Abstract Generated abstract
The paper evaluates the definite integral of sin^m x divided by p plus q cos x over the interval from 0 to pi, for integer m at least 2 and parameters satisfying q not equal to zero and p squared minus q squared nonnegative. By transforming the integral to a beta type integral, analyzing convergence at the endpoint cases, and checking low dimensional examples, it shows that a formula reproduced in standard integral tables gives contradictory behavior. The author derives a recurrence relation connecting Phi_{m+2} and Phi_m, then uses the explicit cases m equals 2 and m equals 3 to obtain a general expression in terms of beta functions, logarithms, and elementary radicals.
Full Text
MATHEMATICS
K. V. BRODOVICH
ON THE INTEGRAL \(\displaystyle \int_0^\pi \frac{\sin^m x}{p+q\cos x}\,dx\)
(Presented by Academician A. N. Kolmogorov, 12 II 1958)
Let us denote
\[ \Phi_m=\int_0^\pi \frac{\sin^m x}{p+q\cos x}\,dx; \tag{1} \]
\(m\geqslant 2\) an integer; \(q\ne 0\); \(p^2-q^2\geqslant 0\).
The substitution \(\cos^2 \frac{x}{2}=z\) brings (1) to the form:
\[ \Phi_m=\frac{2^{m-1}}{q}\int_0^1 \frac{z^{\frac{m-1}{2}}(1-z)^{\frac{m-1}{2}}}{z+\beta}\,dz; \qquad \beta\equiv \frac{p-q}{2q}. \tag{2} \]
It is obvious that for \(p^2-q^2>0\) the integral (1) converges absolutely. For \(p^2-q^2=0\) the denominator of the integrand has one zero point at \(x=0\) or \(x=\pi\). But at the same time \(\sin x\) also vanishes. Thus, in the interval \(0\leq x\leq \pi\), for \(m\geq 2\) the integrand has a pole if \(p^2-q^2<0\), and has none if \(p^2-q^2=0\).
In particular, for \(p=q\), \(\beta=0\),
\[ \Phi_m=\frac{2^{m-1}}{q}\int_0^1 z^{\frac{m-3}{2}}(1-z)^{\frac{m-1}{2}}\,dz = \]
\[ =\frac{2^{m-1}}{q}\, \mathrm{B}\!\left(\frac{m-1}{2},\,\frac{m+1}{2}\right) = \frac{2^{m-1}}{q}\, \frac{\Gamma\!\left(\frac{m-1}{2}\right)\Gamma\!\left(\frac{m+1}{2}\right)} {\Gamma(m)}. \tag{3} \]
The case \(p=-q\) corresponds to \(\beta=-1\) and to a change of sign in the right-hand side of result (3). Using in (1) the substitution \(\pi-x=\alpha\), one can show that \(\Phi_m\) is an even function with respect to the parameter \(q\) and an odd function with respect to \(p\).
Thus, under the given conditions \(\Phi_m\) has a finite and definite value. However, the handbook of I. M. Ryzhik, Tables of Integrals, Sums, Series, and Products (2nd ed., 1948), with reference to the table of integrals in the collection by D. Bierens de Haan (Leiden, 1867),* gives for the integral (1) the result (see formula \(11_{64}\) on p. 180 and I. M. Ryzhik’s note on p. 116, second footnote):
\[ \Phi_m= -\frac{2\sqrt{\pi}}{m\,(p^2-q^2)^{\frac{m+1}{2}}}\, \frac{\Gamma\!\left(\frac{m+1}{2}\right)} {\Gamma\!\left(\frac{m}{2}\right)}. \tag{4} \]
* The integral under consideration is incorrectly evaluated also in the corrected American edition of 1939 (D. Bierens de Haan, Nouvelles tables d’intégrales définies, N. Y., 1939, p. 99, t. 63, No. 15).
As \(p^2-q^2 \to +0\), \(m \ge 2\), here \(\Phi_m \to +\infty\). In addition, solution (4) is represented by an even function with respect to \(p\). This contradicts the expectations following from a preliminary analysis of integral (1), and prompted us to doubt the correctness of formula (4) and of others based on it (for example, formula \(13_{64}\) in the same place), which for 90 years have been included in mathematical handbooks.
Indeed:
\[ \Phi_3=\frac{4}{q}\int_0^1 \frac{z(1-z)}{z+\beta}\,dz =\frac{4}{q}\left[(1+\beta)z-\frac12 z^2-\beta(1+\beta)\ln(z+\beta)\right]_0^1 = \]
\[ =2\,\frac{p}{q^2}+\frac1q\left(1-\frac{p^2}{q^2}\right)\ln\frac{p+q}{p-q}; \tag{5} \]
\[ \Phi_2=\frac{2}{q}\int_0^1 \frac{\sqrt{z(1-z)}}{z+\beta}\,dz =\frac{2p}{q^2}\left[ \sqrt{1-\frac{q^2}{p^2}}\arctg\sqrt{\frac{1-z}{z}\cdot\frac{p-q}{p+q}} \right. \]
\[ \left. -\arctg\sqrt{\frac{1-z}{z}}+\frac{q}{p}\sqrt{z(1-z)} \right]_0^1 =\pi\frac{p}{q^2}\left(1-\sqrt{1-\frac{q^2}{p^2}}\right), \tag{6} \]
and these results contradict solution (4) given in the handbooks.
Noting that
\[ \Phi_{m+2}=\frac{2^{m+1}}{q}\int_0^1 \frac{z^{\frac{m-1}{2}}(1-z)^{\frac{m-1}{2}}}{z+\beta} [(z+\beta)-\beta][(1+\beta)-(z+\beta)]\,dz = \]
\[ =\frac{2^{m+1}}{q}\left[ \int_0^1 (1+\beta)z^{\frac{m-1}{2}}(1-z)^{\frac{m-1}{2}}\,dz \right. \]
\[ \left. -\int_0^1 z^{\frac{m+1}{2}}(1-z)^{\frac{m-1}{2}}\,dz -\beta(1+\beta)\int_0^1 \frac{z^{\frac{m-1}{2}}(1-z)^{\frac{m-1}{2}}}{z+\beta}\,dz \right] = \]
\[ =2^m\frac{p}{q^2}B\left(\frac{m+1}{2},\,\frac{m+1}{2}\right) +\left(1-\frac{p^2}{q^2}\right)\Phi_m, \tag{7} \]
we easily obtain the general solution of the problem under consideration:
\[ \Phi_m = 2^{m-2}\frac{p}{q^2} \sum_{\nu=1}^{k} \left(\frac{p^2-q^2}{-4q^2}\right)^{\nu-1} B\left(\frac{m+1-2\nu}{2},\,\frac{m+1-2\nu}{2}\right) + \left(\frac{p^2-q^2}{-q^2}\right)^k A, \]
where
\[ A= \begin{cases} \dfrac{\pi p}{q^2}\left(1-\sqrt{1-\dfrac{q^2}{p^2}}\right), & \text{for } m=2k+2; \\[1.2em] \dfrac1q\ln\dfrac{p+q}{p-q}, & \text{for } m=2k+1,\ k\ge 1. \end{cases} \]
For \(m=2\) the solution is given by formula (6).
Received
11 X 1957