Complex Analysis to solve this integral? $\int_0^{\pi/2} \frac{\ln(\sin(x))}{\sqrt{1 + \cos^2(x)}}\text{d}x$

Question

Complex Analysis time! I need some help in figuring out how to proceed to calculate this integral:

$$\int_0^{\pi/2} \frac{\ln(\sin(x))}{\sqrt{1 + \cos^2(x)}}\text{d}x$$

I tried to apply what I have been studying in complex analysis, that is stepping into the complex plane. So

$$\sin(x) \to z - 1/z$$ $$\cos(x) \to z + 1/z$$ Obtaining

$$\int_{|z| = 1} \frac{\ln(z^2-1) - \ln(z)}{\sqrt{z^4 + 3z^2 + 1}} \frac{\text{d}z}{i}$$

I found out the poles,

$$z_k = \pm \sqrt{\frac{-3 \pm \sqrt{5}}{2}}$$

But now I am confused: how to deal with the logarithms? Also, what when I have both imaginary and real poles?

I am a rookie in complex analysis so please be patient...

Answer 1

(Too long to post as a comment.)

Using the Fourier series

$$\ln(\sin(x)) = -\ln(2) - \sum_{k=1}^\infty \frac{\cos(2kx)}k$$

and reducing the power of $\cos(x)$ with the identity

$$\cos^2(x) = \frac{1+\cos(2x)}2$$

we have

$$\begin{align*} \int_0^{\pi/2} \frac{\ln(\sin(x))}{\sqrt{1+\cos^2(x)}} \, dx &= -\sqrt2\ln(2) \int_0^{\pi/2} \frac{dx}{\sqrt{3+\cos(2x)}} - \sum_{k=1}^\infty \frac{\sqrt2}k \int_0^{\pi/2} \frac{\cos(2kx)}{\sqrt{3+\cos(2x)}} \, dx \\[1ex] &= -\frac{\ln(2)}{\sqrt2} I_0 - \frac1{\sqrt2} \sum_{k=1}^\infty \frac{I_k}k \end{align*}$$

where for $\alpha\in\{0,1,2,\ldots\}$,

$$I_\alpha = \int_0^\pi \frac{\cos(\alpha x)}{\sqrt{3+\cos(x)}} \, dx$$

each of which apparently (according to Mathematica) evaluate to a linear combination of $E\left(\frac12\right)$ and $K\left(\frac12\right)$ ($K$ and $E$ denote the elliptic integrals of the first and second kind, respectively). Finding a closed form for $I_\alpha$ and in turn for your integral seems unlikely, though.

Answer 2

As already said in answers and comments, using $\sin(x)=t$ $$\int_0^{\frac \pi2} \dfrac{\log(\sin(x))}{\sqrt{1 + \cos^2(x)}}\,dx=\int_0^1 \frac{\log (t)}{\sqrt{(1-t^2)(2-t^2)} }$$

There is a nasty explicit solution in terms of the Gaussian hypergeometric function $$\, _2F_1\left(\frac{3}{4},\frac{3}{4};1;-8\right)$$ and its derivatives with respect to its first, second and third arguments.

Probably easier, would be to expand $$\frac{1}{\sqrt{(1-t^2)(2-t^2)} }=\frac 1{\sqrt 2} \sum_{n=0}^\infty a_n\, t^{2n}$$ where the $a_n$ form the sequence $$\left\{1,\frac{3}{4},\frac{19}{32},\frac{63}{128},\frac{867}{2048},\frac{3069}{ 8192},\frac{22199}{65536},\frac{81591}{262144},\frac{2428451}{8388608},\frac {9119601}{33554432},\frac{68993757}{268435456},\cdots\right\}$$ and use the fact that $$\int_0^1 t^{2n}\,\log(t)\,dt=-\frac{1}{(2 n+1)^2}$$ Using the coefficients listed above, the result is $$-\frac{14645443896353066473}{12947100671567462400 \sqrt{2}}=-0.799862$$ while the exact result is $-0.802496$.

Inverse symbolic calculators do not identify $$-0.80249561860378193216746878737636023438643059976849493589569975292440979\cdots$$ and the proposed closest approximation (for an absolute error of $2.904\times 10^{-8}$) is $$-\frac{ \sqrt{11} \left(19+3 \sqrt{3}\right)}{100}$$

Running my own trigonometric inverse calculator, for an absolute error of $2.529\times 10^{-9}$ $$-\cos \left(\frac{23 \pi }{195}\right) \cos \left(\frac{44 \pi }{259}\right)$$

Answer 3

$$\int_{0}^{1}\frac{\log(t)\,dt}{\sqrt{(1-t^2)(2-t^2)}} \stackrel{t\mapsto\sqrt{t}}{=} \frac{1}{4}\int_{0}^{1}\frac{\log(t)\,dt}{\sqrt{t(1-t)(2-t)}}\stackrel{t\mapsto 1-t}{=}\frac{1}{4}\int_{0}^{1}\frac{\log(1-t)}{\sqrt{t-t^3}}\,dt$$ equals $$ \frac{1}{2}\int_{0}^{1}\frac{\log(1-t^2)}{\sqrt{1-t^4}}\,dt. $$ By replacing $t$ with the inverse function of the primitive of $\frac{1}{\sqrt{1-t^4}}$ we have that this integral, which is also a harmonic-binomial series, is related to the Weierstrass products of the lemniscate elliptic functions.

I am sure Marco Cantarini is working on the subject at the current time, so it might be a good idea to summon him.

Indeed, after some simplification of Glaisher's result (given by the relations between the complete elliptic integral of the first kind and the $\text{AGM}$ mean) we have

$$ \frac{1}{2}\int_{0}^{1}\frac{\log(1-t^2)}{\sqrt{1-t^4}}\,dt =\color{red}{ \frac{\log(2)-\pi}{16\sqrt{2\pi}}\,\Gamma\left(\frac{1}{4}\right)^2}=-\frac{1}{2}\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}\cdot\frac{H_{2n+\frac{1}{2}}}{4n+1}.$$

Similarly

$$ \int_{0}^{1}\frac{\log(t)\,dt}{\sqrt{1-t^4}} = \color{red}{-\sqrt{\frac{\pi}{2}}\,\Gamma\left(\frac{5}{4}\right)^2}=-\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n(4n+1)^2}.$$

Answer 4

Following D'Aurizio hint, we have to deal with $$I=\frac{1}{2}\int_{0}^{1}\frac{\log\left(1-t^{2}\right)}{\sqrt{1-t^{4}}}dt=\frac{1}{2}\int_{0}^{K(i)}\log\left(1-\text{sinlem}\left(t\right)^{2}\right)dt$$ where $\text{sinlem}\left(t\right)$ is the Lemniscate sine function. Using the relation $\text{sinlem}\left(t\right)=\text{sn}\left(t;i\right)$ where $\text{sn}(z;k)$ is the Jacobi Elliptic sine, we get$$I=\frac{1}{2}\int_{0}^{K(i)}\log\left(1-\text{sn}\left(t;i\right)^{2}\right)dt=\int_{0}^{K(i)}\log\left(\text{cn}\left(t;i\right)\right)dt$$ and now it is enough to recall a classical result of Glaisher $$\int_{0}^{K(k)}\log\left(\text{cn}\left(t;k\right)\right)dt=-\frac{1}{4}\pi K^{\prime}\left(k\right)+\frac{1}{2}K\left(k\right)\log\left(\frac{k^{\prime}}{k}\right)$$ and so $$I=-\frac{1}{4}\pi K^{\prime}\left(i\right)+\frac{1}{2}K\left(i\right)\log\left(\frac{\sqrt{2}}{i}\right)$$ $$=\color{blue}{\frac{L}{8}\left(\log\left(2\right)-\pi\right)}=\color{red} {-0.8024956186037819...}$$ where $L/2$ is the Lemniscate constant.

Answer 5

@Hans-André-Marie-Stamm, I hope you don't mind that I was unable to solve this problem using Complex Analysis, but here's a method that relies on the Beta Function and some algebric work.

$$\begin{align}I&=\int_{0}^{\pi/2}\frac{\log\left(\sin(x)\right)}{\sqrt{1+\cos^2(x)}}dx;\ \cos(x)\rightarrow y\\&=\frac{1}{2}\int_{0}^{1}\frac{\log\left(1-y^2\right)}{\sqrt{1-y^4}}dy =\underbrace{\frac{1}{4}\int_{0}^{1}\frac{\log\left(1-y^4\right)}{\sqrt{1-y^4}}dy}_{I_1}+\underbrace{\frac{1}{4}\int_{0}^{1}\frac{\log\left(\frac{1-y^2}{1+y^2}\right)}{\sqrt{1-y^4}}dy}_{I_2}\end{align}$$

$$\begin{align}I_1=&\frac{1}{4}\underbrace{\int_{0}^{1}\frac{\log\left(1-y^4\right)}{\sqrt{1-y^4}}dy}_{y=z^{1/4}}=\frac{1}{16}\int_{0}^{1}z^{1/4-1}\frac{\log\left(1-z\right)}{\sqrt{1-z}}dz\\=&\frac{1}{16}\lim_{t \rightarrow 1/2}\frac{d}{dt}\mathfrak{B}\left(\frac{1}{4},t\right)=\frac{1}{16}\mathfrak{B}\left(\frac{1}{4},\frac{1}{2}\right)\left[\psi^{(0)}\left(\frac{1}{2}\right)-\psi^{(0)}\left(\frac{3}{4}\right)\right]\end{align}$$

$$\begin{align}I_2=&\frac{1}{4}\underbrace{\int_{0}^{1}\frac{\log\left(\frac{1-y^2}{1+y^2}\right)}{\sqrt{1-y^4}}dy}_{y=\sqrt{\frac{1-\sqrt{z}}{1+\sqrt{z}}}}=\frac{1}{32}\int_{0}^{1}z^{1/4-1}\frac{\log\left(z\right)}{\sqrt{1-z}}dz\\=&\frac{1}{32}\lim_{t \rightarrow 1/4}\frac{d}{dt}\mathfrak{B}\left(\frac{1}{2},t\right)=\frac{1}{32}\mathfrak{B}\left(\frac{1}{2},\frac{1}{4}\right)\left[\psi^{(0)}\left(\frac{1}{4}\right)-\psi^{(0)}\left(\frac{3}{4}\right)\right]\end{align}$$

Gathering both results: $$\begin{align}I&=\frac{1}{32}\mathfrak{B}\left(\frac{1}{2},\frac{1}{4}\right)\left[\psi^{(0)}\left(\frac{1}{4}\right)+2\psi^{(0)}\left(\frac{2}{4}\right)-3\psi^{(0)}\left(\frac{3}{4}\right)\right]\\&=\frac{1}{32}\frac{\Gamma\left(1/4\right)\Gamma\left(1/2\right)}{\Gamma\left(3/4\right)}\left[\psi^{(0)}\left(\frac{1}{4}\right)+2\psi^{(0)}\left(\frac{2}{4}\right)-3\psi^{(0)}\left(\frac{3}{4}\right)\right]\end{align}$$

This result can be simplified if one applies Gamma's Reflection Formula, and Digamma's Reflection and Multiplication Formulas, obtaining: $$I=\int_{0}^{\pi/2}\frac{\log\left(\sin(x)\right)}{\sqrt{1+\cos^2(x)}}dx=\frac{\log(2)-\pi}{16\sqrt{2\pi}}\Gamma^2\left(\frac{1}{4}\right)$$