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We are given that $$I(m)=\int_{0}^{\pi} \ln\left(1-2m\cos (x)+m^2\right)\,dx.$$ I could see that there weren't any standard techniques to calculate this integral directly so I concluded that there must be some kind of reduction formula to be derived.

What I did was to apply the King property in definite integrals ie. $$\int_{a}^{b}f(x)\,dx=\int_{a}^{b}f(a+b-x)\,dx.$$

This gave me $$I(m)=\int_{0}^{\pi}\ln\left(1+2m\cos(x)+m^2\right)\,dx,$$ then I added the two expressions for $I(m)$ to get $$2I(m,x)=\int_{0}^{\pi}\ln\left(1-2m^2\cos(2x)+m^4\right)\,dx,$$ i.e.,$$2I(m,x)=I(m^2,2x)$$.

So I thought that this should give us the value of $I(9)/I(3)=2$ and that was the correct answer, too; but what bothers me is that the second expression has $2x$ instead of $x$. So shouldn't that cause a change of limits and thus a problem?

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  • $\begingroup$ It is $$\frac{\pi\log(81)}{\pi\log(9)}$$ $\endgroup$ Jul 15, 2019 at 18:13
  • $\begingroup$ @Dr.SonnhardGraubner How did you calculate the exact values? $\endgroup$ Jul 15, 2019 at 18:14
  • $\begingroup$ $$\pi\log(1-2m+m^2)$$ is the general solution. $\endgroup$ Jul 15, 2019 at 18:17
  • $\begingroup$ Your integral is a function of $m$, not $x$. So what you have is $2I(m)=I(m^2)$. $\endgroup$
    – Anurag A
    Jul 15, 2019 at 18:21
  • $\begingroup$ @AnuragA But from the definition of $I(m)$ it should've been $\cos x$ instead of $cos2x$ $\endgroup$ Jul 15, 2019 at 18:23

1 Answer 1

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\begin{align*} 2I(m)&=\int_{0}^{\pi} \ln\left(1-2m\cos (x)+m^2\right)+\ln\left(1+2m\cos (x)+m^2\right)\,dx\\ &=\int_{0}^{\pi}\ln\left(1-2m^2\cos 2x+m^4\right)\, dx\\ &=\frac{1}{2}\int_{0}^{\color{red}{2\pi}}\ln\left(1-2m^2\cos t+m^4\right)\, dt && (\text{let }t=2x) \end{align*}

Now use the fact that if $f(2a-x)=f(x)$, then $$\int_0^{2a}f(x) \, dx =2\int_0^af(x) \, dx$$

to get \begin{align*} 2I_m & =\frac{1}{2}\int_{0}^{2\pi}\ln\left(1-2m^2\cos t+m^4\right)\, dt\\ & = \int_{0}^{\color{red}{\pi}}\ln\left(1-2m^2\cos t+m^4\right)\, dt\\ &=I(m^2). \end{align*} Thus $$\frac{I(m^2)}{I(m)}=2.$$

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    $\begingroup$ Really beatiful answer. $\endgroup$
    – Dog_69
    Jul 15, 2019 at 18:48
  • $\begingroup$ @Dog_69 Thanks. $\endgroup$
    – Anurag A
    Jul 15, 2019 at 18:48

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