# Calculate $\int_0^\infty\frac{\sin(x)\log(x)}{x}\mathrm dx$.

Calculate $\displaystyle\int_0^\infty\dfrac{\sin(x)\log(x)}{x}\mathrm dx$.

I tried to expand $\sin(x)$ at zero, or use SI(SinIntegral) function, but it did not work. Besides, I searched the question on math.stackexchange, nothing found.

Mathematica tells me the answer is $-\dfrac{\gamma\pi}{2}$, I have no idea how to get it.

Notice that

\begin{align*} \int_{0}^{\infty} \frac{\sin x}{x^s} \, dx &= \frac{1}{\Gamma(s)} \int_{0}^{\infty} \left( \int_{0}^{\infty} t^{s-1}e^{-xt} \, dt \right) \sin x \, dx \\ &= \frac{1}{\Gamma(s)} \int_{0}^{\infty} \left( \int_{0}^{\infty} e^{-tx} \sin x \, dx \right) t^{s-1} \, dt \\ &= \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{t^{s-1}}{1+t^2} \, dt \\ &= \frac{1}{2\Gamma(s)} \beta\left(\frac{s}{2}, 1-\frac{s}{2}\right) \\ &= \frac{\pi}{2\Gamma(s)\sin\left(\frac{\pi s}{2}\right)}. \end{align*}

(This heuristic computation is valid line-by-line on the strip $1 < \Re(s) <2$ in view of Fubini's theorem, and then extends to the larger strip $0 < \Re(s) < 2$ by analytic continuation.) Differentiating both sides, we get

$$\int_{0}^{\infty} \frac{\sin x \log x}{x^s} \, dx \stackrel{(*)}{=} -\frac{d}{ds} \frac{\pi}{2\Gamma(s)\sin\left(\frac{\pi s}{2}\right)} = \frac{\pi}{2\Gamma(s)\sin\left(\frac{\pi s}{2}\right)} \left( \psi(s) + \frac{\pi}{2}\cot\left(\frac{\pi s}{2}\right) \right).$$

Now the answer follows by plugging $s = 1$:

$$\int_{0}^{\infty} \frac{\sin x \log x}{x} \, dx = -\frac{\gamma \pi}{2}.$$

Justification of $\text{(*)}$. Let us prove that

$$F(s) := \int_{0}^{\infty} \frac{\sin x}{x^s} \, dx \tag{1}$$

is analytic on the open strip $S = \{ s \in \Bbb{C} : 0 < \Re(s) < 2 \}$ and its derivative can be computed by the Leibniz's integral rule. A major issue of $\text{(1)}$ is that the defining integral converges only conditionally for $0 < \Re(s) \leq 1$ and thus raises some technical difficulties. In order to circumvent this, we improve the speed of convergence using integration by parts:

$$F(s) = \underbrace{\left[ \frac{1-\cos x}{x^s} \right]_{0}^{\infty}}_{=0} + s \int_{0}^{\infty} \frac{1-\cos x}{x^{s+1}} \, dx.$$

Notice that the resulting integral is absolutely convergent on $S$.

Now we claim that $g(s) := F(s)/s$ is differentiable and its derivative can be computed by the Leibniz's integral rule. Let $\epsilon$ be such that $\bar{B}(s, \epsilon) \subset S$. Then whenever $0 < |h| < \epsilon$,

\begin{align*} &\frac{g(s+h) - g(s)}{h} - \int_{0}^{\infty} \frac{(1-\cos x)(-\log x)}{x^{1+s}} \, dx \\ &\hspace{9em} = \int_{0}^{\infty} \frac{1-\cos x}{x^{1+s}} \left( \frac{x^{-h} - 1}{h} + \log x \right) \, dx. \end{align*}

Notice that the integrand is dominated by

$$\left| \frac{1-\cos x}{x^{1+s}} \left( \frac{x^{-h} - 1}{h} + \log x \right) \right| \leq \frac{1-\cos x}{x^{1+\Re(s)}} (\max\{ x^{\epsilon}, x^{-\epsilon} \} + 1) |\log x|$$

This dominating function is integrable. Hence by the dominated convergence theorem, as $h \to 0$, we have

$$\lim_{h \to 0} \frac{g(s+h) - g(s)}{h} = \int_{0}^{\infty} \frac{(1-\cos x)(-\log x)}{x^{1+s}} \, dx.$$

Plugging this back, we know that $F(s)$ is differentiable with

$$F'(s) = \int_{0}^{\infty} \frac{1-\cos x}{x^{s+1}} \, dx + s \int_{0}^{\infty} \frac{(1-\cos x)(-\log x)}{x^{1+s}} \, dx.$$

Finally, performing integration by part to the latter integral yields the desired conclusion:

\begin{align*} &s \int_{0}^{\infty} \frac{(1-\cos x)(-\log x)}{x^{1+s}} \, dx \\ &\hspace{5em} = \underbrace{\left[ \frac{(1-\cos x)\log x}{x^s} \right]_{0}^{\infty}}_{=0} - \int_{0}^{\infty} \left( \frac{\sin x \log x}{x^s} + \frac{1-\cos x}{x^{1+s}} \right) \, dx \end{align*}

• @JackD'Aurizio, Thank you! – Sangchul Lee Sep 26 '16 at 13:35
• Nothing to thank me for, it is a really well-written answer, credits to you. – Jack D'Aurizio Sep 26 '16 at 13:38
• +1. Not too many people care to write a justification $^{*}$ as you did it ( included me !!! ). – Felix Marin Oct 2 '16 at 8:37


Hereafter the $\ds{\ln}$-branch-cut runs along $\ds{\left(-\infty,0\right]}$ with $\ds{\ln\pars{z}\ \,\mrm{arg}}$ given by $\ds{-\pi < \mrm{arg}\pars{z} < \pi}$.

\begin{align} \mc{J} & \equiv \Im\int_{0}^{\infty}\ln\pars{x}\expo{-\pars{t - \ic}x}\,\dd x = \Im\bracks{{1 \over t - \ic}\int_{0}^{\pars{t - \ic}\infty} \ln\pars{x \over t - \ic}\expo{-x}\,\dd x} \\[5mm] & = -\,\Im\braces{{t + \ic \over t^{2} + 1}\int_{\infty}^{0}\bracks{\ln\pars{x \over \root{t^{2} + 1}} + \arctan\pars{1 \over t}\ic}\expo{-x}\,\dd x} \\[5mm] & = \Im\braces{{t + \ic \over t^{2} + 1}\bracks{% \int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x - {1 \over 2}\ln\pars{t^{2} + 1}+ \arctan\pars{1 \over t}\ic}} \end{align}

Note that
$\ds{\int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x = \left.\partiald{}{\mu}\int_{0}^{\infty}x^{\mu}\expo{-x}\,\dd x\, \right\vert_{\ \mu\ =\ 0} = \Gamma\,'\pars{1}\ =\ \overbrace{\Gamma\pars{1}}^{\ds{=\ 1}}\ \overbrace{\Psi\pars{1}}^{\ds{-\gamma}}\ =\ -\gamma\quad}$ where $\ds{\Gamma}$

and $\ds{\Psi}$ are the Gamma and Digamma Functions, respectively. Then,

\begin{align} \mc{J} & \equiv \bbox[8px,#ffe,border:0.1em groove navy]{% \Im\int_{0}^{\infty}\ln\pars{x}\expo{-\pars{t - \ic}x}\,\dd x} = -\,{\gamma \over t^{2} + 1} - {1 \over 2}\,{\ln\pars{t^{2} + 1} \over t^{2} + 1} + {t\arctan\pars{1/t} \over t^{2} + 1} \\[5mm] & = \bbox[8px,#ffe,border:0.1em groove navy]{-\,{\gamma \over t^{2} + 1} + \totald{}{t}\bracks{{1 \over 2}\,\arctan\pars{1 \over t}\ln\pars{t^{2} + 1}}} \label{2}\tag{2} \end{align}

Note that$\ds{{1 \over 2}\,\arctan\pars{1 \over t}\ln\pars{t^{2} + 1}}$ vanishes out when $\ds{t \to \infty}$ and $\ds{t \to 0^{+}}$.

By replacing \eqref{2} in \eqref{1}: \begin{align} &\color{#f00}{\int_{0}^{\infty}{\sin\pars{x}\ln\pars{x} \over x}\,\dd x} = -\gamma\int_{0}^{\infty}{\dd t \over t^{2} + 1} = \color{#f00}{-\,{1 \over 2}\,\gamma\pi} \end{align}

• (+1) That is how I just did this. Then I looked on MSE to see if it's already been discusses and found this page. Well done my friend! – Mark Viola Apr 23 at 3:36