I came across this expression while solving an integral:
$$\int_0^\frac\pi2(\sin(su))(\cos u)^s du$$
"s" belongs to the set of complex numbers.
Is there a way to simplify this expression? I was thinking of using the De'Moivre's Theorem but the expression becomes too tedious.
This is a partial answer. We must have $\Re s>-1$ for the integral to converge.
For $0<r<1$, let the contour $C_r$ be the boundary of $\{z\in\mathbb{C} : r<|z|<1, 0<\arg z<\pi/2\}$ (with the usual "counterclockwise" orientation), consisting of two quartercircles and two line segments. Then, assuming the principal values of $(\ldots)^s$ taken, we have (by Cauchy's integral theorem) $$0=\int_{C_r}(1+z^2)^s\frac{dz}{z}=\int_r^1\frac{(1+x^2)^s-(1-x^2)^s}{x}\,dx\\+i\int_0^{\pi/2}(1+e^{2i\phi})^s\,d\phi-i\int_0^{\pi/2}(1+r^2 e^{2i\phi})^s\,d\phi.$$ Taking $r\to 0$, and substituting $x^2=t$, we get the "$+$" version of $$2^{s+1}\int_0^{\pi/2}(\cos\phi)^s e^{\pm is\phi}\,d\phi=\pi\pm if(s),\quad f(s)=\int_0^1\frac{(1+t)^s-(1-t)^s}{t}\,dt,$$ and the "$-$" version is obtained similarly. Hence, the given integral is equal to $2^{-s-1}f(s)$.
The equality $f(s)=f(s-1)+2^s/s$ allows to compute $f(s)$ for $s\in\mathbb{Z}_{\geqslant 0}$.