Frullani like Trig integral $$\int_{0}^{\infty}\frac{\left|\cos\left ( x-\frac{\pi}{4} \right ) \right|- \left|\cos\left ( x+\frac{\pi}{4} \right ) \right| }{x}dx=\sqrt{2}\ln(1+\sqrt{2})$$ The above integral seems to look like a frullani type integral and has a closed form in terms of natural log. I tried to indefinitely integrate it. But the closed form are in terms $\text{Si}(x)$ and $\text{Ci}(x)$. I would highly  appreciate if there's any method or unique  substitution  to evaluate  this Integral.

$$I=\int_{0}^{\infty}\frac{\left|\cos\left ( x-\frac{\pi}{4} \right ) \right|- \left|\cos\left ( x+\frac{\pi}{4} \right ) \right| }{x}dx=\int_{0}^{\infty}\frac{\left|\cos\left ( \frac{x}{2}-\frac{\pi}{4} \right ) \right|- \left|\cos\left ( \frac{x}{2}+\frac{\pi}{4} \right ) \right| }{x}dx$$
$$=\frac{1}{\sqrt2}\int_0^\infty\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x}dx=\sqrt 2\int_0^\infty\frac{\sin x}{x}\frac{1}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}dx$$
Using Lobachevsky' integral
$$I=\sqrt 2\int_0^{\pi/2}\frac{dx}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}=\sqrt 2\int_0^{\pi/2}\frac{dx}{\sqrt{1+\cos x}+\sqrt{1-\cos x}}$$
$$\sqrt 2\int_0^{\pi/2}\frac{dx}{\sqrt2\cos\frac{x}{2}+\sqrt2\sin\frac{x}{2}}=\sqrt2\int_0^{\pi/4}\frac{dx}{\sin(x+\frac{\pi}{4})}=\sqrt2\int_{\pi/4}^{\pi/2}\frac{dx}{\sin x}=\sqrt 2\ln\tan\frac{x}{2}\,\bigg|_{\pi/4}^{\pi/2}$$
$$=-\sqrt2\ln\tan\frac{\pi}{8}=-\sqrt 2\ln\sqrt\frac{1-\frac{1}{\sqrt2}}{1+\frac{1}{\sqrt2}}=\frac{1}{\sqrt 2}\ln\frac{\sqrt2+1}{\sqrt2-1}=\sqrt2\ln(\sqrt2+1)$$
