# $\int_0^\infty \frac{\ln\left(1+x-\sqrt{2x}\right)}{1+x^2}\,dx$

I was facing: \begin{align}J=\int_0^\infty \frac{\ln\left(1+x-\sqrt{2x}\right)}{1+x^2}\,dx\end{align}

I want to prove that $$\displaystyle J=0$$, or equivalently, that, \begin{align}\int_0^1 \frac{\ln\left(1+x-\sqrt{2x}\right)}{1+x^2}\,dx=-\dfrac{1}{2}\text{G}\end{align}

$$\text{G}$$ being the Catalan constant.

I know, using so-called Feynman's trick, how to prove this. I would like to obtain a proof, using only integration by parts and change of variable in simple integrals (that is, no multiple integrals) I don't know, if, under these restrictions, such computation is possible.

NB: You're probably wondering what is the link between : $$\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin{2x}})dx$$ and $$J$$.

The link is, for $$x\in\left[0;\frac{\pi}{2}\right]$$, $$(\sin{x}+\cos{x}+\sqrt{\sin{2x}})(\sin{x}+\cos{x}-\sqrt{\sin{2x}})=1$$ and $$\sin(2x)=2\sin x\cos x$$

• Have you made anymore progress with this one? – Zacky Dec 2 '19 at 15:29
• Oh, i have forgotten this one :( – FDP Dec 2 '19 at 15:42

I will focus on the integral $$\int_0^1 \frac{\ln(1+x-\sqrt{2x})}{1+x^2}dx$$.

I am not sure, if it is straightforward enough, but I used only simple substitutions and some of your hints. First split up the integral:

$$\int_0^1 \frac{\ln(1+x-\sqrt{2x})}{1+x^2}dx=\int_0^1 \frac{\ln(\sqrt{x}(1/\sqrt{x}+\sqrt{x}-\sqrt{2})}{1+x^2}dx=$$ $$=\frac{1}{2}\int_0^1\frac{\ln x}{1+x^2}dx + \int_0^1\frac{\ln(\sqrt{x}-\sqrt{2}+1/\sqrt{x})}{1+x^2}dx$$

And investigate the two integrals separately.

1. The first integral is by definition of Catalan constant equal to $$-\frac{1}{2}G$$.

2. The second integral can be broken up further: Substitute $$\arctan x = u$$ we get $$\int_0^{\pi/4}\ln (\sqrt{\tan u}+\sqrt{cotan u}-\sqrt{2})du=$$ $$= \int_0^{\pi/4}\ln\left(\sqrt{\frac{\sin u}{\cos u}} + \sqrt{\frac{\cos u}{\sin u}} - \sqrt{2}\right)du$$ $$= \int_0^{\pi/4}\ln\frac{\sin u + \cos u - \sqrt{\sin 2u}}{\sqrt{\sin u\cos u}}du$$

Next trick is to use the identity (which you mention), i.e., $$(\sin u + \cos u - \sqrt{\sin 2u})(\sin u + \cos u + \sqrt{\sin 2u}) = 1.$$

$$\int_0^{\pi/4}\ln\frac{\sin u + \cos u - \sqrt{\sin 2u}}{\sqrt{\sin u\cos u}}du = -\int_0^{\pi/4}\ln\left(\sqrt{\cos u \sin u}(\sin u + \cos u + \sqrt{\sin 2u})\right)du =$$ $$= -\frac{1}{2}\int_0^{\pi/4}\ln(\sin u\cos u)du - \int_0^{\pi/4}\ln(\sin u + \cos u + \sqrt{\sin 2u})du.$$

The first integral: $$\int_0^{\pi/4}\ln\left(\sin u \cos u\right)du = \int_0^{\pi/4}\ln\sin u du+\int_0^{\pi/4}\ln \cos u du = \int_0^{\pi/4}\ln\sin u du+\int_0^{\pi/4}\ln \sin (\pi/2 - u)du = \int_0^{\pi/4}\ln\sin u du+\int_{\pi/4}^{\pi/2}\ln\sin u du = \int_0^{\pi/2}\ln\sin u du.$$

It is easy to show (e.g. computing-the-integral-of-log-sin-x) that $$\int_0^{\pi/2}\ln\sin u du = -\frac{\pi}{2}\ln 2$$.

The second part of the integral is mentioned in your link above (integral-int-0-frac-pi4-ln-sinx-cosx-sqrt-sin2xdx) and is equal to:

$$\int_0^{\pi/4}\ln(\sin u + \cos u + \sqrt{\sin 2u})du = \frac{\pi}{4}\ln2$$

Putting everything together leads to (as expected):

$$\int_0^1 \frac{\ln(1+x-\sqrt{2x})}{1+x^2}dx=$$ $$=\frac{1}{2}\int_0^1\frac{\ln x}{1+x^2}dx + \int_0^1\frac{\ln(\sqrt{x}-\sqrt{2}+1/\sqrt{x})}{1+x^2}dx = -\frac{1}{2}G + \frac{1}{2}\frac{\pi}{2}\ln 2 - \frac{\pi}{4}\ln 2 = -\frac{1}{2}G.\ \square$$

Sketch of a possible proof: By roots of unity and partial fractions one may write $$I=\int_0^1 \frac{\log{(1+x-\sqrt{2x})}}{1+x^2} dx = 2\int_0^1 \frac{x}{1+x^4} \log{(1+x^2-\sqrt{2}{x} )}$$

$$= 2\int_0^1 \frac{dx \quad x}{(x+w)(x-w)(1+1/w)(x-1/w)}\log{((x-w)(x-1/w))} \quad, \quad w=e^{i\pi/4}$$

$$= \frac{w^2}{(w^2+1)(w^2-1)}\int_0^1 \big(\log{(x-w)} + \log{(x+w)} \big)\Big(\frac{1}{x+w}+\frac{1}{x-w} - \frac{1}{x+1/w} - \frac{1}{x-1/w}\Big)$$

The factor in front of the integral is $$-i/2.$$ Integrals of the form

$$\int dx \frac{ \log{(x-a)} }{x-b} = \log{(x-a)}\log{(1+(x-a)/(b-a))} + \text{Li}_2((x-a)/(b-a))$$

and when $$b=a,$$ the integral is a square of a logarithm, that is, there is no dilogarithm involved. There's a lot of symmetry involved, so maybe the mess can be simplified. There are also dilog identities which can be invoked. I suspect this to work because it is known that $$\text{Li}_2(i) = i G - \pi^2/48.$$ Using dilog identities is probably not an easy way to get the answer.

• I was going to post the same message. ;) – FDP Dec 3 '19 at 8:25
• Yes, I read the bold statement, and the proposed solution does not use multiple integrals. 'Simple integrals' is up to interpretation, for the integral directly above has a simple integrand (log-trigs seem more complicated to me.) By the way, I don't even like the proposed solution; I hope someone else has something simpler, but it's probably different from the original solution. – skbmoore Dec 3 '19 at 19:34