This is related to generating functions for Ls (Log Sine Integrals.) I am trying to calculate $$ \int_{0}^{\pi}\left[2\sin\left(\theta \over 2\right)\right]^{x} {\rm e}^{\theta y}\,{\rm d}\theta. $$ This does have a closed form (it is quite pretty). Note $x,y\in \mathbb{R}$. Thanks. I am not really sure how to calculate this one, because of the term $$ \left[2\sin\left(\theta \over 2\right)\right]^{x}. $$ Thanks.

  • $\begingroup$ @Felix do you like to edit posts purely because of your taste? It seems the changes you make, are personal taste as to how you use parentheses. Also, thanks for your help. $\endgroup$ – Jeff Faraci Apr 12 '14 at 23:38
  • $\begingroup$ Indeed, functions are written with $(\phantom{A})$'s as ${\rm f}\left(x\right)$ but I see many people ignores that. Once we have $\left(\phantom{A}\right)$'s, we continue with $\left[\phantom{A}\right]$'s, next $\left\{\phantom{A}\right\}$'s, etc... Thanks. $\endgroup$ – Felix Marin Apr 12 '14 at 23:41
  • $\begingroup$ @FelixMarin Great, THanks Felix $\endgroup$ – Jeff Faraci Apr 12 '14 at 23:48
  • $\begingroup$ I do not make sure whether dlmf.nist.gov/5.12 really helpful or not, since in that place the $\cos$ one is in $\left[0,\dfrac{\pi}{2}\right]$ while the $\sin$ one is in $[0,\pi]$ , but your question is about the $\sin$ one in $\left[0,\dfrac{\pi}{2}\right]$ . $\endgroup$ – Harry Peter May 8 '14 at 9:41
  • $\begingroup$ Hint: integrals.wolfram.com/… $\endgroup$ – Harry Peter May 8 '14 at 16:37

Since this was not answered, I will go ahead and give my two cents.

Begin with the known result (The proof of which I am sure can be found online under Cauchy's Beta Integral):

$$\int_{-\infty}^{\infty}\frac{1}{(u+ix)^{a}(v-ix)^{b}}dx=\frac{2\pi \Gamma(a+b-1)}{(u+v)^{a+b-1}\Gamma(a)\Gamma(b)}$$

Let $u=v=1$ and make the sub $x=\cot(t), \;\ dx=-\csc^{2}(t)dt$.

$$\int_{0}^{\pi}\frac{1}{\left(1+\frac{\cos(x)}{\sin(x)}i\right)^{a}\left(1-\frac{\cos(x)}{\sin(x)}i\right)^{b}}\cdot \frac{1}{\sin^{2}(x)}dx=\frac{2\pi \Gamma(a+b-1)}{2^{a+b-1}\Gamma(a)\Gamma(b)}$$

rewrite the left side:

$$\int_{0}^{\pi}\frac{\sin^{a+b-2}(x)}{(\sin(x)+i\cos(x))^{a}(\sin(x)-i\cos(x))^{b}}dx=\frac{2\pi \Gamma(a+b-1)}{2^{a+b-1}\Gamma(a)\Gamma(b)}$$

But, $\displaystyle \sin(x)+i\cos(x)=ie^{-ix}, \;\ \sin(x)-i\cos(x)=-ie^{ix}$

so we may write:

$$e^{-\pi (a-b)i/2}\int_{0}^{\pi}\sin^{a+b-2}(x)e^{-i(b-a)x}dx=\frac{2\pi \Gamma(a+b-1)}{2^{a+b-1}\Gamma(a)\Gamma(b)}$$

Let $\displaystyle y=a-b, \;\ x=a+b-2$

$$e^{-\frac{\pi y i}{2}}\int_{0}^{\pi}\sin^{x}(\theta)e^{iy\theta}d\theta=\frac{2\pi \Gamma(x+1)}{2^{x+1}\Gamma\left(1+\frac{x+y}{2}\right)\Gamma\left(1+\frac{x-y}{2}\right)}$$

$$\int_{0}^{\pi}\sin^{x}(\theta)e^{iy\theta}d\theta=\frac{\pi \Gamma(x+1)e^{\frac{\pi y i}{2}}}{2^{x}\Gamma\left(1+\frac{x+y}{2}\right)\Gamma\left(1+\frac{x-y}{2}\right)}$$

Is this similar to the result you were shooting for?. This is a slight variation on your integral.

Here is a link I just found that uses the Beta to derive one of the Cauchy Beta Integrals with cos. With some adjustments one may probably use this method to derive this one.


I made some adjustments to one of these methods to derive this. The form I derived is the sin version of the Beta integral I have commonly encountered

  • $\begingroup$ Despite being a "slight variation", this is excellent and very helpful. Thanks a lot for this, I am quite impressed by the proof. Also, the wikibook is very NICE! +1 $\endgroup$ – Jeff Faraci May 24 '14 at 19:30

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