A integral with polygamma I was doing a integral, the last part is 
$$\int_0^{\frac{\pi}{2}}x^3\csc x\text{d}x$$
I ran this on Maple, it turns into polygammas...How we evaluate this?
I think there should be a way to evaluate manually.
 A: We can use the ideas of this answer.
Integrating by parts three times, we get
$$
\int_0^{\pi/2}x^3\,e^{ikx}\,\mathrm{d}x
=i^{k-1}\frac{\pi^3}{8k}+i^k\frac{3\pi^2}{4k^2}+i^{k+1}\frac{3\pi}{k^3}+\frac6{k^4}\left(1-i^k\right)
$$
Therefore, using $\sin(x)=\dfrac{e^{ix}-e^{-ix}}{2i}$,
$$
\begin{align}
&\int_0^{\pi/2}x^3\csc(x)\,\mathrm{d}x\\
&=\int_0^{\pi/2}x^3\,\frac{2ie^{-ix}\,\mathrm{d}x}{1-e^{-2ix}}\\
&=2i\sum_{k=0}^\infty\int_0^{\pi/2}x^3\,e^{-(2k+1)ix}\,\mathrm{d}x\\
&=2i\sum_{k=0}^\infty(-1)^{k+1}\frac{3\pi^2i}{4(2k+1)^2}+(-1)^k\frac{6i}{(2k+1)^4}\tag{$\ast$}\\
&=\frac{3\pi^2}{2}\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2}-12\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^4}\\
&=\frac{3\pi^2}{2}G-\frac3{64}\left(\zeta(4,1/4)-\zeta(4,3/4)\right)\\
&=\frac{3\pi^2}{2}G-\frac1{128}\left(\psi^{(3)}(1/4)-\psi^{(3)}(3/4)\right)
\end{align}
$$
where G is Catalan's Constant and $\zeta(s,z)$ is the Hurwitz zeta function and $\psi^{(k)}(z)$ is a polygamma function.
Only the real parts of $(\ast)$ are retained.
A: Playing with Fourier series I found earlier (formally since the series is divergent!) :
$$\sum_{k=0}^\infty \cos(((2k+1)q-p)x)=\Re\left(\frac{e^{-ipx}}{e^{\,iqx}-e^{-iqx}}\right)=\frac{\sin(px)}{2\sin(qx)}$$
so that (for $a:=\frac pq$ and after integration) we get the general (and convergent) :
$$\int_0^u \frac{\sin(at)}{\sin(t)}dt=\frac{\pi}2-2\sum_{k=0}^\infty \frac{\sin((2k+1+a)u)}{2k+1+a}$$
Setting $u:=\frac{\pi}n$ and using the repetition gives :
$$f_n(a):=\int_0^{\frac{\pi}n} \frac{\sin(at)}{\sin(t)}dt=\frac{\pi}2+\frac 1n\sum_{k=0}^{n-1} \sin\left((2k+1+a)\frac{\pi}n\right)\,\psi\left(\frac{2k+1+a}{2n}\right)$$
with $\psi$ the digamma function.
Let's apply this to the case $n=2$ :
$$f_2(a)=\int_0^{\frac{\pi}2} \frac{\sin(at)}{\sin(t)}dt=\frac{\pi}2+\frac 12\cos\left(\frac{\pi}2 a\right)\left( \psi\left(\frac{a+1}4\right)-\psi\left(\frac{a+3}4\right)\right)$$ 
rewritten using the function $\ \displaystyle B(x):=\sum_{k=0}^\infty \frac{(-1)^k}{x+k}=\frac 12\left(\psi\left(\frac{x+1}2\right)-\psi\left(\frac x2\right)\right)$ as :
$$f(a):=\int_0^{\frac{\pi}2} \frac{\sin(at)}{\sin(t)}dt=\frac{\pi}2-\cos\left(\frac{\pi}2 a\right)\,B\left(\frac{a+1}2\right)$$ 
Now we need only to expand everything in Taylor series in $a$ as $a\to 0$ :
$$f_2(a)=\frac {\pi}2-B\left(\frac 12\right)-B'\left(\frac 12\right)\frac a2-\left(B''\left(\frac 12\right)-\pi^2 B\left(\frac 12\right)\right)\frac {a^2}8-\left(B'''\left(\frac 12\right)-3\pi^2 B'\left(\frac 12\right)\right)\frac{a^3}{48}+O(a^4)$$
But we have too 
$$f_2(a)=\int_0^{\frac{\pi}2} \frac t{\sin(t)}a-\frac 1{3!}\frac {t^3}{\sin(t)}a^3+O(a^5)\,dt$$
Since the Dirichlet $\beta$ function is defined by $\ \displaystyle\beta(n):=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^n}$ we get following successive derivatives :
$$B^{(n-1)}\left(\frac 12\right)=-(-2)^n\,(n-1)!\,\beta(n)$$
and conclude (with $\beta(2)$ the Catalan constant) :
$$\int_0^{\frac{\pi}2} \frac t{\sin(t)}dt=2\,\beta(2)$$
$$\boxed{\displaystyle\int_0^{\frac{\pi}2} \frac {t^3}{\sin(t)}dt=-12\,\beta(4)+\frac 32\pi^2\,\beta(2)}$$
$$\int_0^{\frac{\pi}2} \frac {t^5}{\sin(t)}dt=240\,\beta(6)-30\pi^2\,\beta(4)+\frac 58\pi^4\,\beta(2)$$
and so on... (of course the even terms disappear)
There is probably a more direct derivation in your case but this method allows some general results.
