fun-looking log-cos integral

[galactus]

Show that:

$$\int_{0}^{\frac{\pi}{2}}\log\left(\frac{1+\cos^{2}\theta}{\sqrt{1+1/8\cos^{2}\theta}}\right)d\theta = \frac{\pi}{4}\ln(2)$$

[Tolaso J Kos]

Show that:

$$\int_{0}^{\frac{\pi}{2}}\log\left(\frac{1+\cos^{2}\theta}{\sqrt{1+1/8\cos^{2}\theta}}\right)d\theta = \frac{\pi}{4}\ln(2)$$

Hello C,

For starters, let us consider the function:

$$f(a)=\int_{0}^{\pi/2}\log \left ( 1+a \cos^2 \theta \right )\, {\rm d}\theta, \;\; a\geq 0$$

We differentiate $f(a)$ once , hence:

\begin{align*}
f'(a) &=\int_{0}^{\pi/2}\frac{\cos^2 \theta}{1+a\cos^2 \theta}\, {\rm d}\theta \\
&=\int_{0}^{\pi/2} \frac{{\rm d}\theta}{\frac{1}{\cos^2 \theta}+a} \\
&=\int_{0}^{\pi/2}\frac{{\rm d}\theta}{\tan^2 \theta +1 +a} \\
&=\frac{\pi}{2}\cdot \frac{1}{a+\sqrt{a+1}+1}
\end{align*}

Integrating back we see that:

\begin{equation} f(a)= \pi \log \left ( \sqrt{a+1}+1 \right ) - \pi \log 2 \end{equation}

Hence, for our integral, we have successively:

\begin{align*}
\int_{0}^{\pi/2}\log \left ( \frac{1+\cos^2 \theta}{\sqrt{1+\frac{1}{8}\cos^2 \theta}} \right )\, {\rm d}\theta &=\int_{0}^{\pi/2} \log \left ( 1+\cos^2 \theta \right )\, {\rm d}\theta - \frac{1}{2}\int_{0}^{\pi/2}\log \left ( 1+ \frac{1}{8}\cos^2 \theta \right )\, {\rm d}\theta \\
&\!\overset{(1) \Rightarrow a=1, \; a= \frac{1}{8}}{=\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\!} \; \pi \log \left ( \sqrt{2}+1 \right ) - \pi \log 2 -\frac{1}{2}\left [ \pi \log \left ( \sqrt{\frac{9}{8}}+1 \right ) - \pi \log 2 \right ] \\
&=\frac{\pi \log 2}{4}
\end{align*}

as claimed.

[Tolaso J Kos]

C, have a look at a fascinating result a failed attempt brought up:

\begin{align*}
f(1) &=\pi \log \left ( \sqrt{2}+1 \right ) -\pi \log 2 \\
&=\int_{0}^{\pi/2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1} \cos^{2n}\theta}{n} \, {\rm d}\theta \\
&= \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\int_{0}^{\pi/2} \cos^{2n} \theta \, {\rm d}\theta\\
&= \frac{\sqrt{\pi}}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\frac{\Gamma \left ( n+\frac{1}{2} \right )}{n!}\\
&=\frac{\pi}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}\frac{\left ( 2n \right )!}{4^n (n!)^2} \\
&= \frac{\pi}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{4^n n^2}\binom{2n}{n}
\end{align*}

Hmm.. I am pretty confident that a closed form of the general case:

$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}\binom{2n}{n}x^n $$

exists. Is anyone aware of?

[r9m]

Hmm.. I am pretty confident that a closed form of the general case:

$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}\binom{2n}{n}x^n $$

exists. Is anyone aware of?

We may start with $\displaystyle \sum\limits_{n=0}^{\infty} \binom{2n}{n}x^n = \frac{1}{\sqrt{1-4x}}$ dividing by $x$ on both sides and integrating gives us: $$\sum\limits_{n=1}^{\infty} \binom{2n}{n}\frac{x^n}{n} = -2\log \left(\frac{1+\sqrt{1-4x}}{2}\right)$$

and repeating the process and making the change of variable $y = \sqrt{1-4x}$:

\begin{align*}\sum\limits_{n=1}^{\infty} \binom{2n}{n}\frac{x^n}{n^2} &= -2\int_0^{x}\frac{1}{x}\log \left(\frac{1+\sqrt{1-4x}}{2}\right)\,dx\\&= 2\int_{1}^{\sqrt{1-4x}} \frac{2y}{1-y^2}\log \left(\frac{1+y}{2}\right)\,dy\\&= 2\int_1^{\sqrt{1-4x}}\frac{1}{1-y}\log \left(\frac{1+y}{2}\right)\,dy - \int_1^{\sqrt{1-4x}}\frac{2}{1+y}\log\left(\frac{1+y}{2}\right)\,dy\\&= -2\int_0^{1-\sqrt{1-4x}}\frac{1}{y}\log \left(1-\frac{y}{2}\right)\,dy - \left[\log^2 \left(\frac{1+y}{2}\right)\right]_1^{\sqrt{1-4x}}\\&= 2\operatorname{Li}_2 \left(\frac{1-\sqrt{1-4x}}{2}\right) - \log^2 \left(\frac{1+\sqrt{1-4x}}{2}\right)\end{align*}

Making the change of variable $x \mapsto -x$ should give us the desired series.

[Tolaso J Kos]

Now that we know the identity:

$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}\binom{2n}{n}x^n = \log^2 \left ( \frac{1+\sqrt{1+4x}}{2} \right ) - 2\operatorname{Li}_2 \left ( \frac{1- \sqrt{1+4x}}{2} \right ) \tag{2}$$

we can evaluate the value of $\displaystyle \operatorname{Li}_2 \left ( \frac{1-\sqrt{2}}{2} \right )$. Indeed, setting $x=\frac{1}{4}$ back at $(2)$ we have that:


$$\operatorname{Li}_2 \left ( \frac{1-\sqrt{2}}{2} \right ) = \frac{1}{2}\log^2 \left ( \frac{1+\sqrt{2}}{2} \right ) - \log \left ( \sqrt{2}+1 \right )+\log 2 $$


Interesting...