Definite integral of $x\sin^n(x)$, recurrence relations, $\zeta(3)$

Question

This is my first ever post on this website, so bear with me.

I'm trying to find a general solution to the following integral:

$$\int_{0}^{\pi/2}x\sin^s(x)dx$$ With respect to "$s$"

So naturally I do integration by parts and differentiate "$x$".

It boils down to [$x\int\sin^s(x)$ from $0$ to $\pi/2$] minus $\int_{0}^{\pi/2}\int\sin^s(x)dxdx$

The first thing I need to evaluate, which is [$x\int\sin^s(x)$ from $0$ to $\pi/2$] follows this recurrence relation

$$I(s)=\pi/2\cdot (\frac{s-1}{s})\cdot I(s-2)$$ $$I(0)=\pi/2, I(1)=0$$

The second thing I must evaluate, which is $\int_{0}^{\pi/2}\int\sin^s(x)dxdx$ follows this recurrence relation

$$J(s)= -\frac{1}{s^2}+(\frac{s-1}{s})J(s-2)$$ $$J(0)=\frac{\pi^2}{4}, J(1)= -1$$

Can anyone please help me find an explicit formula for these two recurrence relations?

I think generating functions might help. This will greatly help me on my journey to finding $\zeta(3)$

Answer 1

$$ \begin{aligned} I_n=&\int_{0}^{\frac{\pi}{2}}x\sin^n(x)dx \\ =&\int_{0}^{\frac{\pi}{2}}(xsin^{n-1}(x))sin(x)dx \\=&-(xsin^{n-1}(x))cos(x)|_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}((n-1)xsin^{n-2}(x)cos(x)+sin^{n-1}(x))cos(x)dx \\=&\int_{0}^{\frac{\pi}{2}}((n-1)xsin^{n-2}(x)cos^2(x)dx+\int_{0}^{\frac{\pi}{2}}sin^{n-1}(x)cos(x)dx \\=&\int_{0}^{\frac{\pi}{2}}((n-1)xsin^{n-2}(x)dx-\int_{0}^{\frac{\pi}{2}}((n-1)xsin^n(x)dx+\frac{1}{n} \\=&(n-1)I_{n-2}-(n-1)I_n+\frac{1}{n} \\=&\frac{(n-1)}{n}I_{n-2}+\frac{1}{n^2} \end{aligned} $$

Answer 2

For the question in title $$I(s)=\int_{0}^{\frac \pi 2}x\sin^s(x)\,dx$$ I think that you are facing an hypergeometric function.

Using Mathematica, after simplifications, the result is $$\color{blue}{ I(s)=\frac{\pi ^{\frac 3 2}}{4}\,\,\frac{\Gamma \left(\frac{s+1}{2}\right)}{\Gamma\left(\frac{s+2}{2}\right)}-\frac{\pi }{2^{s+2}}\,\,\frac{\Gamma (s+1)}{\Gamma \left(\frac{s+3}{2}\right)^2}\ {}_3F_2\left(\left.{\frac{1}{2},\frac{s+1}{2},\frac{s+2}{2}\atop \frac{s+3}{2},\frac{s+3}{2}}\right|1\right)}$$ which is valid for any $s>0$.

It seems that it even works for complex values of $s$.