Logarithmic – Trigonometric integral

Let $\gamma$ denote Euler – Mascheroni’s constant. Prove that

$\int_{0}^{\infty} \frac{\ln x \left ( 1 - \cos t \right )}{t^2} \, \mathrm{d}t = \frac{\pi}{2} \left ( 1 - \gamma \right )$

Solution

We are applying a powerful technique; using Laplace and Inverse Laplace transformation. Hence,

$\begin{align*} \int_{0}^{\infty} \frac{\log t \left ( 1-\cos t \right )}{t^2} \, {\rm d}t &= \int_{0}^{\infty} \frac{\log t \left ( 1-\cos t \right )}{t \cdot t} \, {\rm d}t \\ &= -\frac{1}{2}\int_{0}^{\infty} \left ( \log x + \gamma \right ) \log \left ( \frac{1}{x^2}+1 \right ) \, {\rm d}x \\ &=- \frac{1}{2} \int_{0}^{\infty} \log x \log \left ( \frac{1}{x^2}+1 \right ) \, {\rm d}x - \\ & \quad \quad \quad -\frac{1}{2} \gamma \int_{0}^{\infty} \log \left ( \frac{1}{x^2} +1 \right ) \, {\rm d}x \\ &= \frac{\pi}{2} - \frac{\pi \gamma}{2}\\ &= \frac{\pi}{2} \left ( 1-\gamma \right ) \end{align*}$

since

$\begin{align*} \int_{0}^{\infty} \log x \log \left ( \frac{1}{x^2}+1 \right ) \, {\rm d}x &= \int_{0}^{1} \log x \log \left ( \frac{1}{x^2}+1 \right ) \, {\rm d}x + \\ &\quad \quad \quad + \int_{1}^{\infty} \log x \log \left ( \frac{1}{x^2}+1 \right )\, {\rm d}x\\ &\!\!\!\!\!\overset{u=1/x}{=\! =\! =\! =\! =\!} \int_{0}^{1}\log x \log \left ( \frac{1}{x^2}+1 \right ) \, {\rm d}x - \\ &\quad \quad \quad - \int_{1}^{0} \log \left ( \frac{1}{x} \right ) \log \left ( x^2+1 \right ) \frac{1}{x^2} \, {\rm d}x \\ &=\int_{0}^{1} \log x \log \left ( \frac{1}{x^2}+1 \right ) \, {\rm d}x - \\ &\quad \quad \quad - \int_{0}^{1} \frac{\log x \log \left ( 1+x^2 \right )}{x^2} \, {\rm d}x \\ &=-2\mathcal{G} - \frac{\pi}{2} - \log 2 - \left ( -2\mathcal{G} + \frac{\pi}{2} - \log 2\right )\\ &=- \pi \cancel{- 2\mathcal{G} - \log 2 + 2\mathcal{G} + \log 2} \\ &= -\pi \end{align*}$

The integrals $\displaystyle \int_{0}^{1} \log x \log \left ( \frac{1}{x^2}+1 \right )\, {\rm d}x$ , $\displaystyle \int_{0}^{1}\frac{\log x \log \left ( 1+x^2 \right )}{x^2} \, {\rm d}x$ are quite easy to calculate. For instance:

$\begin{align*} \int_{0}^{1} \log x \log \left ( \frac{1}{x^2}+1 \right ) \, {\rm d}x &= \int_{0}^{1} \log x \log \left ( \frac{x^2+1}{x^2} \right ) \, {\rm d}x \\ &= \int_{0}^{1} \log x \left ( \log \left ( x^2+1 \right ) - 2 \log x \right ) \, {\rm d}x\\ &=\int_{0}^{1} \log x \log \left ( x^2+1 \right ) \, {\rm d}x - 2 \int_{0}^{1} \log^2 x \, {\rm d}x \\ &= \int_{0}^{1} \log x \log \left ( x^2+1 \right ) \, {\rm d}x -4 \\ &= -2\mathcal{G} - \frac{\pi}{2} -\log 2 + 4 -4 \\ &= -2\mathcal{G} - \frac{\pi}{2} - \log 2 \end{align*}$

where $\mathcal{G}$ denotes the Catalan’s constant. The other integral is done similarly. Throughout the solution it was used the fact that

$\mathcal{L}\left \{ \frac{1-\cos t}{t} \right \} = \int_{0}^{\infty} e^{-st} \left ( \frac{1-\cos t}{t} \right ) \, {\rm d}t= \frac{1}{2} \log \left ( \frac{1}{s^2}+1 \right )$

Logarithmic – Trigonometric integral

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