4
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.