Yesterday my friend emailed me a math question, which perplexed me greatly when I first read it. I've been working on it, but I can't solve it.
Show that
$$\int_{0}^{\frac{\pi}{4}}\ln\left(\sqrt{\sin^3{x}}+\sqrt{\cos^3{x}}\right)dx=\dfrac{G}{12}-\dfrac{5\pi}{16}\ln{2}+\dfrac{\pi}{8}\ln{(2-\sqrt{2})}-\dfrac{\pi}{8}\ln{(2+\sqrt{2})}-\dfrac{\pi}{3}\ln{(\sqrt{3}-1)}+\dfrac{\pi}{3}\ln{(1+\sqrt{3})}$$
where $G$ is Catalan's constant
I hope someone can help me. Thank you.
We can attempt this using some results of questions that have been asked lately. I will lay out some partial results that hopefully someone else can use. First, note that \begin{align} \int_{0}^{\pi/4}\ln{(\sqrt{\sin^3{x}}+\sqrt{\cos^3{x}})}\ \mathrm{d}x & = \int_{0}^{\pi/4}\ln{(\tan^{3/2}(x)+1)}\ \mathrm{d}x + \frac{3}{2}\int_{0}^{\pi/4}\ln{\cos(x)}\ \mathrm{d}x\\ \end{align} The second integral can be found here, giving us that $$ \int_{0}^{\pi/4}\ln{\cos(x)}\ \mathrm{d}x = \frac{1}{4} (2G-\pi\log 2), $$ where $G$ is Catalan's constant. So you can see we are at least recovering some pieces.
For the first integral, we try a substitution $z = \tan^{1/2}(x)$, such that $$ \int_{0}^{\pi/4}\ln{(\tan^{3/2}(x)+1)}\ \mathrm{d}x = \int_{0}^{1} \frac{2z\ln{(1+z^3)}}{1+z^4}\ \mathrm{d}z. $$ Now, integrals like this are usually amenable to series expansion, like in this recently resolved question, so expanding, \begin{align} \int_{0}^{1} \frac{2z\ln{(1+z^3)}}{1+z^4}\ \mathrm{d}z &= 2\int_{0}^{1} \displaystyle \sum_{k=1}^{\infty} \frac{(-1)^{k+1}z^{3k+1}}{k} \displaystyle \sum_{n=0}^{\infty} (-1)^n z^{4n}\ \mathrm{d}z \\ &= \displaystyle \sum_{k=1}^{\infty}\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^{k+n+1}}{k(4n+3k+2)} \end{align} Unfortunately, as in the link that @Paramanand Singh posted, these double sums seem very difficult to evaluate. Unless there is some other integral transform for your special case, it appears we may have to use those number theoretic techniques.
Note that \begin{align} I&=\int_{0}^{\frac{\pi}{4}}\ln{\left(\sqrt{\sin^3{x}}+\sqrt{\cos^3{x}}\right)}dx\\ &=\frac12 \int_{0}^{\frac{\pi}{2}}\ln{\left(\sqrt{\sin^3{x}}+\sqrt{\cos^3{x}}\right)}dx\\ &=\frac34 \int_0^{\frac\pi2}\ln (\cos x)dx+\frac12 \int_0^{\frac\pi2}\ln \left(\sqrt{\tan^3x}+1\right)\overset{t^2=\tan x}{dx}\\ &=-\frac{3\pi}8\ln2+\int_0^\infty \frac{t\ln(t^2+1)}{t^4+1}dt +\int_0^\infty \frac{t\ln(t+1)}{t^4+1}dt + \int_0^\infty \frac{t\ln\frac{t^2-t+1}{t^2+1}}{t^4+1}dt \tag1 \end{align} Evaluate \begin{align} &\int_0^\infty \frac{t\ln(t+1)}{t^4+1}dt \\ =&\int_0^\infty \int_0^1 \frac{t^2}{(t^4+1)(yt+1)}dy\ dt\\ =& \int_0^1\left( \frac{y\ln y}{1+y^4} +\frac\pi4\frac{y^3}{1+y^4} +\frac\pi{2\sqrt2}\frac{1-y^2}{1+y^4}\right)dy\\ =&-\frac14G+\frac\pi{16}\ln2+\frac\pi4\ln(\sqrt2+1)\tag2\\ \\ &\int_0^\infty \frac{t\ln\frac{t^2-t+1}{t^2+1}}{t^4+1}dt \\ =&\int_0^\infty \int_0^{-\frac\pi6} \frac{2t^2\cos y}{(t^4+1)(t^2+2t\sin y+1)}dy\ dt\\ =& \int_0^{-\frac\pi6} \left( \frac\pi{\sqrt2}\frac{\cos y}{\cos2y} -\frac\pi4\tan2y+\frac{2y-\pi}{2\cos2y}+\right)dy\\ =& -\frac\pi2\ln(\sqrt2+1)-\frac\pi8\ln2 +\left(\frac{2\pi}{3}\ln\frac{\sqrt3+1}{\sqrt2}-\frac16G\right)\tag3\\ \end{align} where $\int_0^{\frac\pi{12}}\ln(\tan x)dx=-\frac{2G}{3}$ is used in evaluating the third integral in (3). Substitute (2), (3) and $\int_0^\infty \frac{t\ln(t^2+1)}{t^4+1}dt =\frac12G+\frac\pi8\ln2$ into (1) to obtain $$I=\frac1{12}G+\frac{2\pi}3\ln(\sqrt3+1)-\frac\pi4\ln(\sqrt2+1)-\frac{31\pi}{48}\ln2 $$
