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I want to evaluate the integral $\int_0^{\frac{\pi}{2}} \ln|\sin^2x-a^2|dx$ where $a^2\leqslant 1$.

And I have the Euler’s integral $\int_0^{\frac{\pi}{2}} \ln\sin x dx=-\frac{\pi}{2}\ln{2}$, but it seems that I have difficulties in substitution.

闫嘉琦
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3 Answers3

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Let $$ I(t)=\int_0^{\frac{\pi}{2}} \ln|t-\sin^2x|dx, t\ge1. $$ Then \begin{eqnarray} I'(t)&=&\int_0^{\frac{\pi}{2}} \frac{1}{t-\sin^2x}dx\\ &=&\int_0^{\frac{\pi}{2}} \frac{\csc^2x}{t\csc^2x-1}dx\\ &=&-\int_0^{\frac{\pi}{2}} \frac{1}{t(\cot^2x+1)-1}d\cot x\\ &=&-\int_{\infty}^{0} \frac{1}{t(u^2+1)-1}du\\ &=&\frac1t\int^{\infty}_{0} \frac{1}{u^2+\frac{t-1}t}du\\ &=&\frac1t\sqrt{\frac t{t-1}}\arctan(u\sqrt{\frac t{t-1}})\bigg|^{\infty}_{0}\\ &=&\frac{\pi}{2t}\sqrt{\frac t{t-1}}. \end{eqnarray} Now it is easy to obtain (I omit the detail) $$ I(t)-I(1)=\int_1^t\frac{\pi}{2t}\sqrt{\frac t{t-1}}dt=\text{arcsinh}(\sqrt{t-1}) $$ and hence $$ I(a^2)=I(1)+\text{arcsinh}(a^2-1)=-\pi\ln2+\text{arcsinh}(\sqrt{a^2-1}). $$

xpaul
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  • If $a\in(-1,1)$ then $\sqrt{a^2-1}$ is purely imaginary and $\text{arcsinh}(\sqrt{a^2-1})$ is purely imaginary too. Something is odd, since this is the outcome of a real integral. I got that $I(a^2)$ does not really depend on $a$ if $a\in(-1,1)$, and numerical experiments agree. – Jack D'Aurizio Nov 22 '17 at 18:34
  • @JackD'Aurizio, I thought that $a^2\ge1$. – xpaul Nov 22 '17 at 21:32
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You may just use Riemann sums.

For any $n\in\mathbb{N}^+$: $$ x^{2n}-1 = \prod_{k=1}^{2n}\left(x-\exp\frac{\pi i k}{n}\right) = (x^2-1)\prod_{k=1}^{n-1}\left(x^2+1-2x\cos\frac{\pi k}{n}\right).$$ If we choose $x\in\mathbb{C}$ in such a way that $\frac{x^2+1}{2x}=a$ holds, for instance through $x=a+i\sqrt{1-a^2}$,
we get: $$ \int_{0}^{\pi}\log(a-\cos\theta)\,d\theta = \lim_{n\to +\infty} \frac{\pi}{n}\log\prod_{k=1}^{n-1}\left(a-\cos\tfrac{\pi k}{n}\right) = \pi \log\left(\frac{a+i\sqrt{1-a^2}}{2}\right).$$ We may write the LHS as $\int_{0}^{\pi/2}\log(a-\cos\theta)+\log(a+\cos\theta)\,d\theta $ and the RHS as $-\pi\log 2 + i\pi\arccos(a)$. With some care in managing the determinations of the complex logarithm, $$ \int_{0}^{\pi/2}\log\left|a^2-\cos^2\theta\right|\,d\theta = \int_{0}^{\pi/2}\log\left|a^2-\sin^2\theta\right|\,d\theta = \color{blue}{-\pi\log 2} $$ for any $a\in(-1,1)$ is a straightforward consequence. This can be seen as a consequence of the properties of the Poisson kernel, too: have a look at page 55 of my notes.

Jack D'Aurizio
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  • Great as always. Just skimmed so far... I will give this a more thorough reading later. If it might ask, where do Riemann Sums come into this? Are you implicitly using the fact that logarithms convert products to sums? – Brevan Ellefsen Nov 22 '17 at 20:05
  • @BrevanEllefsen: in the second equation I am computing $\int \log$ through its definition as a Riemann integral, and yes, I am expoiting $\log\prod = \sum\log$, of course. – Jack D'Aurizio Nov 22 '17 at 20:09
  • Why the downvote? – Jack D'Aurizio Nov 25 '17 at 15:26
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For $a^2\le1$, let $a=\sin t$ where $t\in[0,\pi/2]$. Then \begin{eqnarray} I&=&\int_0^{\frac{\pi}{2}} \ln|\sin^2t-\sin^2x|dx\\ &=&\int_0^{\frac{\pi}{2}} \ln|(\sin t-\sin x)(\sin t-\sin x)|dx\\ &=&\int_0^{\frac{\pi}{2}} \ln|\sin(x-t)\sin(x+t)|dx\\ &=&\int_0^{\frac{\pi}{2}} \ln|\sin(x-t)|dx+\int_0^{\frac{\pi}{2}} \ln|\sin(x+t)|dx\\ &=&\int_{-t}^{\frac{\pi}{2}-t} \ln|\sin(x)|dx+\int_{t}^{\frac{\pi}{2}+t} \ln|\sin(x)|dx. \end{eqnarray} Let $$ I(t)=\int_{-t}^{\frac{\pi}{2}-t} \ln|\sin(x)|dx+\int_{t}^{\frac{\pi}{2}+t} \ln|\sin(x)|dx. $$ Clearly $$ I'(t)=-\ln|\sin(\frac{\pi}{2}-t)|+\ln|\sin(-t)|+\ln|\sin(\frac{\pi}{2}+t)|-\ln|\sin(t)|=0 $$ and hence $$ I(t)=I(0)=-\pi\ln 2.$$

xpaul
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