I couldn't evaluate this integral. Could you please help me?
$$\frac{1}{2\pi}\int _{-\pi}^\pi \frac{e^{-i\varphi}}{1-k\cos (\varphi)} \, \mathrm{d}\varphi,\text{ where $k$ is a constant}$$
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1What did you try? – Loki Clock Mar 09 '13 at 17:36
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When the integral exists it should be zero – Dominic Michaelis Mar 09 '13 at 17:44
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$|k|<1$ for convergence, but I don't get zero. – GEdgar Mar 09 '13 at 17:50
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1have you tried to use substitution $u = \tan(\dfrac{\phi}{2})$? Since you can cancel the imaginary part due to $\sin$ is odd, and $\cos$ is even. You problem turns out to be $\int_{-\pi}^{\pi}\dfrac{\cos(\phi)}{1-k\cos(\phi)}d\phi$ – Yimin Mar 09 '13 at 18:10
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yes i could't solve this part of integral(real part) – mary Mar 09 '13 at 18:14
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Just integrate $\int_0^{\pi} \dfrac{1}{1-k\cos(x)}dx$ – Yimin Mar 09 '13 at 18:16
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I've posted one of (currently) three answers, but so far I'm the only one who's up-voted the question. – Michael Hardy Mar 09 '13 at 18:31
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A typesetting oddity: \operatorname{d}\varphi results in extra space between the $\mathrm{d}$ and the $\varphi$. That's just how \operatorname ought to work. I changed it to \mathrm{d}\varphi. – Michael Hardy Mar 09 '13 at 18:32
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A related problem. – Mhenni Benghorbal Mar 09 '13 at 19:12
3 Answers
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$$|k|<1:$$
$$\begin{align*}\int_{-\pi}^{\pi}\frac{e^{-i\phi}}{1-k\cos \phi}d\phi &=\int_{-\pi}^{\pi}\frac{\cos \phi}{1-k\cos\phi}d\phi-i\underbrace{\int_{-\pi}^{\pi}\frac{\sin\phi}{1-k\cos\phi}d\phi}_{=\,0}\\&=2\int_{-\infty}^{\infty}\frac{1-t^2}{(1+t^2)(1+t^2-k(1-t^2))}dt\\[7pt]&=\frac{2}{k}\left(\int_{-\infty}^{\infty}\frac{1}{(k+1)t^2-k+1}dt-\int_{-\infty}^{\infty}\frac{1}{t^2+1}dt\right)\\[7pt]&=\frac{2}{k}\left(\left[\frac{1}{\sqrt{1-k^2}}\arctan\left(t\cdot\sqrt{\frac{1+k}{1-k}}\right)\right]_{-\infty}^{\infty}-\pi\right)\\[7pt]&=\frac{2\pi}{k}\left(\frac{1}{\sqrt{1-k^2}}-1\right)\end{align*}$$
Hence
$$\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{e^{-i\phi}}{1-k\cos \phi}d\phi=\frac{1}{k}\left(\frac{1}{\sqrt{1-k^2}}-1\right)$$
L. F.
- 8,498
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Let $z=e^{-i \phi}$ and use the residue theorem. The integral becomes, assuming $|k|<1$:
$$-\frac{1}{i 2 \pi} \frac{2}{k}\oint_{|z|=1} dz \frac{z}{z^2-\frac{2}{k}z+1}$$
when you use the fact that $\cos{\phi} = \frac{1}{2}(e^{i \phi}+e^{-i \phi})$.
This may be evaluated with the residue theorem. The poles of the integrand are at
$$z_{\pm} = \frac{1}{2} \left ( \frac{1}{k} \pm \sqrt{\frac{1}{k^2}-1}\right )$$
Note that only $z_-$ is inside the unit circle. By the residue theorem, the integral is equal to
$$-\frac{2}{k} \lim_{z \rightarrow z_-} \left[(z-z_-) \frac{z}{z^2-\frac{2}{k}z+1}\right]= -\frac{2}{k} \frac{z_-}{z_--z_+} = \frac{1}{k} \left ( \frac{1}{\sqrt{1-k^2}}-1 \right )$$
Ron Gordon
- 138,521
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Following up on Yimin's suggestion: $$ \int_{-\pi}^\pi \frac{e^{-i\varphi}}{1-k\cos\varphi} \, d\varphi = \int_{-\pi}^\pi \frac{\cos\varphi}{1-k\cos\varphi} \, d\varphi - i\int_{-\pi}^\pi \frac{\sin\varphi}{1-k\cos\varphi} \, d\varphi. $$ Since the second integral is that of an odd function over an interval symmetric about $0$, it is $0$, and we only need to work on the first integral.
The Weierstrass substitution is \begin{align} x & = \tan\frac\varphi2 \tag{1}\\[8pt] \frac{1-x^2}{1+x^2} & = \cos\varphi \tag{2}\\[8pt] \frac{2\,dx}{1+x^2} & = d\varphi\tag{3} \end{align} How to get $(2)$ and $(3)$ from $(1)$ using trigonometric identities could be discussed here (I wonder if anyone's ever posted that as a question?). There is also the identity $\dfrac{2x}{1+x^2}=\sin\varphi$, but we won't need that here.
We get $$ \int_{-\pi}^\pi \frac{\cos\varphi}{1-k\cos\varphi} \, d\varphi = \int_{-\infty}^{\infty}\frac{\frac{1-x^2}{1+x^2}}{1-k\frac{1-x^2}{1+x^2}} \cdot\frac{2\,dx}{1+x^2} $$ $$ = \int_{-\infty}^\infty \frac{1-x^2}{(1+x^2)-k(1-x^2)}\cdot\frac{2\,dx}{1+x^2} $$ $$ = \int_{-\infty}^\infty \frac{1-x^2}{(1-k)+(1+k)x^2}\cdot\frac{2\,dx}{1+x^2} $$ $$ = \int_{-\infty}^\infty \left( \frac{Ax+B}{(1-k)+(1+k)x^2} + \frac{Cx+D}{1+x^2} \right) \, dx $$ etc. The antiderivative should involve logarithms and/or arctangents.
