# Integral $\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin{2x}})dx$

Prove that $$\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin{2x}})dx =\frac{\pi}{4} \ln2$$

I tried to use King's rule and to scale by $$2$$ and then to add the integrals, to get product of terms and use the result $$\int_{0}^{\frac{\pi}2} \ln(\sin{x})dx=\int_{0}^{\frac{\pi}2} \ln(\cos{x})dx=-\frac{\pi}2\ln2$$ but it didnt work. Any help?

• @Zacky i was experimenting with mathematica ! – Grentouce Jul 25 '19 at 12:02
• Standard techniques will work. – Axion004 Jul 25 '19 at 12:02
• @Axion004 like what?that sqaure root is annoying...help plz – Grentouce Jul 25 '19 at 12:03

$$I=\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin(2x)})dx =\frac12 \int_0^\frac{\pi}{2} x'\ln(\sin x+\cos x+\sqrt{\sin (2x)})dx$$ $$\overset{IBP}=\frac12 \int_0^\frac{\pi}{2}x\,\frac{\sin x-\cos x}{\sqrt{\sin(2x)}}dx\overset{x=\arctan t}=\frac{1}{2\sqrt 2}\int_0^\infty \frac{\arctan t}{1+t^2}\frac{t-1}{\sqrt t}dt$$

$$I(a)=\int_0^\infty \frac{\arctan(at)}{1+t^2}\frac{t-1}{\sqrt t}dt\Rightarrow I'(a)=\int_0^\infty \frac{(t-1)\sqrt t}{(1+a^2 t^2)(1+t^2)}dt$$ $$\overset{t=x^2}=\frac{2}{1-a^2}\int_0^\infty \frac{1+a^2 x^2}{1+a^2 x^4}dx-\frac{2}{1-a^2}\int_0^\infty \frac{1+x^2}{1+x^4}dx=\frac{\pi}{\sqrt 2}\frac{1-\sqrt a}{\sqrt a (1+a)(1+\sqrt a)}$$

$$\Rightarrow I=\frac{\pi}{4}\int_0^1 \frac{1-\sqrt a}{\sqrt a (1+a)(1+\sqrt a)}da\overset{\sqrt a=x}=\frac{\pi}{2}\int_0^1 \frac{1-x}{(1+x^2)(1+x)}dx=\frac{\pi}{4}\ln 2$$

• Thanks a lot!!! – Grentouce Jul 25 '19 at 12:07
• Yeah its clear! – Grentouce Jul 25 '19 at 12:08
• @TheCoolDrop if we consider $f(x)=\ln(\sin x+\cos x+\sqrt{\sin(2x)})$ then notice that: $$\int_0^\frac{\pi}{2} f(x)dx=\int_0^\frac{\pi}{4} f(x) dx +\int_\frac{\pi}{4}^\frac{\pi}{2} f(x) dx$$ Now let $\frac{\pi}{2}-x =t$ and notice that $f(\pi/2-t)=f(t)$ to get: $$\int_\frac{\pi}{4}^\frac{\pi}{2} f(x) dx =\int_0^\frac{\pi}{4} f(t) dt\Rightarrow \int_0^\frac{\pi}{2} f(x)dx=2\int_0^\frac{\pi}{4} f(x) dx$$ – Zacky Jul 25 '19 at 12:50
• Very elegant solution, for sure. – Claude Leibovici Jul 25 '19 at 12:59
• With the substitution $\frac{\pi}{2}-x=t\Rightarrow dx=-dt$ we get: $$\int_\frac{\pi}{4}^\frac{\pi}{2}f(x)dx=\int_\frac{\pi}{4}^0 f\left(\frac{\pi}{2}-t\right)(-dt)=\int_0^\frac{\pi}{4} f\left(\frac{\pi}{2}-t\right)dt$$ – Zacky Jul 25 '19 at 14:09

Here is a different solution using complex method. We can notice that

\begin{align*} I = &\int_0^{\frac \pi 4} \ln\left(1+\tan x + \sqrt{2\tan x} \right) dx + \int_0^{\frac\pi 4} \ln(\cos x) dx \\ = &I_1 + I_2. \end{align*} Evaluation of $$I_1$$ : Let $$\displaystyle\omega = e^{\frac {\pi i} 4} =\frac{1+i}{\sqrt 2}$$ and make substitution $$\tan x= y^2$$ to find that \begin{align*} I_1 =& \int_0^1 \log\left((1+\omega y)(1+\bar \omega y)\right)\frac{2y }{1+y^4}dy \\ =& 4 \Re \int_0^1 \frac{y \log\left(\frac{1+\omega y}2 \right)}{1+y^4} dy +\frac {\pi\ln 2}2. \end{align*} Let $$\omega y = z$$. Then using $$\omega^2 = i, \omega^4 = -1$$, \begin{align*} \Re \int_0^1 \frac{y \log\left(\frac{1+\omega y}2 \right)}{1+y^4} dy =& \Re \int_0^\omega \frac{ \frac{z}{\omega} \log\left(\frac{1+z}{2}\right)}{1+ (\frac z \omega)^4} \frac{dz}\omega \\ =& \Im \int_0^\omega \frac{z \log\left(\frac{1+z}{2}\right)}{1-z^4} dz \\ =& \frac 1 2 \Im \int_{-\omega}^\omega \frac{z \log\left(\frac{1+z}{2}\right)}{1-z^4} dz. \end{align*} Note that $$\displaystyle f(z) = \frac{z \log\left(\frac{1+z}{2}\right)}{1-z^4}$$ is analytic in $$|z|<1$$, and continuous in $$|z|\le 1$$ except at points $$\displaystyle -1, \pm i$$. So by choosing a contour $$z = e^{i\theta}, -\frac \pi 4\le \theta \le \frac \pi 4$$ and using $$\log(1+e^{i\theta}) = \log\left(2\cos\left(\frac \theta 2\right)\right) + \frac {i\theta}{2}$$, etc, we get \begin{align*} \Im \int_{-\omega}^\omega \frac{z \log\left(\frac{1+z}{2}\right)}{1-z^4} dz =&\Im \int_{-\frac \pi 4}^{\frac \pi 4} \frac{ie^{i2\theta}\left(\log(\cos (\theta/2)) + i\theta/2\right)}{1-e^{i4\theta}} d\theta\\ =&-\Im \int_{-\frac \pi 4}^{\frac \pi 4} \frac{\log(\cos (\theta/2)) + i\theta/2}{2\sin(2\theta)} d\theta\\ =& -\frac 1 4\int_{-\frac \pi 4}^{\frac \pi 4} \frac{\theta}{\sin(2\theta)} d\theta\\ =& -\frac 1 4 \int_{0}^{\frac \pi 4} \frac{\theta}{\sin\theta \cos \theta} d\theta\\ =:& -\frac 1 4 J \end{align*} (In fact $$J = G$$, the Catalan's number.) This gives $$I_1 = -\frac 1 2 J +\frac {\pi \ln 2}2.$$

Evaluation of $$I_2$$ : By integrating by parts, we have \begin{align*} I_2 =& x\ln(\cos x)|^{\frac \pi 4}_0 +\int_0^{\frac \pi 4} \frac {x \sin x}{\cos x} dx \\ =&-\frac{\pi \ln 2}{8} +K. \end{align*} We observe that \begin{align*} K-J = & -\int_0^{\frac \pi 4} \frac{x\cos x}{\sin x} dx\\ =& -x\log(\sin x)|^{\frac \pi 4}_0 + \int_0^{\frac \pi 4} \log(\sin x) dx\\ =& \frac{\pi \ln 2}{8} + \int_{\frac \pi 4}^{\frac\pi 2} \log(\cos x) dx \tag{\frac \pi 2 - x\mapsto x}\\ =&-\frac{3\pi \ln 2}{8} -I_2 \end{align*} thus $$\displaystyle K = -\frac{3\pi \ln 2}{8} -I_2 +J$$ and $$I_2 = -\frac{\pi \ln 2}{2} -I_2 +J \Longrightarrow I_2 = \frac 1 2 J -\frac{\pi \ln 2}{4}.$$ Therefore, we have $$I = I_1 + I_2 = \left(-\frac 1 2 J +\frac{\pi\ln 2}2\right)+\left(\frac 1 2 J -\frac{\pi \ln 2}{4}\right) = \frac {\pi \ln 2}4.$$

• Wow! Amazing one! Thanks a lot! – Grentouce Jul 27 '19 at 7:30
• I hope it was helpful! :) – Song Jul 29 '19 at 3:50
• It was helpful :) – Grentouce Jul 29 '19 at 4:25