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I am trying to find, or better calculate the following integral:

$\int_{0}^{\pi} \sin (nx)\arctan_2(1-r\cos x,r \sin x) dx$

Could someone help me out. For $r<1$, the $\arctan_2$ is just the normal $\arctan\frac{r \sin(x)}{(1-r \cos(x)}$, the argument is in this case always in the first or fourth quadrant. This is a tabulated standard integral and evaluates to $\frac{\pi}{2 n}r^n$. For $r>1$, I have not found it and was not successful in calculating it. Does someone have an idea what the answer is and how it is calculated. Thanks!

Quanto
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Thomas
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2 Answers2

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Note that, for $r>1$, the argument of $1-r\cos x +i r\sin x$ is $$g(x,r) = \left\{ \matrix{\sin nx \left(\pi+\tan^{-1}\frac{r \sin x}{1-r \cos x} \right)&\>\>\>\>\>\>0<x\le \cos^{-1}\frac1r\\ \sin nx \tan^{-1}\frac{r \sin x}{1-r \cos x} &\>\>\>\>\>\>\cos^{-1}\frac1r <x<\pi}\right. $$ which is continuous and differentiable. The integral is then $$I(r)= \int_{0}^{\pi} \sin nx \ g(x,r)\ dx$$

$$I’(r)=\int_{0}^{\pi} \frac{\sin nx \sin x}{1-2r \cos x +r^2} dx= \frac\pi{2r^{n+1}} $$ Thus $$I(r) = I(1)+\int_1^r I’(t)dt =\frac\pi{2n}+\frac\pi2\int_1^r \frac1{t^{n+1}}dt = \frac\pi{n}\left(1-\frac1{2r^n}\right) $$

Quanto
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It is a nice application of Feynman's trick $$I(r)=\int_0^{\pi } \sin (n x) \tan ^{-1}\left(\frac{r \sin (x)}{1-r \cos (x)}\right) \, dx$$ $$I'(r)=\int_0^{\pi }\frac{\sin (x) \sin (n x)}{r^2+1-2 r \cos (x)}\, dx=\frac 12\int_0^{\pi }\frac {\cos ((n-1) x)-\cos ((n+1) x) }{r^2+1-2 r \cos (x)}\, dx$$

Now, look here to see the role of the sign of $r$.

Claude Leibovici
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