2
$\begingroup$

Hi I was wondering if someone can help me evaluate the following integral.

Show that if $-1 < x < 1$, then

$$\int_{0}^{\pi} \frac{\log{(1+x\cos{y})}}{\cos{y}}dy= \pi \arcsin{x} $$

thank you in advance.

$\endgroup$
3
$\begingroup$

Probably the most direct way to do this is by infinite series: expanding the logarithm, and interchanging the sum and integral, we have $$ I := \int_0^{\pi} \frac{\log{(1+x\cos{y})}}{\cos{y}} \, dy = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^n}{n} \int_0^{\pi} \cos^{n-1}{y} \, dy. $$ Firstly, note that $n-1$ has to be even for the integral to be nonzero, since $\cos{(\pi-y)}=-\cos{y}$. Therefore the sum is actually $$ \sum_{k=0}^{\infty} \frac{x^{2k+1}}{2k+1} 2\int_0^{\pi/2} \cos^{2k}{y} \, dy $$ Now we have to do the integral. Thankfully, Wallis was there before us, and the answer is $$ 2\int_0^{\pi/2} \cos^{2k}{y} \, dy = \frac{(2k)!}{2^{2k}(k!)^2} \pi. $$

Finally, identification of the power series, $$ \pi \sum_{k=0}^{\infty} \frac{x^{2k+1}}{2k+1} \frac{(2k)!}{2^{2k}(k!)^2} $$ is required. Well, what does the series for $\arcsin{x}$ look like? It's the integral of the series of $(1-x^2)^{-1/2}$, so should contain coefficients of the form $$ \frac{(-1)^k}{2k+1} \binom{-1/2}{k}, $$ so we just have to check that $$ (-1)^k \binom{-1/2}{k} = \frac{(2k)!}{2^{2k}(k!)^2} \tag{1} $$ to show that the series we have is that of $\arcsin{x}$. At this point I'm going to cheat a bit more: first, note that both sides of (1) are equal (to $1$) if $k=0$. Now look at the ratio of the $k$th to $(k+1)$th terms: $$ \frac{(-1)^{k+1} \binom{-1/2}{k+1}}{(-1)^k \binom{-1/2}{k}} = \frac{-(-1/2-k)}{k+1} = \frac{2k+1}{2(k+1)}, \\ \frac{(2(k+1))!/(2^{2(k+1)}((k+1)!)^2)}{(2k)!/(2^{2k}(k!)^2)} = \frac{(2k+2)(2k+1)}{(2(k+1))^2} = \frac{2k+1}{2(k+1)}, $$ and so by induction they are equal for all $k>0$, and hence $$ I = \pi \sum_{k=0}^{\infty} \frac{x^{2k+1}}{2k+1} \frac{(2k)!}{2^{2k}(k!)^2} = \pi \arcsin{x}. $$

$\endgroup$
  • $\begingroup$ Thank you for the response. I have some questions for clarifications, how did you show the first step involving the infinite series. Did you expand the infinite series of log and how did you deal with 1+ x*cos y and also 1/cos y? $\endgroup$ – Seth Mai Apr 27 '15 at 1:02
  • 1
    $\begingroup$ I used $\log{(1+z)} = \sum_{n=1}^{\infty} (-1)^{n-1}z^n/n$, with $z=x\cos{y}$. Notice that the $\cos{y}$ in the denominator then cancels a $\cos{y}$ from each term in the series. $\endgroup$ – Chappers Apr 27 '15 at 1:09
  • $\begingroup$ Oh my god. That is so cool. It feels such like a dirty trick when you evaluate integral with infinite series. Thank you so much. I am practicing for an exam on calculus on manifolds. This is one of the question from past exam. I tried plugging into maple and it did not work out well. $\endgroup$ – Seth Mai Apr 27 '15 at 1:11
3
$\begingroup$

Setting $$ f(x)=\int_0^\pi\frac{\log(1+x\cos y)}{\cos y}\,dy, \quad x\in (-1,1), $$ we have $f(0)=0$, and $$ f'(x)=\int_0^\pi\frac{\cos y}{(1+x\cos y)\cos y}\,dy=\int_0^\pi\frac{1}{1+x\cos y}\,dy\quad \forall x\in (-1,1) $$ Setting $$ t=\tan\frac{y}{2}, $$ we have $$ y=2\arctan t,\, \frac{1}{1+x\cos y}=\frac{1}{1+x\frac{1-t^2}{1+t^2}}=\frac{1+t^2}{1+x+(1-x)t^2} \,\mbox{ and } \, dy=2\frac{dt}{1+t^2}. $$ Now, for every $x\in (-1,1)$ we get: \begin{eqnarray} f'(x)&=&\int_0^\infty\frac{1+t^2}{1+x+(1-x)t^2}\cdot\frac{2}{1+t^2}\,dt=\frac{2}{1-x}\int_0^\infty\frac{1}{(1+x)(1-x)^{-1}+t^2}\,dt\\ &=&\frac{2}{\sqrt{1-x^2}}\arctan\left(\frac{t}{\sqrt{(1+x)(1-x)^{-1}}}\right)\Big|_0^\infty=\frac{2}{\sqrt{1-x^2}}\cdot\frac\pi2=\frac{\pi}{\sqrt{1-x^2}}. \end{eqnarray} Since $f(0)=0$, we deduce that $$ \int_0^\pi\frac{\log(1+x\cos y)}{\cos y}\,dy=f(x)=f(0)+\int_0^xf'(\xi)\,d\xi=\int_0^x\frac{\pi}{\sqrt{1-\xi^2}}\,d\xi=\pi\arcsin x. $$

$\endgroup$
  • $\begingroup$ Thank you for the alternative solution. $\endgroup$ – Seth Mai Apr 27 '15 at 5:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.