I need to a evaluate the following integral
$$I=\int_0^\infty\frac{\ln(x+4)}{\sqrt{x\,(x+3)\,(x+4)}}dx.$$
Both Mathematica and Maple failed to evaluate it in a closed form, and lookups of the approximate numeric value $4.555919963334436…$ in ISC+ and WolframAlpha did not return plausible closed form candidates either. Does anybody by any chance have an idea if a closed form exists for this integral, and what it could be?
Best Answer
Here is my solution:
Part 1
Let $$I(a)=\int_0^\infty\frac{\log(1+x^2 )}{\sqrt{(1+x^2 )(a^2+x^2 )}}dx\tag{1}$$ for $a>1$. Then rewrite the integral as $$ \begin{align*} I(a) &= \frac{1}{2}\int_1^\infty\frac{\log x}{\sqrt{x(x-1)(x+a^2-1)}}dx\quad x \mapsto \sqrt{x-1} \\ &= \frac{1}{2}\text{Re}\int_0^\infty\frac{\log x}{\sqrt{x(x-1)(x+a^2-1)}}dx \\ &= 2 \text{Re}\int_0^\infty \frac{\log x}{\sqrt{(x^2-1)(x^2+a^2-1)}}dx\quad x\mapsto x^2 \\ &\stackrel{\color{blue}{[1]}}{=} \text{Re} \left( -i K(a)\log\left(i\sqrt{a^2-1}\right)\right) \\ &= \text{Re} \left( K(a)\left( \frac{\pi}{2}-\frac{i}{2}\log(a^2-1)\right)\right) \end{align*} $$ where $K(k)$ denotes the complete elliptic integral of the first kind. Furthermore, it is known that $$K(k)=\frac{K\left(\frac{1}{k} \right)+iK'\left(\frac{1}{k} \right)}{k} \quad k>1$$ Thus, we obtain $$ \begin{align*} I(a)&= \text{Re} \left\{ \frac{K\left(\frac{1}{a} \right)+iK'\left(\frac{1}{a} \right)}{a}\left( \frac{\pi}{2}-\frac{i}{2}\log(a^2-1)\right)\right\} \\ &= \frac{\pi}{2a}K\left(\frac{1}{a} \right)+\frac{\log(a^2-1)}{2a}K'\left(\frac{1}{a}\right)\tag{2} \end{align*} $$
Part 2
Now, we turn our attention back to the original problem.
$$ \begin{align*} \int_0^\infty\frac{\log(x+4)}{\sqrt{x\,(x+3)\,(x+4)}}dx &=2 \int_0^\infty \frac{\log(4+x^2)}{\sqrt{(3+x^2)(4+x^2)}}dx \quad x\mapsto x^2 \\ &= 4\int_0^\infty \frac{\log(4+4x^2)}{\sqrt{(3+4x^2)(4+4x^2)}}dx \quad x\mapsto 2x \\ &= 2\int_0^\infty \frac{2\log(2)+\log(1+x^2)}{\sqrt{(4x^2+3 )(1+x^2)}}dx \tag{3} \end{align*} $$ The first integral is straightforward and it's value is $$\int_0^\infty \frac{1}{\sqrt{(4x^2+3)(1+x^2)}}dx\stackrel{\color{blue}{[2]}}{=} \frac{1}{2}K\left(\frac{1}{2} \right)$$ And the second integral can be evaluated as follows: $$ \begin{align*} &\; \int_0^\infty \frac{\log(1+x^2)}{\sqrt{(4x^2+3 )(1+x^2)}}dx \\ &= \int_0^\infty \frac{\log(1+x^2)-2\log(x)}{\sqrt{(4+3x^2 )(1+x^2)}}dx\quad x\mapsto 1/x \\ &\stackrel{\color{blue}{[1]}}{=}\frac{1}{\sqrt{3}}\int_0^\infty \frac{\log(1+x^2)}{\sqrt{\left(x^2+\frac{4}{3} \right)(1+x^2)}}dx+\frac{1}{2}\log\left(\frac{\sqrt{3}}{2} \right)K\left( \frac{1}{2}\right) \\ &= \frac{1}{\sqrt{3}}I\left(\frac{2}{\sqrt{3}} \right)+\frac{1}{2}\log\left(\frac{\sqrt{3}}{2} \right)K\left( \frac{1}{2}\right) \\ &\stackrel{\color{blue}{\text{eq }(2)}}{=} \frac{1}{\sqrt{3}}\left(\frac{\pi\sqrt{3}}{4}K\left(\frac{\sqrt{3}}{2}\right) -\frac{\log(3)\sqrt{3}}{4}K\left(\frac{1}{2}\right)\right)+\frac{1}{2}\log\left(\frac{\sqrt{3}}{2} \right)K\left( \frac{1}{2}\right) \\ &= \frac{\pi}{4}K\left(\frac{\sqrt{3}}{2}\right)-\frac{\log 2}{2}K\left(\frac{1}{2} \right) \end{align*} $$ Substituting everything in equation $(3)$, we get $$\int_0^\infty\frac{\log(x+4)}{\sqrt{x\,(x+3)\,(x+4)}}dx=K\left(\frac{\sqrt3}2\right)\frac\pi2+K\left(\frac12\right)\log(2)$$
Explanations
$\color{blue}{[1]}$ Refer to equation $(7.13)$ of this paper.
$\color{blue}{[2]}$ In general, we have $$\int_0^\infty \frac{1}{\sqrt{(1+x^2)(a^2+x^2)}}dx=\begin{cases}\displaystyle\frac{1}{a}K' \left(\frac{1}{a} \right)\quad \text{if }a>1 \\ K'(a) \quad \text{if } 0<a<1\end{cases}$$