Logarithm integral in terms of beta function

Question

Question : Evaluate $$I =\int_{0}^{\infty} \frac{\ln(2x)}{\sqrt{x(x+1)(2x+1)}} \mathbb{d}x$$ in terms of beta function.

My approach :

Put $2x=z$ $$I=\int_{0}^{\infty} \frac{\ln(z)}{\sqrt{z(z+1)(z+2)}} \mathbb{d}z$$ then by $z \to \frac{1}{z}$ we get $$I = \int_{0}^{\infty} \frac{-\ln(z)}{\sqrt{z(z+1)(2z+1)}} \mathbb{d}z$$ so by this result and the original integral expression we get $$2I = \int_{0}^{\infty} \frac{\ln(2)}{\sqrt{x(x+1)(2x+1)}} \mathbb{d}x$$ Now put $x=z^2$ $$I=\ln(2) \int_0^{\infty} \frac{1}{\sqrt{(z^2+1)(2z^2+1)}} \mathbb{d}z$$ Now I was thinking to write as half of $\int_{-\infty}^{\infty}$ and then try to use semicircular contour but that is turning out to be quite messy, so how to proceed further...

Answer 1

Not using the beta function.

$$I=\int \frac{dz}{\sqrt{(z^2+1)(2z^2+1)}}$$ $$z=\frac 1{\sqrt 2}\sin(t) \quad \implies \quad I=\int \frac{\cos (t)}{\sqrt{\sin ^4(t)+3 \sin ^2(t)+2}}\,dt$$

Using elliptic integrals $$I=-i\, F\left(\left.i \,\text{csch}^{-1}\left(\sqrt{2} \csc (t)\right)\right|2\right)$$ Back to $z$ $$I=-i F\left(\left.i \,\text{csch}^{-1}\left(\frac{1}{z}\right)\right|2\right)$$

Using the bounds $$I=\int_0^\infty \frac{dz}{\sqrt{(z^2+1)(2z^2+1)}}=K(-1)$$

Answer 2

I found one way as follows, continuing from the form $$2I = \int_{0}^{\infty} \frac{\ln(2)}{\sqrt{x(x+1)(2x+1)}} \mathbb{d}x$$ Using $x \to \frac{1}{x}$ $$I = \frac{\ln(2)}{2} \int_0^{\infty} \frac{1}{\sqrt{x(x+1)(x+2)}} \mathbb{d}x$$ Now putting $x+1=z^2$ we get $$I = \ln(2) \int_1^{\infty} \frac{1}{\sqrt{z^4-1}}\mathbb{d}z=\ln(2) \int_1^{\infty} z^{-2}\left(1-z^{-4}\right)^{-\frac{1}{2}}\mathbb{d}z$$ putting $z^{-4}=t$ $$I=\frac{\ln(2)}{4}\int_{0}^{1} t^{-\frac{3}{4}}\left(1-t\right)^{-\frac{1}{2}}\mathbb{d}t$$ Hence we get $$\fbox{$I = \frac{\ln2}{4} B\left(\frac{1}{2},\frac{1}{4}\right)$}$$