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.
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.
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.
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.
Substitute $t=e^{i x}$
\begin{align} &\int_{0}^{\frac{\pi}{2}} x^3 \csc x\>dx\\ =&\int_i^1 \frac{2i \ln^3 t}{1-t^2}dt =2i \int_0^1 \frac{\ln^3 t}{1-t^2}dt +2i\int_0^1 \frac{\ln^3 (it)}{1+t^2}dt \\ =&-\frac{3\pi^2}2\int_0^1 \frac{\ln t}{1+t^2}dt+2 \int_0^1 \frac{\ln^3t}{1+t^2}dt \end{align} where only the real parts are retained. Substitute $\int_0^1 \frac{\ln t}{1+t^2}dt=-G$ and $\int_0^1 \frac{\ln^3 t}{1+t^2}dt=-6 \beta(4) $ to obtain $$\int_{0}^{\frac{\pi}{2}} x^3 \csc x\>dx=\frac{3\pi^2}2G-12\beta(4) $$