Integral Iteration $\int_{0}^{2\pi}x^{2m}\cos 2nx dx$

Question

everyone.

Calculate $\displaystyle\int_{0}^{2\pi}x^{2m}\cos 2nx dx~~~$ where $m$,$n$ are positive integers.
Could you tell me the formula and proof ? Thanks in advance.

Answer 1

$$I_m=\int_0^{2\pi} x^{2m}\cos{(2nx)}dx$$ $$=\frac{1}{2n}\sin(2nx)x^{2m}\big|_0^{2\pi}-\frac{m}{n}\int_0^{2\pi}x^{2m-1}\sin(2nx)dx$$ $$=\frac{m}{2n^2}(\cos(2nx))x^{2m-1}\big|_0^{2\pi}-\frac{m(2m-1)}{2n^2}I_{m-1}$$ Can you take it from here?

Answer 2

I am slightly afraid that the general solution is given by an hypergeometric function $$I_{m,n}=\int_{0}^{2\pi}x^{2m}\cos (2nx)\, dx=\frac{(2 \pi )^{2 m+1}}{2 m+1}\, _1F_2\left(m+\frac{1}{2};\frac{1}{2},m+\frac{3}{2};-4 \pi ^2 n^2\right)$$

Here are some values $$\left( \begin{array}{ccc} m & n &I_{m,n} \\ 1 & 1 & \pi \\ 1 & 2 & \frac{\pi }{4} \\ 1 & 3 & \frac{\pi }{9} \\ 1 & 4 & \frac{\pi }{16} \\ 1 & 5 & \frac{\pi }{25} \\ 1 & 6 & \frac{\pi }{36} \\ 2 & 1 & -3 \pi +8 \pi ^3 \\ 2 & 2 & -\frac{3 \pi }{16}+2 \pi ^3 \\ 2 & 3 & -\frac{\pi }{27}+\frac{8 \pi ^3}{9} \\ 2 & 4 & -\frac{3 \pi }{256}+\frac{\pi ^3}{2} \\ 2 & 5 & -\frac{3 \pi }{625}+\frac{8 \pi ^3}{25} \\ 2 & 6 & -\frac{\pi }{432}+\frac{2 \pi ^3}{9} \\ 3 & 1 & \frac{45 \pi }{2}-60 \pi ^3+48 \pi ^5 \\ 3 & 2 & \frac{45 \pi }{128}-\frac{15 \pi ^3}{4}+12 \pi ^5 \\ 3 & 3 & \frac{5 \pi }{162}-\frac{20 \pi ^3}{27}+\frac{16 \pi ^5}{3} \\ 3 & 4 & \frac{45 \pi }{8192}-\frac{15 \pi ^3}{64}+3 \pi ^5 \\ 3 & 5 & \frac{9 \pi }{6250}-\frac{12 \pi ^3}{125}+\frac{48 \pi ^5}{25} \\ 3 & 6 & \frac{5 \pi }{10368}-\frac{5 \pi ^3}{108}+\frac{4 \pi ^5}{3} \\ 4 & 1 & -315 \pi +840 \pi ^3-672 \pi ^5+256 \pi ^7 \\ 4 & 2 & -\frac{315 \pi }{256}+\frac{105 \pi ^3}{8}-42 \pi ^5+64 \pi ^7 \\ 4 & 3 & -\frac{35 \pi }{729}+\frac{280 \pi ^3}{243}-\frac{224 \pi ^5}{27}+\frac{256 \pi ^7}{9} \\ 4 & 4 & -\frac{315 \pi }{65536}+\frac{105 \pi ^3}{512}-\frac{21 \pi ^5}{8}+16 \pi ^7 \\ 4 & 5 & -\frac{63 \pi }{78125}+\frac{168 \pi ^3}{3125}-\frac{672 \pi ^5}{625}+\frac{256 \pi ^7}{25} \\ 4 & 6 & -\frac{35 \pi }{186624}+\frac{35 \pi ^3}{1944}-\frac{14 \pi ^5}{27}+\frac{64 \pi ^7}{9} \end{array} \right)$$

Answer 3

Here's an alternative way involving derivatives. Observe \begin{align} \int^{2\pi}_0 x^{2m}\cos(2n x)\ dx =&\ \operatorname{Re}\left(\int^{2\pi}_0x^{2m}e^{i2n x}\ dx \right)= \operatorname{Re}\left(\int^{2\pi}_0\frac{(-1)^m}{4^{m}}\frac{d^{2m}}{ds^{2m}}e^{2 xis}\ dx\bigg|_{s=n} \right)\\ =&\ \frac{(-1)^m}{4^m}\frac{d^{2m}}{ds^{2m}} \operatorname{Re}\left( \int^{2\pi}_0 e^{2xis}\ dx\right)\bigg|_{s=n}\\ =&\ 2\pi\frac{(-1)^m}{4^m}\frac{d^{2m}}{ds^{2m}}\left(\frac{\sin(4\pi s)}{4\pi s}\right) \bigg|_{s=n}.\\ \end{align}