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.

  • $\begingroup$ Could you show up the integral (and the conditions on $s$)? $\endgroup$ – metamorphy Oct 9 '19 at 8:27
  • $\begingroup$ @metamorphy Actually the conditions and integral is w.r.t. to u. Here it is : integral of (sin(su))(cos u)^s du from 0 to pi/2 $\endgroup$ – Rohan Asif Oct 9 '19 at 10:31
  • $\begingroup$ @metamorphy s belongs to set of complex numbers $\endgroup$ – Rohan Asif Oct 9 '19 at 10:39
  • $\begingroup$ Just a comment as I plugged it into mathematica, but the integral seems to evaluate to $-2^{-s-1} \left(\pi \cot (\pi s)+e^{i \pi s} B_{-1}(-s,s+1)\right)$ where $B$ represents the incomplete beta function (en.wikipedia.org/wiki/Beta_function#Incomplete_beta_function) $\endgroup$ – QC_QAOA Oct 9 '19 at 11:24
  • $\begingroup$ @NickGuerrero if B is the incomplete beta function, I have very little knowledge about that so I can't use this result. If it can be expressed in other functions it would be helpful. $\endgroup$ – Rohan Asif Oct 9 '19 at 11:26

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 $s\in\mathbb{Z}_{\geqslant 0}$.


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