pars{\theta} \equiv \sum_{k = 1}^{\infty}{\sin\pars{k\theta} \over k^{2u + 1}}$$

\begin{align} {\cal J}_{u}\pars{\theta}&={\cal J}_{u}\pars{\theta} - {\cal J}_{u}\pars{0}= \int_{0}^{\theta}{\cal J}_{u}'\pars{\theta - t}\,\dd t = {\cal J}'_{u}\pars{0}\theta + \int_{0}^{\theta}{\cal J}_{u}''\pars{\theta - t}t \,\dd t \\[3mm]&= {\cal J}'_{u}\pars{0}\theta + \half\int_{0}^{\theta}{\cal J}_{u}'''\pars{\theta - t} t^{2}\,\dd t \\[3mm]&= {\cal J}'_{u}\pars{0}\theta + {1 \over 2.3}\,{\cal J}'''_{u}\pars{0}\theta^{3} +{1 \over 2.3}\int_{0}^{\theta}{\cal J}_{u}^{\tt\pars{IV}}\pars{\theta - t}t^{3}\,\dd t \\[3mm]&=\cdots\cdots\cdots\cdots\cdots\cdots \\[3mm]&=\sum_{n = 0}^{N} {{\cal J}_{u}^{\tt\pars{2n + 1}}\pars{0} \over \pars{2n + 1}!}\,\theta^{2n + 1} + {1 \over \pars{2N + 1}!} \int_{0}^{\theta}{\cal J}_{u}^{\tt\pars{2N + 2}}\pars{\theta - t}t^{2N + 1}\,\dd t \end{align}

Also, \begin{align} {\cal J}_{u}^{\tt\pars{2n + 1}}\pars{0} &= \pars{-1}^{n + 1} \zeta\pars{2u - 2n}\quad\mbox{with}\quad \pars{~2u - 2n \geq 2\quad\imp\quad n \leq u + 1~} \\[3mm]&\mbox{and}\quad {\cal J}_{u}^{\tt\pars{2n}}\pars{\theta} =\pars{-1}^{n}{\cal J}_{u - n}\pars{\theta} \end{align}

Then \begin{align} {\cal J}_{u}\pars{\theta}&= \sum_{n = 0}^{N}\pars{-1}^{n + 1}\, {\zeta\pars{2u - 2n} \over \pars{2n + 1}!}\,\theta^{2n + 1} + {\pars{-1}^{N + 1} \over \pars{2N + 1}!} \int_{0}^{\theta}{\cal J}_{u - N - 1}\pars{\theta - t}t^{2N + 1}\,\dd t \end{align} With $\ds{N = u - 1}$: \begin{align} {\cal J}_{u}\pars{\theta}&= \sum_{n = 0}^{u - 1}\pars{-1}^{n}\, {\zeta\pars{2u - 2n} \over \pars{2n + 1}!}\,\theta^{2n + 1} + {\pars{-1}^{u} \over \pars{2u - 1}!} \int_{0}^{\theta}{\cal J}_{0}\pars{\theta - t}t^{2u - 1}\,\dd t \end{align} Since the OP reports that $\ds{{\cal J}_{0}\pars{\theta} = \half\pars{\pi - \theta}}$: \begin{align}\!\!\! \color{#00f}{\sum_{k = 1}^{\infty}{\sin\pars{k\theta} \over k^{2u + 1}}}&= \color{#00f}{\sum_{n = 0}^{u - 1}\pars{-1}^{n}\, {\zeta\pars{2u - 2n} \over \pars{2n + 1}!}\,\theta^{2n + 1}} + \color{#00f}{{\pars{-1}^{u}\pi \over 4u\pars{2u - 1}!}\,\theta^{2u} + {\pars{-1}^{u + 1} \over 4u\pars{2u + 1}\pars{2u - 1}!}\,\theta^{2u + 1}} \\[3mm]&\mbox{with}\quad u \geq 1 \end{align}

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