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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.

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  • $\begingroup$ If you have Maple, why not ask Maple to evaluate this? Otherwise there are lots of numerical libraries (e.g. gsl) out there, that can evaluate the polygamma function. Just google for it. $\endgroup$ – Elmar Zander Feb 13 '13 at 12:00
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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.

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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.

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