A curious logarithmic and trigonometric integral

Let \gamma denote the Euler – Mascheroni constant. Prove that

    \[\int_{0}^{\infty}{\cos(x^n)-\cos(x^{2n})\over x}\ln x {\rm d}x=\frac{12\gamma^2-\pi^2}{ 2(4n)^2}\]


Let us consider the following

    \[\mathcal{J}\left ( \alpha \right )=\int_{0}^{\infty }x^{\alpha-1}\left ( \cos x-\cos x^{2} \right )\mathrm{d}x\]

It is known that \mathcal{J}(0) = -\frac{\gamma}{2} . Also, the Mellin transform of \cos x^b is known to be

\displaystyle\int_{0}^{\infty}x^{a-1}\cos\left(x^{b}\right)\,\mathrm{d}x=\frac{1}{b}\cos\left(\frac{\pi a}{2b}\right)\Gamma \left(\frac{a}{b} \right), \;\ b>a>1


\displaystyle \mathcal{J}\left ( \alpha \right )=\Gamma \left ( a \right )\cos\left ( \frac{a\pi }{2} \right )-\frac{1}{2}\Gamma \left ( \frac{a}{2} \right )\cos\left ( \frac{a\pi }{4} \right )

On the other hand

\displaystyle \Gamma \left ( x \right )\sim \frac{1}{x}-\gamma +\frac{6\gamma ^{2}+\pi ^{2}}{12}x+o(x)

What we are seeking is \mathcal{J}'(0).

\begin{aligned} \mathcal{J}'(0) &= \lim_{a\rightarrow 0} \frac{\mathcal{J}(a) - \mathcal{J}(0)}{a -0} \\ &= \lim_{a\rightarrow 0} \frac{-\frac{1}{2} \cos \left ( \frac{a \pi}{4} \right ) \Gamma \left ( \frac{a}{2} \right )+ \cos \left ( \frac{a \pi}{2} \right ) \Gamma(a) + \frac{\gamma}{2}}{a} \\ &=\frac{3 \gamma^2}{8} - \frac{\pi^2}{32} \end{aligned}

and the result follows.

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