This question is a reference to Evaluation of $\displaystyle \int_{0}^{\pi}\ln(5-4\cos x)dx$ which was posted earlier by Juantheron. I tried to work it out in a slightly "easier" way than Eric did in his answer, but couldn't get exactly to the right answer. Here is what I did: I considered the integral $\displaystyle \int_{0}^{\pi}\ln(a-4\cos x)dx$ and took derivative wrt $a$: $I'(a)=\displaystyle \int_{0}^{\pi}\frac{dx}{a-4\cos x}$ Then I applied "Weierstrass sub" to get $$I'(a)= \int_{0}^{\infty}\frac{2dt}{(4+a)t^2+a-4}$$ To get this ready for the "arctan" I pull out $\frac{1}{4+a}$ to get $$I'(a)= \frac{1}{4+a}\int_{0}^{\infty}\frac{2dt}{t^2+\frac{a-4}{a+4}}=\frac{2}{4+a}\sqrt{\frac{a+4}{a-4}}\arctan\sqrt{\frac{a+4}{a-4}}t$$ ,from $0$ to infinity which gives: $\frac{\pi}{\sqrt{a^2-16}}=I'(a)$ So now $I(a)={\pi}\ln|a+\sqrt{a^2-16}|$ For $a=5$ I get ${\pi}\ln8$ as the answer. But the actual answer is ${\pi}ln4$. I don't know what I have done wrong. The only thing I can come up with, this that by integrating from $I'(a)$ to $I(a)$ there is a constant $C$ that might contribute? Can anyone help me if my approach is valid and how to get the right answer with my approach? This would be a great learning experience for me. Thanks!
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Yes. If you integrate $I'(a)$ you get $\pi\ln(a+\sqrt{a^2-16})+C$, so you still have to find $C$. That means evaluating $I(a_0)$ for some $a_0\ge4$. – almagest May 10 '16 at 16:07
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I am afraid then that this $C$ is not zero, so how do I do that? – imranfat May 10 '16 at 16:09
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Yes, $C\ne0$. Indeed, you know from the earlier question that $C=-\pi\ln2$. – almagest May 10 '16 at 16:12
1 Answers
Your solution was good up to $$I(a)=\pi\ln(a+\sqrt{a^2-16})+C$$ But you still have to evaluate $C$ and it's trickier than when $b$ was the variable because in that case you knew the value of the integral for $b=0$. This time there are no values of $a$ for which you know the value of the integral, but you still can get to $C$ via the slant asymptote. We propose to start with the approximation $I(a)\approx m\cdot\ln a+b$. To get $m$, note that $$\begin{align}m&=\lim_{a\rightarrow\infty}\frac{I(a)}{\ln a}=\lim_{a\rightarrow\infty}\frac{\int_0^{\pi}\ln(a-4\cos x)dx}{\ln a}\\ &=\lim_{a\rightarrow\infty}\frac{\int_0^{\pi}\left(\ln a+\ln(1-\frac{4\cos x}a)\right)dx}{\ln a}\\ &=\lim_{a\rightarrow\infty}\frac{\pi\ln a+\int_0^{\pi}\ln(1-\frac{4\cos x}a)dx}{\ln a}\\ &=\pi+\lim_{a\rightarrow\infty}\frac{\int_0^{\pi}\ln(1-\frac{4\cos x}a)dx}{\ln a}=\pi\end{align}$$ Because $$\lim_{a\rightarrow\infty}\ln\left(1-\frac{4\cos x}a\right)=0$$ Now we can find $b$ as $$\begin{align}b&=\lim_{a\rightarrow\infty}\left(I(a)-\pi\ln a\right)\\ &=\lim_{a\rightarrow\infty}\left(\pi\ln a+\int_0^{\pi}\ln\left(1-\frac{4\cos x}a\right)dx-\pi\ln a\right)\\ &=\lim_{a\rightarrow\infty}\int_0^{\pi}\ln\left(1-\frac{4\cos x}a\right)dx=0\end{align}$$ For the same reason. Thus we require $$\begin{align}\pi&=\lim_{a\rightarrow\infty}\frac{\pi\ln(a+\sqrt{a^2-16})+C}{\ln a}\\ &=\lim_{a\rightarrow\infty}\frac{\pi\ln a+\pi\ln\left(1+\sqrt{1-\frac{16}{a^2}}\right)+C}{\ln a}\\ &=\pi+\lim_{a\rightarrow\infty}\frac{\pi\ln\left(1+\sqrt{1-\frac{16}{a^2}}\right)+C}{\ln a}=\pi\end{align}$$ OK, but also $$\begin{align}0&=\lim_{a\rightarrow\infty}\pi\ln(a+\sqrt{a^2-16})+C-\pi\ln a\\ &=\pi\ln a+\pi\ln\left(1+\sqrt{1-\frac{16}{a^2}}\right)+C-\pi\ln a\\ &=\pi\ln2+C\end{align}$$ So now we have established that $C=-\pi\ln2$, so $$I(a)=\pi\ln(a+\sqrt{a^2-16})-\pi\ln2$$ And $$I(5)=\pi\ln(5+3)-\pi\ln2=\pi\ln4$$
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I really appreciate the work. But this makes me think that my solution wasn't all that quick at all. Finding the $C$ is harder than the actual problem, seems like. So I think I can't use my approach in my classroom I'm afraid. But at least I got one worry less at night :) – imranfat May 10 '16 at 16:21