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Has anyone encountered the following integral ?: $$ I = \int_{0}^{\pi/2}\frac{\sin\left(ax\right)}{\sin\left(bx\right)}\,{\rm d}x $$ Besides the numerical computation, is there a general analytic solution ?.

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  • $\begingroup$ For many cases, e.g. $a=2,b=3$, WA says the integral doesn't converge. (I wonder which are such cases, generally?) $\endgroup$
    – r.e.s.
    Commented Jun 20 at 15:32
  • $\begingroup$ @r.e.s. It won't converge if the denominator is zero when the numerator is not, and this is probably an equivalence. $\endgroup$
    – J. S.
    Commented Jun 20 at 15:41
  • $\begingroup$ @J.S. Right, I should have said "for which $(a,b)$ is this the case?" -- i.e., for which the denominator is zero when the numerator isn't. $\endgroup$
    – r.e.s.
    Commented Jun 20 at 15:47
  • $\begingroup$ Unless $a,b$ are special values, the integrand will not be related to the upper limit $\pi/2$. So this definite integral will likely be as difficult as the indefinite integral. $\endgroup$
    – GEdgar
    Commented Jun 20 at 15:49
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    $\begingroup$ @pstall : You really should edit the constraints on $a$ and $b$ in terms of $c$ into your Question since that's your actual question. $\endgroup$ Commented Jun 20 at 17:21

3 Answers 3

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If $a$ and $b$ are arbitrary, there's nothing special about $\pi/2$, and you're basically asking for an antiderivative of $\sin(ax)/\sin(bx)$. This probably does not have a "closed form": it certainly is not elementary.

On the other hand, if $a$ is a positive integer and $b=1$, you can use $$\sin(ax)/\sin(x) = U_{a-1}(\cos(x))$$ where $U_n$ are Chebyshev polynomials of the second kind. If $a$ is odd, the integral is $\pi/2$, if even it seems to be

$$ \int_{0}^{\pi/2} \frac{\sin(2nx)}{\sin(x)} \; dx = \sum_{i=0}^{n-1} \frac{2 (-1)^i}{2i+1} $$

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lets use Euler's formula So $$ I(a,b)=2i \int_0^{\frac{\pi}{2}} \frac{e^{-bix}}{1-e^{-2bix}} \sin (ax)dx$$ So the converge of integral is when $b\in (-2,2) , b\ne 0$ So $$ I(a,b)=2i\sum_{k=0}^\infty \int_0^{\frac{\pi}{2}} e^{-bi(2k+1)x} \sin (ax)dx$$ and since $I(a,b)$ is real So we can take real part only So $$ I(a,b)=2\sum_{k=0}^\infty \int_0^{\frac{\pi}{2}} \sin(b(2k+1)x) \sin (ax)dx$$ now if $a\ne n b , n\in N$ we have $$ I(a,b)=2\sum_{k=0}^\infty \frac{b \sin(\frac{\pi}{2}a)\cos(\frac{\pi}{2}b(2k+1))-a \sin(\frac{\pi}{2}b(2k+1))\cos(\frac{\pi}{2}a)}{a^2-b^2(2k+1)^2}$$ I don't have ideas to simplify this series but maybe we can use some special functions to get the closed form for that

but for $a=nb , n\in N=\{1,2,3,...\}$ we have $$ I(nb,b)=\int_0^{\frac{\pi}{2}} \frac{1-e^{-2bnix}}{1-e^{-2bix}} e^{(nb-b)ix} dx=\sum_{k=0}^{n-1} \int_0^{\frac{\pi}{2}} e^{-2bk ix} e^{(nb-b)ix} dx$$ So by taking real part $$ I(nb,b)=\sum_{k=0}^{n-1} \int_0^{\frac{\pi}{2}} \cos\left(bx(n-2k-1)\right)dx$$ So $$ I(nb,b)=\frac{1}{b}\sum_{k=0}^{n-1} \frac{\sin\left(\frac{\pi}{2}b (n-2k-1)\right)}{n-2k-1}$$ note : take limit when $k=\frac{n-1}{2}$

for example if $b=1 , a=2n$ we get $$ I(2n,1)=\sum_{k=0}^{2n-1} \frac{\sin\left(\frac{\pi}{2} (2n-2k-1)\right)}{2n-2k-1}=\sum_{k=0}^{2n-1} \frac{(-1)^{n-k-1}}{2n-2k-1}=\sum_{k=0}^{n-1} \frac{2(-1)^{k}}{2k+1}$$ which is same results given in the answers

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    $\begingroup$ T.Y. for your result. I will evaluate the series as far as reasonable with some specific values for a and b. $\endgroup$
    – pstall
    Commented Jun 20 at 22:21
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From OP's comments to the question:

Let $0<c<1$. Then (via Mathematica 12.3), \begin{align*} \int_0^{\pi/2}\; \frac{\sin(c x)}{\sin((1-c)x)} \,\mathrm{d}x &= \frac{-1}{2-2c} \left( (2-2c) {}_2 F_1\left(1, \frac{1}{2-2c}; 1+\frac{1}{2-2c}; -\mathrm{e}^{\mathrm{i}\pi c} \right) - \right. \\ & \left. \mathrm{i}\pi \cot\left(\frac{\pi}{2-2c}\right)+\left(-\mathrm{e}^{\mathrm{i}\pi c}\right)^{1/(2-2c)} \mathrm{B}_{-\mathrm{e}^{\mathrm{i}\pi c}} \left( 1 - \frac{1}{2-2c}, 0 \right) \right) \text{,} \end{align*} where $B_z(a,b)$ is the incomplete beta function and ${}_2 F_1(\vec{a}; \vec{b}; z)$ is a generalized hypergeometric series and \begin{align*} \int_0^{\pi/2}\; \frac{\sin((2-c) x)}{\sin((1-c)x)} \,\mathrm{d}x &= \frac{-1}{2-2c} \left( \left(-\mathrm{e}^{\mathrm{i}\pi c} \right)^{1/(2-2c)} \left( \left(-\mathrm{e}^{\mathrm{i}\pi c}\right)^{1/(c-1)} \mathrm{B}_{-\mathrm{e}^{\mathrm{i}\pi c}} \left( 1+\frac{1}{2-2c}, 0 \right) + \right.\right. \\ &\left.\left. \mathrm{B}_{-\mathrm{e}^{\mathrm{i}\pi c}} \left(\frac{-1}{2-2c}, 0 \right) \right) - \mathrm{i}\pi \cot\left( \frac{\pi}{2-2c} \right) \right) \end{align*}

It's plausible that there are only a finite number of typo's in the above.

I have no idea how to generate these results by hand. I suspect Mathematica used a contour integration based on the appearance of $\mathrm{e}^{\mathrm{i}\pi c}$ and the $\mathrm{i}\pi\cot(\dots)$s.

P.S. Although these are "analytic solutions", it's not clear that they are in any way useful. These were generated via

Assuming[{0<c<1},
  FullSimplify[
    Integrate[
      Evaluate[FullSimplify[
        Sin[a x]/Sin[b x] /.{a -> c, b -> 1-c}]],
      {x, 0, Pi/2}]]
]

and then the same thing, but with the rewrite rules changed to

... /. {a -> 2-c, b -> 1-c}]],
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  • $\begingroup$ @E.T. T.Y. for your input. These particular integrals arise in an entropy calculation I'm pursuing through the Boltzmann formulation. I do not have access to Mathematica, but would like to evaluate your result. I think the second vector input to the generalized hypergeometric function is not complete. Can you supply the correction? $\endgroup$
    – pstall
    Commented Jun 20 at 21:59
  • $\begingroup$ @pstall : It's a ${}_2F_1$, so $\vec{a}$ has two elements and $\vec{b}$ has one -- and I see "two elements, semicolon, one element, semicolon, argument", so the vectors are complete. $\endgroup$ Commented Jun 20 at 23:03
  • $\begingroup$ @pstall : Worth understanding dlmf.nist.gov/16.2#iii to know that this ${}_2F_1$ is convergent for all $0<c<1$, but convergence would fail if $c$ were an odd integer. Analytic continuation can give different values at the same input, so it's possible that the expression in this Answer gives values from a branch different than the one you want. If you want to take the limit $c \rightarrow 1$, start here. $\endgroup$ Commented Jun 20 at 23:14
  • $\begingroup$ @pstall you can use Mathematica free online $\endgroup$
    – Andrew
    Commented Jun 21 at 1:12

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