$\int_0^\infty \frac{\log x \sin x}{x}\, dx$

[Tolaso J Kos]

Prove that:

$$\int_0^\infty \frac{\log x \sin x}{x}\, {\rm d}x = - \frac{\gamma \pi}{2}$$

where $\gamma$ stands for Euler - Mascheroni constant.

[r9m]

We might begin with the observation that, for $\mathfrak{Re}(s) < 1$, the integral $\displaystyle \int_{\gamma_r} z^{s-1}e^{iz}\,dz$ vanishes over the contour $\gamma_r = [0,r] \cup re^{i[0,\pi/2]}\cup [ir,0]$.

Since, \begin{align*}\left|\int_{[re^{i[0,\pi/2]}]} z^{s-1}e^{iz}\,dz\right| &\le \int_0^{\pi/2} r^{\mathfrak{Re}(s)}e^{|\mathfrak{I}(s)|\pi/2}e^{-r\sin \theta}\,d\theta\\& \le r^{\mathfrak{Re}(s)}e^{|\mathfrak{I}(s)|\pi/2}\int_0^{\pi/2} e^{-2r\theta/\pi}\,d\theta \qquad \text{ (Jordan's Inequality) }\\& = \frac{\pi}{2}r^{\mathfrak{Re}(s)-1}e^{|\mathfrak{I}(s)|\pi/2}(1-e^{-r})\end{align*}

Therefore, the integrals over the arc: $r[0,\pi/2]$ vanishes as $r \to 0^{+}$ and $r \to +\infty$.

Hence, $$\int_0^{\infty} z^{s-1}e^{iz}\,dz - \int_0^{i \infty} z^{s-1}e^{iz}\,dz = 0$$

or, $$\int_0^{\infty} z^{s-1}e^{iz}\,dz = e^{i\pi s/2}\int_0^{\infty} z^{s-1}e^{-z}\,dz = e^{i\pi s/2}\Gamma(s)$$

Thus, \begin{align*}&\int_0^{\infty} x^{s-1}\sin x\,dx &=& \sin \left(\frac{\pi s}{2}\right)\Gamma (s)\\ \implies & \int_0^{\infty} x^{s-1}\sin x \log x\,dx &=& \frac{\partial}{\partial s}\left(\sin \left(\frac{\pi s}{2}\right)\Gamma (s)\right)\\ \implies & \int_0^{\infty} \frac{\sin x \log x}{x}\,dx &=& \lim\limits_{s \to 0^{+}} \frac{\frac{\pi}{2}\cos \frac{\pi s}{2} + \psi (s)\sin \frac{\pi s}{2}}{s}\\&&=& \frac{\pi}{2}\lim\limits_{s \to 0^{+}} \left(\frac{1}{s}+\psi(s)\right)\\&&=& -\frac{\gamma \pi}{2}\end{align*}

[Tolaso J Kos]

Nicely done. Here is another approach using the MAZ identity which is:

MAZ identity

Let $g, h$ be two functions and let $G, H$ be their Laplace Transforms respectively. Then the equation:

$$\int_0^\infty g(x) H(x) \, {\rm d}x = \int_0^\infty G(x) h(x) \, {\rm d}x$$

holds.

Applying the above identity it is easy to note that $\mathcal{L}^{-1} \left\{\frac{\log x}{x}\right\}= -\log y -\gamma$ and we already know that $\mathcal{L}\{\sin x\}= \frac{1}{y^2+1}$. Hence:

$$\int_0^\infty \frac{\log x \sin x}{x}\, {\rm d}x = -\int_0^\infty \frac{\log y +\gamma}{y^2+1}\, {\rm d}y =-\frac{\gamma \pi}{2}$$

since $\displaystyle \int_0^{\infty} \frac{\ln x}{x^2+1}\, {\rm d}x =0$. (easy to prove)