# How find this integral $\int_{0}^{\frac{\pi}{4}}\ln{(\sqrt{\sin^3{x}}+\sqrt{\cos^3{x}})}dx$

Yesterday my friend had sent me by email a math question, which when I first read, perplexed me a great deal. I've been working on it, but I can't solve it. I hope someone can help me. Thank you

show that

$$\int_{0}^{\frac{\pi}{4}}\ln{(\sqrt{\sin^3{x}}+\sqrt{\cos^3{x}})}dx=\dfrac{G}{12}-\dfrac{5\pi}{16}\ln{2}+\dfrac{\pi}{8}\ln{(2-\sqrt{2})}-\dfrac{\pi}{8}\ln{(2+\sqrt{2})}-\dfrac{\pi}{3}\ln{(\sqrt{3}-1)}+\dfrac{\pi}{3}\ln{(1+\sqrt{3})}$$

where $G$ is Catalan's constant

• Have you broken down the $a^3+b^3$ form and separated based on properties of $\ln$? – abiessu Oct 14 '13 at 9:55
• I have try it,But not usefull.Thank you – math110 Oct 14 '13 at 10:16
• Note that the answer can be simplified to $$\dfrac{G}{12}-\dfrac{5\pi}{16}\ln{2}+\dfrac{\pi}{8}\ln{(3-2\sqrt{2})}+\dfrac{\pi}{3}\ln{(2+\sqrt{3})}$$ – Bennett Gardiner Oct 14 '13 at 13:05
• After AstroSharp reduction, we can probably use the method shown here math.stackexchange.com/questions/426325/… – Paramanand Singh Oct 21 '13 at 6:22
• where did this integral come from? – cactus314 Oct 27 '13 at 17:05

## 1 Answer

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.

• That last integral is evaluated here. – Shobhit Bhatnagar Dec 24 '13 at 14:23
• Brilliant! Great work. – Bennett Gardiner Dec 27 '13 at 0:43