Closed-form expressions for Fourier integrals

Question

Define the following two Fourier integrals $$ C(k) = \int_0^{\pi/2} \big( \sin x\big)^{2\alpha}\cos (2kx) dx,\quad S(k) = \int_0^{\pi/2} \big( \sin x\big)^{2\alpha}\sin (2kx) dx, $$ for some constant $\alpha>0$ and integer $k$. Using symbolic software like MAPLE, we can get \begin{align} C(1) &= -\frac{\sqrt{\pi}\Gamma(\frac{1}{2}+\alpha)}{\Gamma(\alpha)}\frac{1}{2(1+\alpha)},\cr C(2) &= -\frac{\sqrt{\pi}\Gamma(\frac{1}{2}+\alpha)}{\Gamma(\alpha)}\frac{1-\alpha}{2(1+\alpha)(2+\alpha)},\cr C(3) &= -\frac{\sqrt{\pi}\Gamma(\frac{1}{2}+\alpha)}{\Gamma(\alpha)}\frac{(1-\alpha)(2-\alpha)}{2(1+\alpha)(2+\alpha)(3+\alpha)},\cr C(4) &= -\frac{\sqrt{\pi}\Gamma(\frac{1}{2}+\alpha)}{\Gamma(\alpha)}\frac{(1-\alpha)(2-\alpha)(3-\alpha)}{2(1+\alpha)(2+\alpha)(3+\alpha)(4+\alpha)},\cr &\ \ \vdots \end{align} It is easy to guess that (seems right, though no proof) $$ C(k) = -\frac{\alpha\sqrt{\pi}}{2} \frac{\Gamma(\frac{1}{2}+\alpha)\Gamma(k-\alpha)}{\Gamma(1-\alpha)\Gamma(k+1+\alpha)},\quad k=1,2,\cdots. $$

But the other integral $S(k)$ seems quite different. Here is the first few ones: \begin{align} S(1) &= \frac{1}{1+\alpha},\cr S(2) &= -\frac{2\alpha}{(1+\alpha)(2+\alpha)},\cr S(3) &= \frac{3\alpha^3-\alpha+2}{(1+\alpha)(2+\alpha)(3+\alpha)},\cr S(4) &= -\frac{4{\alpha}^{3}-4{\alpha}^{2}+16\alpha}{(1+\alpha)(2+\alpha)(3+\alpha)(4+\alpha)},\cr S(5) &= \frac{5\,{\alpha}^{4}-10\,{\alpha}^{3}+67\,{\alpha}^{2}-14\,\alpha+24}{(1+\alpha)(2+\alpha)(3+\alpha)(4+\alpha)(5+\alpha)},\cr S(6) &= -\frac{6\,{\alpha}^{5}-20\,{\alpha}^{4}+202\,{\alpha}^{3}-124\,{\alpha}^{2}+ 368\,\alpha }{(1+\alpha)(2+\alpha)(3+\alpha)(4+\alpha)(5+\alpha)(6+\alpha)},\cr S(7) &= \frac{7\,{\alpha}^{6}-35\,{\alpha}^{5}+497\,{\alpha}^{4}-601\,{\alpha}^{3}+ 2736\,{\alpha}^{2}-444\,\alpha+720 }{(1+\alpha)(2+\alpha)(3+\alpha)(4+\alpha)(5+\alpha)(6+\alpha)(7+\alpha)},\cr S(8)&=-\frac{8\,{\alpha}^{7}-56\,{\alpha}^{6}+1064\,{\alpha}^{5}-2120\,{\alpha}^{4} +13712\,{\alpha}^{3}-6464\,{\alpha}^{2}+16896\,\alpha }{(1+\alpha)(2+\alpha)(3+\alpha)(4+\alpha)(5+\alpha)(6+\alpha)(7+\alpha)(8+\alpha)},\cr &\ \ \vdots \end{align} Since the numerator of $S(k)$ can not be further factorized (except the obvious factor $\alpha$ when $k$ is even), it is unlikely that $S(k)$ has a single close-form expression as $C(k)$ using Gamma functions. My question is whether there is one with two or more terms?

Answer 1

These integrals can be expressed on closed form thanks to the hypergeometric function $_2F_1$

In case of the integral $C(k)$ , there is a relationship between the hypergeometric function of argument $1$ and the Gamma function which allows a simpler closed form.

In case of the integral $S(k)$ which involves the hypergeometric function of argument $-1$ , as far as I know there is no similar relationship. Nether the less, there is a relationship with the Incomplete Beta function, but in the complex domaine since the argument is $-1$ : Formula above.

Done with the invaluable help of WolframAlpha :

$ _2F_1(-2a,k-a;-a+k+1;-1)=(-1)^k(-1)^a(k-a)\text{B}_{-1}(k-a,2a+1)$

$\text{B}_{-1}(k-a,2a+1) \equiv \text{B}(-1;k-a,2a+1)\quad$ is the Incomplete Beta function with argument $-1$.

http://mathworld.wolfram.com/IncompleteBetaFunction.html

Answer 2

Here's a sum expression for $S(k)$.

We have $$ S(k) = \int_0^{\pi/2} \big( \sin x\big)^{2\alpha}\sin (2kx) dx $$ Integrating by parts gives $$ S(k) = \frac{(-1)^{(k+1)}}{2 k} + \frac{\alpha}{k} \int_0^{\pi/2} \big( \sin x\big)^{(2\alpha -1)}\cos (2kx)\cos (x) dx $$ We can use the expansion (documented e.g. in the manuscripts of H. W. Gould) $$ \cos (2kx) = \sum_{r=0}^k (-1)^r \frac{k}{r + k} {r + k \choose 2r } 4^r \big(\sin x \big)^{2r} $$ to obtain $$ S(k) = \frac{(-1)^{(k+1)}}{2 k} + \frac{\alpha}{k} \sum_{r=0}^k (-1)^r \frac{k}{r + k} {r + k \choose 2r } 4^r \int_0^{\pi/2} \big( \sin x\big)^{(2(r +\alpha) -1)}\cos (x) dx $$ Now the integral can be easily evaluated to give $\big( \sin x\big)^{(2(r +\alpha))}/(2(r + \alpha))$ where the definite integral becomes $1/(2(r + \alpha))$ . Hence $$ S(k) = \frac{(-1)^{(k+1)}}{2 k} + \sum_{r=0}^k (-1)^r \frac{\alpha}{r + k} {r + k \choose 2r } \frac{4^r}{2(r + \alpha)} $$ or $$ S(k) = \frac{1 + (-1)^{(k+1)}}{2 k} + \sum_{r=1}^k (-1)^r \frac{\alpha}{r + k} {r + k \choose 2r } \frac{4^r}{2(r + \alpha)} $$ which is the desired expession. Due to the general (not: integer) $\alpha$ this generally cannot be summed any further. Writing this as one fraction, the common denominator is $\prod_{r=1}^{k} (r + \alpha)$ as in the examples in the task description. $\quad \quad \qquad \Box$