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Evaluate the following integral: $$I=\int_0^{\pi} \ln(a^2-2a\cos x+1) \,dx$$

My attempt: Letting $t=\pi-x$

$$I=\int_0^{\pi} \ln(a^2+2a\cos t+1) \,dt$$ Adding both equations, $$2I=\int_0^{\pi} \ln(a^2+2a\cos x+1)((a^2-2a\cos x+1)) \,dx$$ $$\implies 2I=\int_0^{\pi}\ln(a^4-2a^2\cos 2x+1)dx$$

I don't think applying integration by parts will yield an answer and no substitution seems to be evident here. The final answer given at the end of the worksheet is $2\pi\ln a$. How do I go about evaluating this integral? Is there a clever method to find the primitive of this function?

Cognoscenti
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    https://math.stackexchange.com/questions/2573554/calculating-int-0-pi-log1-2a-cos-xa2-dx?noredirect=1 and https://math.stackexchange.com/questions/650513/computing-int-0-pi-ln-left1-2a-cos-xa2-right-dx – Svyatoslav Jan 30 '24 at 15:00
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    Ah, this is a duplicate, my bad. thanks a lot for pointing that out @Svyatoslav. – Cognoscenti Jan 31 '24 at 05:14

1 Answers1

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Suppose $a>0$. If $a>1$, $$\begin{eqnarray} I&=&\int_0^{\pi} \ln(a^2-2a\cos x+1) \,dx\\ &=&\frac12\int_0^{2\pi} \ln(a^2-2a\cos x+1) \,dx\\ &=&\frac12\int_0^{2\pi} \ln((a-\cos x)^2+\sin^2x) \,dx\\ &=&\int_0^{2\pi} \ln|a-(\cos x+i\sin x)| \,dx\\ &\overset{z=e^{ix}}=&\Re\int_{|z|=1} \ln(a-z) \frac1{iz}dz\\ &=&\Re\bigg[2\pi\text{Res}\bigg(\ln(a-z),z=0\bigg)\bigg]\\ &=&2\pi\ln a. \end{eqnarray}$$ If $a\in(0,1)$, set $b=\frac1a$ and hence $$\begin{eqnarray} I&=&\int_0^{\pi} \ln(a^2-2a\cos x+1) \,dx\\ &=&\int_0^{\pi} \ln\bigg[a^2(1-2b\cos x+b^2)\bigg] \,dx\\ &=&2\pi\ln a+\int_0^{\pi} \ln(b^2-2b\cos x+1) \,dx\\ &=&2\pi\ln a+2\pi\ln b\\ &=&0. \end{eqnarray}$$ If $a=1$, $$ I=\int_0^{\pi} \ln(2-2\cos x) \,dx=0. $$ You can treat the same way when $a<0$.

xpaul
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  • Uh, I got lost in the step after you converted it into the Riemann sum. What exactly do you do in the fifth and sixth line in your answer? – Cognoscenti Jan 31 '24 at 05:10
  • @Cognoscenti he converted it to a contour integral about the unit circle, and then applied the residue theorem. (Have you learned complex analysis yet? If not then rip) – Captain Chicky Jan 31 '24 at 09:22
  • I unfortunately haven't but I think I'll read up a bit on it and try my best to understand! – Cognoscenti Jan 31 '24 at 10:40