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It is known that $$\int_0^\frac{\pi}{2}\cos^{a-1}x\cos bx~dx=\frac{\pi}{2^aa\mathrm{B}\left(\dfrac{a+b+1}{2},\dfrac{a-b+1}{2}\right)}$$ as https://dlmf.nist.gov/5.12 stated.

How about $\int_0^\frac{\pi}{2}\cos^{a-1}x\sin bx~dx$ , $\int_0^\frac{\pi}{2}\sin^{a-1}x\sin bx~dx$ , $\int_0^\frac{\pi}{2}\sin^{a-1}x\cos bx~dx$ ?

Or how about $\int_0^\pi\cos^{a-1}x\cos bx~dx$ , $\int_0^\pi\cos^{a-1}x\sin bx~dx$ , $\int_0^\pi\sin^{a-1}x\sin bx~dx$ , $\int_0^\pi\sin^{a-1}x\cos bx~dx$ ?

I notice someone can evaluate $\int_0^\pi\sin^\alpha t\cos kt~dt$ for integer $k$ : How to do $\int_{0}^{\pi}(\sin t)^{\alpha}\cos (k t)dt$ please?

metamorphy
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Harry Peter
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1 Answers1

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Let's recall (and modify) one of the known approaches to get the first of these integrals.

Assume (temporarily) that $0<\Re a<\Re b$. Let $\gamma_+$ be the boundary of $$\{z:|z|<1,0<\arg z<\pi/2\}$$ (consisting of the segment $[0,1]$ of the real line, the quartercircle $\{e^{ix}:0\leqslant x\leqslant\pi/2\}$ and the segment $[i,0]$ of the imaginary line), and similarly $\gamma_-$ be the boundary of $$\{z:|z|<1,-\pi/2<\arg z<0\}.$$ Then, by Cauchy's integral theorem (the singularity at $z=0$ taken as a limit) $$0=\int_{\gamma_+}z^{b-a}(1+z^2)^{a-1}~dz=\int_{\gamma_-}z^{b-a}(1+z^2)^{a-1}~dz.$$ Writing explicitly the integrals along each piece of the contours, we get $$0=F_+(a,b)+2^{a-1}i\int_0^{\pi/2}e^{ibx}\cos^{a-1}x~dx-ie^{i(b-a)\pi/2}F_-(a,b),\\0=F_+(a,b)-2^{a-1}i\int_0^{\pi/2}e^{-ibx}\cos^{a-1}x~dx+ie^{i(a-b)\pi/2}F_-(a,b),\\\text{where }F_\pm(a,b):=\int_0^1 x^{b-a}(1\pm x^2)^{a-1}~dx.$$ Observe that $F_-(a,b)=\frac{1}{2}\mathrm{B}\left(a,\frac{b-a+1}{2}\right)$. Thus, taking the difference of the preceding equalities, we get rid of $F_+$, and using the reflection formula for $\Gamma\left(\frac{b-a+1}{2}\right)$, we obtain the first formula in the question (which then holds, by analytic continuation, for any $a,b$ with $\Re a>0$). Taking the sum instead, we get $$2^{a-1}\int_0^{\pi/2}\cos^{a-1}x\sin bx~dx=F_+(a,b)+F_-(a,b)\sin\frac{(b-a)\pi}{2}.$$ This time, the $F_+$ is retained, and there's no simple formula for this integral. (Of course, $F_+(a,b)$ can be expressed in terms of incomplete beta functions, or evaluated in closed form in some particular cases, say when $a$ and $b$ are both (half-)integers, or when $(b-a-1)/2$ is an integer.)

The upper limit of $\pi$ makes it a different story. For $0<\Re a<\Re b$, integrating $z^{b-a}(1-z^2)^{a-1}$ over the boundary of $\{z : |z|<1,0<\arg z<\pi\}$ (and using the reflection formula for $\Gamma$ again), we get $$\int_0^\pi e^{ibx}\sin^{a-1}x~dx=\frac{\pi e^{i\pi b/2}}{2^{a-1}a\mathrm{B}\left(\frac{a+b+1}{2},\frac{a-b+1}{2}\right)}$$ (which, again by analytic continuation, holds for any $a,b$ with $\Re a>0$).

metamorphy
  • 39,111